Contents · 2020-06-25 · Grade 11 Trigonometry Notes 2020 7 | P a g e Reduction Formulae The trig...
Transcript of Contents · 2020-06-25 · Grade 11 Trigonometry Notes 2020 7 | P a g e Reduction Formulae The trig...
Grade 11 Trigonometry Notes 2020
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Contents Caps References .................................................................................................................... 2
Trig Ratios in 90 Triangles .................................................................................................. 3
Ratios and Co-Ratios ......................................................................................................... 3
Application of These Ratios .............................................................................................. 3
The Trig Ratios on The Cartesian Plane............................................................................. 4
Application: Using Diagrams to Solve Problems ............................................................... 5
Special Angles and Reduction Formulae ............................................................................... 6
Special Angles .................................................................................................................. 6
Reduction Formulae .......................................................................................................... 7
Very Big Positive or Negative Angles ............................................................................. 10
Trigonometry equations................................................................................................... 11
Applying Special Angles and the Reduction Formulae .................................................... 12
Trigonometric Identities ...................................................................................................... 14
How to prove identities ................................................................................................... 14
Mixed Exercise 1 ................................................................................................................ 16
Trigonometric Equations and the General Solution .............................................................. 16
tan θ = k .......................................................................................................................... 16
cos θ = k .......................................................................................................................... 18
sin θ = k .......................................................................................................................... 21
Further Examples ............................................................................................................ 23
Trigonometric Equations General Solutions .................................................................... 24
Trigonometric equations in the form: a sin θ ± b cos θ = 0.............................................. 25
Trigonometric equations for which factorising is necessary ............................................. 26
Trigonometric equations: when identities are invalid ....................................................... 26
Solving Trigonometric Equations without the use of a Calculator .................................... 28
Mixed exercise 2 ................................................................................................................. 29
Trigonometric Functions ..................................................................................................... 31
Difference between Grade 10 and Grade 11 Functions .................................................... 31
Solving Problems in Two Dimensions ................................................................................. 39
The Sine, Cosine and Area Rules..................................................................................... 39
Tips for Solving 2D Problems ......................................................................................... 41
Using the Sine, Cosine and Area Rules ............................................................................ 41
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Caps References Progression in terms of concepts and skills
GRADE 10 GRADE 11 GRADE 12
1. Define the trigonometric
ratios sin θ, cos θ and tan
θ, using right-angled
triangles.
2. Extend the definitions of
sin θ, cos θ and tan θ for
0 ≤ θ ≤ 360.
3. Define the reciprocals of
the trigonometric ratios
cosec θ, sec θ and cot θ,
using right-angled
triangles (these three
reciprocals should be
examined in Grade 10
only).
4. Derive values of the
trigonometric ratios for
the special cases
(without using a
calculator)
θ ∈ {0; 30; 45; 60; 90}.
5. Solve two-dimensional
problems involving
right-angled triangles.
6. Solve simple
trigonometric equations
for angles between 0
and 90.
7. Use diagrams to
determine the numerical
values of ratios for
angles from 0 to 360.
1. Derive and use the
identities
, k an
odd integer; and
2. Derive and use reduction
formulae to simplify the
following expressions:
( ); ( );
( );
( );
( );
( );
( );
( );
( );
( );
( );
3. Determine for which values
of a variable an identity
holds.
4. Determine the general
solutions of trigonometric
equations. Also, determine
solutions in specific
intervals.
5. Prove and apply the sine,
cosine and area rules.
6. Solve problems in two
dimensions using the sine,
cosine and area rules.
1. Proof and use of the
compound angle
identities:
( )
( )
2. Solve problems in two
and three dimensions.
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Trig Ratios in 90 Triangles
Ratios and Co-Ratios
When we start with trig ratios, we use right-angled triangles to start off. The ratios are sin ,
cos and tan while the reciprocals are cot , sec and cosec .
When dealing with the trig ratios in a right-angled triangle, the sides of the triangle are
named as follows:
The side opposite the angle we are working with is y
The side opposite the 90 angle (hypotenuse) is r
The remaining side is x
OR
For the triangles above:
Trig Ratio Reciprocal
Application of These Ratios
Solve KLM:
C
A
B 𝜽
r
x
y
A
B C
r
x
y
𝜽
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The Trig Ratios on The Cartesian Plane
(NOTE: In the diagram below, is in the standard position)
Trig Ratio Reciprocal
Please note – that reciprocal ratios is not
important after Grade 10
You can use the theorem of Pythagoras:
To calculate if you have x and y.
To calculate if you have x and r.
To calculate if you have y and r.
NOTE:
r is always positive, since r represents the distance from the origin to the ordered
number pair on the terminal arm of the angle .
Since x and y are coordinates and represent the position of a point, they can either be
negative or positive.
The signs of coordinates x and y depend on the position of the angle.
Does the angle lie in quadrant I, II, III or IV?
Therefore, the signs of the trig ratios depend on the signs of x and y.
Since r is always positive, it has no influence on the sign of the trig ratio.
The CAST diagram is a diagram of positives – it indicates which trig ratio is
positive in which quadrant:
All ratios are positive in quadrant I (A)
Only (and ) is positive in quadrant II (S)
Only tan (and ) is positive in quadrant III (T)
Only cos (and ) is positive in quadrant IV (C)
𝑥 ( ; )
𝑃(𝑥 ; 𝑦)
𝑦
𝑥
𝑦 r
A
C
S
T
( - ; - ) ( + ; - )
( + ; + ) ( - ; + )
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Application: Using Diagrams to Solve Problems
Worked Example 1
Given: and .
Make use of a diagram and determine the value of:
a) tan b) sin c) cos
SOLUTION
a)
b)
=
c) cos
=
We will revisit questions like this once reduction formula and co-ratios have been addressed.
Do the Revision exercise on page 137 to make sure your Grade 10 knowledge is up to date!
Given information:
𝜽
𝑦
𝑥 ? 𝑥
𝑦
To determine the quadrant:
𝑠𝑖𝑛𝜃
(sin = negative)
𝑐𝑜𝑠𝜃 (cos = positive)
Because of quadrant 4 x is +12
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Special Angles and Reduction Formulae
Special Angles
The trig ratios of the ‘special angles’ are used frequently, especially in problems where a
calculator may not be used. The special angles are: 0; 30; 45; 60; 90; 180; 270;
360
The ratios of the special angles must be memorised by making use of
one of the methods below. The method chosen depends on the preference of the teacher.
ON THE CARTESIAN PLANE
r = 2
USING THE CASIO CALCULATOR
The CASIO Calculator will automatically give you the answers to the special angles. The
question in the exam will say “without using a calculator”. So even if we use the calculator
we need to ensure that we show all steps
Example
Without using a calculator find the value of sin 30ᵒ×tan 60ᵒ + cos 30°. Show all steps.
3
2
32
2
3
2
3
2
33
2
1
30cos60tan30sin
.
We will revisit questions like this once reduction formula and co-ratios have been addressed.
𝟏 ; 𝟑
𝟐 ; 𝟐
𝟑 ; 𝟏
𝟔𝟎
𝟒𝟓
𝟑𝟎
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Reduction Formulae
The trig ratio of every single obtuse or reflex angle can be rewritten as the trig ratio of an
acute angle.
(180 – ); (180 + ) and (360 – )
Use your calculator and determine the value of sin 20°, sin 160°, sin 200° and sin 340°.
What do you notice?
Use your calculator and determine the value of cos 40°, cos 140°, cos 220° and cos 320°.
What do you notice?
Use your calculator and determine the value of tan 88°, tan 92°, tan 268° and tan 272°
What do you notice?
You should have notice a similarity in the answer of the four trigonometry ratios asked in
each question.
Think back to the trigonometric graphs you drew in Grade 10. Remember that the shape
stayed the same? Once we worked through the first 90° the curve is just transformed (by
reflections in either a vertical line of the x – axis. The means that the values on the y-axis is
repeated over and over.
So what is the link between 20°, 160°, 200° and 340°? Or 40°, 140°, 220° and 320°? Or
88°, 92°, 268° and 272°? Or between all 12 angles?
The obvious and easy link to find is that 20°, 40° and 88° are all in the first quadrant; 160°,
140° and 92° are all in the second quadrant; 200°, 220° and 268° are all in the third
quadrant and 340°, 320° and 272° are all in the fourth quadrant.
Looking closer we can see that:
160° = 180° – 20° 200° = 180° + 20° 340° = 360° – 20°
140° = 180° – 40° 220° = 180° + 40° 320° = 360° – 40°
92° = 180° – 88° 268° = 180° + 88° 272° = 360° – 88°
When the original (first) angle is deviated from the x – axis – although the angles become
bigger, the trigonometry ratios stay the same except for, in two case each, the sign changes.
You already know that the x and y values repeat themselves BUT in different signs. Refer
to our work in the Cartesian plane.
You should remind yourself that the y value of the Cartesian Plane is positive in the first
and second quadrant; hence sin α is positive in the first and second quadrant (sin 20° and
sin 160°), but the y value of the Cartesian Plane is negative in the third and fourth
quadrant; hence sin α is negative in the third and fourth quadrant (sin 200° and sin 340°).
You should also remember r
ysin when we work in the Cartesian Plane; r is always
positive. Hence whether sin α is positive depends on the y – value.
Similar: You should remind yourself that the x value of the Cartesian Plane is positive in
the first and fourth quadrant; hence cos α is positive in the first and fourth quadrant (cos40°
and cos 320°), but the x value of the Cartesian Plane is negative in the second and third
quadrant; hence cos α is negative in the second and third quadrant (cos 140° and cos 220°).
You should also remember r
xcos when we work in the Cartesian Plane; r is always
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positive. Hence whether sin α is positive depends on the x – value.
When we work with tan α we have to start with the fact that x
ytan . Consider the
Cartesian Plane. Both x and y can be positive or negative. In the first quadrant, x and y are
both positive and hence tan α is positive (tan 88°). In the third quadrant, x and y are both
negative and hence tan α is positive (tan 278°). In the second quadrant, x is negative and y
is positive so tan α will be negative (tan 92°) and finally in the fourth quadrant, x is
positive and y is negative so tan α will be negative (tan 272°).
We also have to address (360° + α). If α is an acute angle – this places the angle in the first
quadrant; which means all three the trigonometric ratios will be positive.
What about (–α)? Where will we find this angle on the Cartesian Plane? Is we rewrite this
as (0° – α) it is easy to see that we are in the fourth quadrant
How can we generalise all of this?
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
The CAST diagram plays a key role when using the reduction formulae. It is important
that you understand the ‘logic’ of the CAST diagram to get to answers ‘logically’.
However, you will have to memorise the diagram itself plus a few useful TIPS that will
make this section of trig easy.
Examples
Rewrite the following as the trigonometry ratio of a positive, acute angle:
𝟎 / 𝟑𝟔𝟎
𝟗𝟎
𝟏𝟖𝟎
𝟐𝟕𝟎
A
𝜽
I
T
(𝟏𝟖𝟎 𝜽)
III
C
(𝟑𝟔𝟎 𝜽)
IV
S
(𝟏𝟖𝟎 𝜽)
II
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a) ( ) b) ( ) c) ( ) d) ( ) e) ( ) see this as ( )
Do Exercise 1 on page 141
(90 – ), (90 + ), (270 – ) and (270 + )
According to the CAPS document only ( ) should be covered.
( ) works the same as ( ), just in different quadrants.
You might find ( ) useful and time-saving when solving problems. (This a
nice to know and you should please not worry about this)
Use your calculator again.
Compare sin 60° and cos 30°. Both answers in 2
3. We can say:
sin 60° = sin(90° – 30°) (because 90° – 30° = 60°)
= cos 30°
Compare cos 74° and sin 16°. Both answers, rounded off to 4 decimals, is 0,2756. We can
say:
cos 74° = cos(90° – 16°) (because 90° – 16° = 74°)
= sin 16°
I II
III IV
A S
T C
𝟎 / 𝟑𝟔𝟎
𝟗𝟎
𝟏𝟖𝟎
𝟐𝟕𝟎
(𝟗𝟎 𝜽) (𝟗𝟎 𝜽)
(𝟐𝟕𝟎 𝜽) (𝟐𝟕𝟎 𝜽)
𝜽
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TIP: The name of the trig ratio changes to the name of the co-function ONLY WHEN WE
WORK with ( ) [or with ( )], i.e. the VERTICAL axis.
NOTE: It is only when the question is asking (forcing) you to work with a co-ratio that you
do that.
( )
( )
( )
( ) ( ) ( )
( ) ( )
THE SECRET IS …
When working with reduction formulae and co-ratios, you must always consider three
things in your answer……
The sign (+ or –)
The trig ratio
An acute angle
Do Exercise 2 on page 144
Very Big Positive or Negative Angles
You may add or subtract as many 360o’s as you like. The angles are named ‘co terminal’
angles. The value of the trig ratios of co terminal angles is equal.
Consider: cos 1 456 o
Repeatedly subtract 360o until you get an
angle less than 360
1 456 o – 360
o = 1 096
o
1 096 o – 360
o = 736
o
736 o – 360
o = 376
o
376 o – 360
o = 16
o
Thus: cos 1456 o = cos 16
o
Consider: cos (– 568)o
Repeatedly add 360o until you get an
angle that is greater than 0 but less than
360
– 568o+ 360
o = –208
o
– 208o + 360
o = 152
o
So, cos 152 o
= cos (180 o – 28
o)
= -cos 28 o
Thus: cos (-568o) = cos 152
o
= – cos 28 o
Do Exercise 3 on page 145
Do Exercise 4 on page 148
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Trigonometry equations
Worked Example 1
If ( ) determine the value(s) of the following in terms of t:
a) ( ) b) ( ) c) ( ) d) ( )
a) ( ) ( ) ( )
b) ( ) ( ) ( ( ) ( ))
( ) = ( )
Make use of a diagram and determine the value of:
c) ( )
and
Using Pythagoras, you can calculate
( )
=
d) ( ) ( ) ( )
Do Exercise 5 on page 150
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Applying Special Angles and the Reduction Formulae
Worked Example 1
a) Show that: ( )
( ) ( )
b) Hence, calculate without the use of a calculator, the value of:
( )
(Leave your answer in surd form)
Please note: The word “Hence” is often used in a follow up question. It implies that
you have to apply the first part in answering the second part. If this was the exam
and you could not do part (a) you can still use the identity to answer part (b).
SOLUTION
a) ( )
( ) ( )
( ( )) ( )
( ) ( )
( ) ( )
( ) ( )
= RHS
b)
( )
( )
( ) ( )
(using (a) when )
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Worked Example 2
Evaluate without a calculator and showing all steps: ( )
( )
SOLUTION: ( )
( )
( )
( )
(
) (
)
Worked Example 3
If 08sin10 and 0tan calculate without the use of a calculator and with the aid of
a diagram the value of:
a) (sin + cos )2
b) 180tan
c) 90cos
5
4sin
10
8sin
8sin10
3
9
1625
5)4(
2
2
222
222
x
x
x
x
ryx
870° – 2(360°) = 150°
–1020° + 3(360°) = 60°
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a) (sin + cos )2
25
49
25
241
5
4
5
321
coscos.sin2sin 22
b) 180tan
3
4
3
4
tan
c) 90cos
5
4
5
4
sin
Do Exercise 6 on page 153
Trigonometric Identities
NOTE:
There are two main identities taught in Grade 11:
In Grade 12 you will be introduced to the compound angle identities. Double angle
identities are derived from the compound angle identities.
We can use these identities to prove other identities.
How to prove identities
There are always two sides to prove equal when dealing with identities.
Start with simplifying the least complicated side. (And at least get a mark)
Look for squared terms and see if you can use substitution to simplify.
Where possible, change everything to sine and cosine.
Trigonometric expressions can often need to be simplified like in algebraic
expressions, so check to see if the expression has like terms, fractions or factors
before substituting by an identity.
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Worked Examples
Prove the following identities:
a)
= 1 + sin x
b)
22
23
sin1
cos
270sin
810cos
cos.tan
cos.sinsin
SOLUTION (a)
Note: The RHS cannot be simplified so I move to the LHS
x
xxxLHS
sin1
cossincos 422
RHS = 1 + sin x
x
xxx
sin1
cossincos 222
= x
x
sin1
)1(cos2
= x
x
sin1
sin1 2
=
x
xx
sin1
sin1sin1
= xsin1
RHSLHS
SOLUTION (b)
Note: The RHS can be simplified so I start there before I move to the LHS:
RHS:
cos
1
cos
cos
sin1
cos
2
2
LHS:
RHS
cos
1
1
0
cos
1
1
90cos
cossin
1sin
)1(
720810cos
1
cos
cos
sin
cossinsin
270sin
810cos
cos.tan
cos.sinsin
2
22
2
23
Do Exercise 7 on page 156
Do Exercise 8 on page 157
Do Exercise 9 on page 159
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Mixed Exercise 1
1. ; and ( ; ) are points on the
Cartesian plane, as shown in the diagram
below. and .
Determine, without the use of a calculator,
the value of:
1.1
1.2 ( ) 1.3
2. If and , determine the value of where
and ∈ [ ; ].
3. Simplify the following:
)cos()720cos()90cos()180sin(
)360cos()180tan(
xxxx
xx
4. Simplify: ( ) ( ) ( ) ( )
5. Prove, without the use of a calculator, that:
Trigonometric Equations and the General Solution
Before dealing with this topic it is absolutely essential for you to ensure that your calculator
is on “degrees”.
tan θ = k
We are going to do a short investigation:
Find the numerical value of the following:
tan 45º, tan (–135º), tan 225º, tan (–315º)
The function value for the four different angles gives the same numerical value.
The inverse operation – finding the angle – only gives one answer.
Find the angle if the numerical value is given:
tan θ = 1 451tan 1 but what about the other values (–135º,
225º or –315º)?
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We need to look at the graph of y = tan θ to understand this phenomenon.
Points A, B, C and D have a y-value of 1 – this corresponds with tan 45º = 1, tan (–135º) = 1,
tan 225º = 1 and tan (–315º) = 1. When the calculator has to get the inverse answer, it seems
as though it uses the answer closes to the y – axis, i.e. 45º. It is important to note at this point
that each point differs 180º from the next point.
Find the numerical value of the following:
tan 150º, tan (–30º), tan 330º, tan (–210º)
Find the angle if the numerical value is given:
tan θ = 3
3
30
3
3tan 1 but what about the other values
(150º, 330º or –210º)?
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Points E, F, G and H have a y-value of 3
3 – this corresponds with tan 150º =
3
3 ,
tan (–30º) = 3
3 , tan 330º =
3
3 and tan (–210º) =
3
3 . When the calculator has to get
the inverse answer, it again uses the answer closes to the y – axis, i.e. -30º. It is important to
note, again, that each point differs 180º from the next point.
The formula to solve a “tan θ” – equation is
If tan θ = k Rk
then Znnk ;180.)(tan 1
EXAMPLE 12 (of textbook done easier)
(c) Solve for θ if 2 tan θ = –9, 360;0 . Round answers off to two decimal places.
180..........47111,77
;180.2
9tan
2
9tan
9tan2
1
n
Znn
To find the applicable answers in the range, substitute n with integers:
n = 0 θ = –77,47º
n = 1 θ = 102,53º ✓
n = 2 θ = 282,53º ✓
n = 3 θ = 362,53º ×
Hence the applicable answers is θ = 102,53º or θ = 282,53º
cos θ = k
Find the numerical value of the following:
cos 55º, cos (–55º), cos 305º, cos (–305º)
The function value for the four different angles gives the same numerical value.
The inverse operation – finding the angle – only gives one answer.
Find the angle if the numerical value is given:
cos θ = 0,5769 555769,0cos 1 but what about the other values (–55º,
305º or –305º)?
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We need to look at the graph of y = cos θ to understand this phenomenon.
Points A, B, C and D have a y-value of 0,5736 – this corresponds with cos 55º = 0,5736,
cos (–55º) = 0,5736, cos 225º = 0,5736 and cos (–315º) = 0,5736. When the calculator has to
get the inverse answer, it seems as though it uses the positive answer closes to the y – axis,
i.e. 55º. It is important to note at this point that
There are always a positive and a negative value of angles with the same function
value(B and C; A and D).
From A to C and from B to D there is a 360º difference.
Find the numerical value of the following:
cos 120º, cos (–120º), cos 240º, cos (–240º)
Find the angle if the numerical value is given:
cos θ = 2
1
120
2
1cos 1 but what about the other values
(–120º, 240º or –240º)?
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Points E, F, G and H have a y-value of 2
1 – this corresponds with cos 120º =
2
1 ,
cos (–120º) =2
1 , cos 240º =
2
1 and cos (–240º) =
2
1 . When the calculator has to get the
inverse answer, it again uses the positive answer closes to the y – axis, i.e. 120º. It is
important to note, again, that
There are always a positive and a negative value of angles with the same function
value (E and H; F and G).
From E to G and from F to H there is a 360º difference.
The formula to solve a “cos θ” – equation is
If cos θ = k 1;1k
then Znnk ;360.)(cos 1
EXAMPLE 12
(b) Solve for θ if cos θ = –7
5, 360;0 . Round answers off to two decimal places.
...5846914,135
;360.7
5cos
7
5cos
1
Znn
To find the applicable answers in the range, substitute n with integers:
θ = 135,58º
n = 0 θ = 135,58º ✓
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n = 1 θ = 495,58º ×
θ = –135,58º
n = 0 θ = –135,58º ×
n = 1 θ = 224,42º ✓
n = 2 θ = 584,42 ×
Hence the applicable answers is θ = 135,58º or θ = 224,42º
sin θ = k
Find the numerical value of the following:
sin 55º, sin 125º, sin (–235º), sin (–305º)
The function value for the four different angles gives the same numerical value.
The inverse operation – finding the angle – only gives one answer.
Find the angle if the numerical value is given:
sin θ = 0,8192 558192,0sin 1 but what about the other values (125º,
–235º or –305º)?
We need to look at the graph of y = sin θ to understand this phenomenon.
Points A, B, C and D have a y-value of 0,8192 – this corresponds with sin 55º = 0,8192,
sin 125º = 0,8192, sin (–235º) = 0,8192 and sin (–315º) = 0,8192. When the calculator has to
get the inverse answer, it seems as though it uses the positive answer closes to the y – axis,
i.e. 55º. It is important to note at this point that
There is always an acute angle (C), the angle given by the calculator.
The next angle (D) has to be calculated by using the formula (180º – the acute angle).
From A to C and from B to D there is a 360º difference.
Find the numerical value of the following:
sin (–30º), sin (–150º), sin 210º, sin 330º
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Find the angle if the numerical value is given:
sin θ = 2
1
30
2
1sin 1 but what about the other values
(–150º, 210º or 330º)?
Points E, F, G and H have a y-value of 2
1 – this corresponds with sin (–150º) =
2
1 ,
sin (–30º) =2
1 , sin 210º =
2
1 and sin 330º =
2
1 . When the calculator has to get the
inverse answer, it again uses the answer closes to the y – axis, i.e. –30º. It is important to
note, again, that
There is always a negative acute angle (F), the angle given by the calculator.
The next angle (G) has to be calculated by using the formula (180º – the acute angle).
From F to H and from G to E there is a 360º difference.
The formula to solve a “sin θ” – equation is
If sin θ = k 1;1k
then Znnk
k
;360.sin180
sin
1
1
PLEASE NOTE the two answers.
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EXAMPLE 12
(a) Solve for θ if sin θ = 2
1; 360;0
360.150
30
;360.
2
1sin180
2
1sin
2
1sin
1
1
n
Znn
To find the applicable answers in the range, substitute n with integers:
θ = 30º
n = 0 θ = 30º ✓
n = 1 θ = 390º ×
θ = 150º
n = 0 θ = 150º ✓
n = 1 θ = 410 º ×
Hence the applicable answers is θ = 30º or θ = 150º
Do Exercise 10 on page 161
Further Examples
EXAMPLE 13
(a) Solve for A is sin (A – 24º) = –0,7 and A 360;0 rounded off to two decimals.
360.43,248
43,20
360.427004,224
427004,4424
;360.)7,0(sin180
)7,0(sin24
7,024sin
1
1
nA
nA
ZnnA
A
A = 339,57º or A = 248,43º
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(b) Solve for A if cos 2A = –0,867; 360;0A , rounded off to two decimals.
180.....0555,75
360...1118697,1502
;360.)867,0(cos2
867,02cos
1
nA
nA
ZnnA
A
A = 75,06º or A = 255,06º or A = 104,94º or A = 284,94º
Do Exercise 11 on page 162
Trigonometric Equations General Solutions
EXAMPLE 14
(a) Determine the general solution of sin θ = 2
1
(b) Find θ if 360;360
Solutions
(a) sin θ = 2
1
360.150
30
;360.
2
1sin180
2
1sin
1
1
n
Znn
(b) θ = 30º or θ = – 330º or θ = 150º or θ = –210º
EXAMPLE 15
(a) Determine the general solution of tan 2θ = –4
(b) Find θ if 180;90 , rounded off to two decimals.
(a) tan 2θ = –4
90...98187,37
180....9637,752
;180.)4(tan2 1
n
n
Znn
(b) θ = –37,98º or θ = 52,02º or θ = 142,02º
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Summarised notes on finding the general solution
If tan θ = k, Znnk ;180.)(tan 1
If cos θ = k, Znnk ;360.)(cos 1
If sin θ = k, Znnk
k
;360.sin180
sin
1
1
Do Exercise 12 on page 163
Trigonometric equations in the form: a sin θ ± b cos θ = 0
EXAMPLE 16
(a) Find the general solution of the equation 3 sin θ = 2 cos θ.
(b) Solve for θ if sin θ + cos θ = 0; 360;0
Solutions
(a) The first step is to divide both sides by cos θ because you would like to work with a
single trigonometric ratio and
tan
cos
sin .
180.69,33
;180.3
2tan
3
2tan
2tan3
cos
cos2
cos
sin3
cos2sin3
1
n
Znn
(b) sin θ + cos θ = 0
180.45
;180.1tan
1tan
cos
cos
cos
sin
cossin
1
n
Znn
θ = 135º or θ = 315º
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Trigonometric equations for which factorising is necessary
EXAMPLE 17
(a) Find the general solution for 1 + sin θ = cos2 θ
(b) Solve for θ, if 2cos2 θ + cos θ = 3
Solutions
(a) As you can see there is a sine and a cosine ratio in this equation. In this regard,
dividing by cos θ will not simplify the equation. The 1 as well as the cos2 θ creates
a problem.
We will choose to work with only sine in this example because it is easier to
write cos2 θ in terms of sin θ than writing sin θ in terms of cos θ.
1 + sin θ = cos2 θ
1 + sin θ = 1 – sin2θ
sin2θ + sin θ = 0
sin θ(sin θ + 1) = 0
sin θ = 0 or sin θ + 1 = 0
sin θ = –1
360.180
0
;360.0sin180
0sin
1
1
n
Znn
360.270
90
;360.1sin180
1sin
1
1
n
Znn
(b) 2cos2 θ + cos θ = 3
2cos2 θ + cos θ – 3 = 0
(2cos θ + 3)(cos θ – 1) = 0
2 cos θ + 3 = 0 cos θ – 1 = 0
solution No
2
3cos
3cos2
360.0
;360.1cos
1cos
1
n
Znn
Do Exercise 13 on page 166
Trigonometric equations: when identities are invalid
An identity is a statement of equality that is true for all values (except those values for which
the identity is not defined). Previously, equations containing fractions were solved. When
solving these types of equations, we always stated the restrictions to prevent division by zero.
We will now determine the restrictions for a trigonometric identity. In other words, the
value(s) for which the identity will be undefined.
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EXAMPLE 18
(a) Consider the identity 1sin
cos.tan
x
xx.
For which values of x will the identity above be invalid? State the general
solution.
(b) Consider the identity
cos
1
sin1
costan
.
For which values of θ will the identity above be invalid for the interval
180;180 ?
Solutions
(a) First you consider the tangent ratio:
You will recall the graph of y = tan x has asymptotes at x = ±90º; ±270º……
We can summarise this as:
x = 90º + n.180º; Zn
Secondly consider division by zero:
The identity will be invalid if any denominator equals zero. In this example it
means if sin x = 0.
360.180
0
;360.)0(sin180
)0(sin
0sin
1
1
nx
Znnx
x
(b) Tangent ratio:
θ = 90º + n.180º; Zn
Division by zero:
1 + sin θ = 0 cos θ = 0
sin θ = –1
360.270
90
;360.)1(sin180
)1(sin
1
1
n
Znn
360.90
;360.)0(cos 1
n
Znn
Hence the three solutions are giving us similar answers and the identity will be
undefined for θ = 90º and θ = –90º
Do Exercise 14 on page 168
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Solving Trigonometric Equations without the use of a Calculator
In this section a knowledge of special angles will be useful. The special angles are 0 º; 30 º;
45 º; 60 º; 90 º; 180 º; 270 º; 360 º.
The diagram for special angles is provided below.
EXAMPLE 19
Solve for θ if 2
2cos and 360;0
Solution
Znn
;360.135
2
2cos
θ = 135º or θ = 225º
Do Exercise 15 on page 169
Up until this point all the work, we have done will be ONE questions in the exam of about
25 marks.
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Mixed exercise 2
Answer the following questions:
1. If determine the value of each of the following in terms of WITHOUT USING A CALCULATOR.
1.1
1.2 cos 56°
1.3 cos 34 °
1.4 tan 34 °
2. Determine the value of the following expression: )720cos().180(sin
sin).90cos(2
3. If 900 < A < 360
0 and tan A = , determine without the use of a calculator.
3.1 sin A
3.2 cos2A – sin
2A
4. Given that sin x = t, express the following in terms of t, without the use of calculator.
4.1 cos (x – 900)
4.2 tan x
5. Calculate without the use of a calculator:
332cos.28sin.118tan
208cos2
6. If and [ ; ] determine by using a sketch, the value of
, without the use of a calculator.
7. Simplify to a single trigonometric ratio of x: ( ) ( )
( )
8. Simplify: xx
xxx 22
sin)180sin(
)90(sin.)360tan(.)180cos(
9. Simplify:
)450cos().sin(
)360tan().180cos().180sin(
xx
xxx
3
2
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10. In the diagram below, P (–15 ; m) is a point in the third quadrant and 17cos β + 15=0.
WITHOUT USING A CALCULATOR, determine the value of the following:
10.1 m
10.2 sin β + tan β
10.3 34.cos(180° – β) + 17.cos(90° + β)
11. Evaluate, without using a calculator:
β
.
P (–15; –m)
x
y
O
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Trigonometric Functions
Difference between Grade 10 and Grade 11 Functions
In Grade 10, you plotted the basic graphs of xy sin ; xy cos ; xy tan where
∈ [ ; ]. In Grade 11, you plot graphs within the interval [ ; ]
qpxkay
qpxkay
qpxkay
)(tan
)(cos
)(sin
In Grade 10 you need to know the effect of a and q
In Grade 11 you need to know the effect of k and p
The parameters a; p; q and k affect cos x and sin x in the same way. The tan x graph
behaves differently to both sin x and cos x, because of the asymptotes and the difference
in range.
The function for sin x and cos x are wave- like shapes whereas tan x is a repeated curve
shape.
Because of the wave-shape of the graphs of sin x and cos x, these two graphs have an
amplitude (a). The amplitude is the height from the rest value q to the maximum or the
minimum.
All the three functions have a period which depends on the value of k. The period is the
length required for the graph to make one complete shape.
Knowing the features and the characteristics of the function will help in finding the
equation and interpreting the graph. NB
xy tan has asymptotes, and they should not be part of the domain i.e. kx 18090
Make sure you can use the calculator to draw the functions as it will help you to save
time during the exams.
First make sure that your table mode works only with f(x) and not f(x) and g(x)
qwR41
Above is for the CASIO fx-82ZA
For the CASIO fx-991ZA we use
qwR51
When you work in table mode, we always work in steps of 15 when we are drawing trig
graphs
Worked Example 1
Draw the following sets of graphs on the same set of axes and investigate the effect of parameter a;
p; q and k on the graphs.
a) ( ) , ( ) , ( ) ; for ∈ [ ; ] b) ( ) ; ( ) ; ( ) ; for ∈ [ ; ] c) ( ) ; ( ) ( ); for ∈ [ ; ] d) ( ) ; ( ) ( ); for ∈ [ ; ]
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SOLUTIONS
a)
The effect of k on the graph:
Value of k Period
( ) , so 1 complete shape within 360
( ) , so 2 complete shapes within 360
( ) , so 3 complete shapes within 360
b)
The effect of a on the graphs:
( )
( )
( ) The effect of q on the graphs: Vertical shift
The graph of ( ) is the shift of f(x) by 2 units up,
If q is + the basic graph will move upward
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If q is - the basic graph will move downward
c)
d)
The effect of p on the graph: Horizontal Shift
The graph of m(x) is the shift of the basic graph f(x) by to the left.
The graph of j(x) is the shift of the basic graph f(x) by 45 to the right with the amplitude of 2.
– p shifts the graph to the right.
+ p shifts the graph to the left.
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Worked Example 2
Draw the graphs of ( ) ; ( ) and ( ) for
∈ [ ; ]on the same set of axes.
Hence investigate the effects of the parameters a and q on the graphs of f(x) and g(x) and then
on h(x).
a widens or narrows the graph y = tan x. When a becomes bigger the graph narrows
closer to the line , and when it becomes smaller the graph widens moving away
from The graph of y = tan x has asymptotes which sin x and cos x graphs do not have. The
asymptote is a value where the graph is undefined. It is usually indicated by a dotted
line on the graph.
The effect of the parameters a and q on the graph ( ) is the same as on
the graph of sin x as discussed in the previous Worked Example.
The effect of q on the graph y = tan x is the same as on the graphs of y = cos x and y = sin
x.
Worked Example 3
On the same set of axes, draw sketch graphs of sin 1f x x and cos 2g x x for
90 ;270x . Then answer the questions that follow:
a. Give the range of f.
b. For which values if x, 90 ;270x , is:
i. 0xg
ii. f x g x
iii. . 0f x g x
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The drawing of the graphs are simply done by using your calculator to develop the table and
carefully plotting the graphs. There are letter values on the graphs to assist in the explanation
of question (b). In the exam you will not really do that – unless specifically asked.
a. Range: 2;0y
b. There are three types of questions asked here. Question (i) involves only the graph of
g(x), whereas question (ii) and (iii) involves both graph.
i. The question is asking for which x – values is the graph of g(x) > 0. We need to
remember that g(x) represents the y –values. The y – axis is positive above the x – axis,
so basically this question is asking “where is the graph of g(x) above the x – axis. The
points involved are E, F, G and H.
So in English we will answer “between E and F and again between G and H”.
In Mathematics we will write:
-45° < x < 45° or 135° < x < 225°.
ii. In this question we have to compare the two graphs and find out where the y – values of
f(x) are smaller than the y – values of g(x).
We can see that the y – values of the two graphs are equal at A, B, C and D. The y –
values of f(x) is lying lower than those of g(x) between A and B as well as between C
and D. If the graph is lower, the y – values will be smaller. The Mathematical answer
will be:
-30° ≤ x ≤ 0° or 180° ≤ x ≤ 210°
iii. Although this question involves both graphs, the question is not asking you to compare
the two graphs, but when will the product of the two graphs’ y – values be negative.
The product of two values is negative of one is positive and the other negative. So in
general this means that the graphs must NOT be on the same side of the x – axis. If both
graphs are below the x – axis, the y – values are both negative and the product of the
two will be positive.
For this specific question, the values of f(x) are always positive, so the answer depends
on the y – value of g(x).
-90° ≤ x ≤ -45° or 45° ≤ x ≤ 135° or x ≥ 225°.
Grade 11 Trigonometry Notes 2020
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Exercise on Trigonometry Functions
1. xxf 2sin)( for 90180 x is shown in the sketch below:
1.1 Write down the range of f. (2)
1.2 Determine the period of
xf
2
3. (2)
1.3 Draw the graph of )30cos()( xxg for 90180 x on the same set of
axes. Clearly label ALL x- intercepts and turning points. (4)
1.4 Determine the general solution of )30cos(2sin xx (5)
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2. The graphs of the functions xaxf tan)( and xbxg cos)( for 2700 x are
shown in the diagram below. The point ( )2;225 lies on f. The graphs intersect at points
P and Q.
2.1 Determine the numerical values of a and b. (4)
2.2 Determine the minimum value of g(x) +2. (2)
2.3 Show that if the x-coordinate of P is , then the x-coordinate of Q is ( )
3
3.1 Write down the solutions for 13sin x on the interval ]180;90[ (2)
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3.2 Give the maximum value of h of h(x) = f(x) – 1 (2)
3.3 Draw the graph of g(x) = 3cos x for ∈ [ ; ] on the same set of axis. (3)
3.4 Use the graphs to determine how many solutions there are to the equation
on the interval [ ; ]
(2)
3.5 Use the graphs to solve: f(x).g(x) < 0 (4)
Do Exercise 1 on page 188
Do Exercise 2 on page 192
Do Exercise 3 on page 194
Do Revision Exercise on page 195
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Solving Problems in Two Dimensions
Any triangle can be solved, if THREE properties of the triangle are given/known, of which
one must be a side length, by using:
The trig ratios in RIGHT-ANGLED triangles
The sine or cosine rule
The Sine, Cosine and Area Rules
TYPES OF QUESTIONS:
Numeric (calculations) problems
Non-numeric (prove type) problems
According to the CAPS document, you must be able to:
Establish (prove) the rules
Apply the rules in solving 2D problems
When to use what?
If the triangle is
RIGHT-ANGLED
If the triangle is NOT
right-angled
Use the trig RATIOS
Sine or Cosine Rule:
COSINE RULE if ……
3 sides of the triangle are given
2 sides and an included angle of the
triangle is given
SINE RULE if ……
Any condition that does NOT satisfy
the cosine rule
AREA RULE if ……
…. “area” is mentioned (only)
Remember:
3 properties of a triangle, of
which at least one is a side, must
be given in a triangle in order to
work in that triangle
Grade 11 Trigonometry Notes 2020
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In any ABC the rules are applied as follow:
A
B C
b
A
b
B C
c
a
c
a
Sine rule
If an ANGLE is asked
𝑠𝑖𝑛𝐴
𝑎
𝑠𝑖𝑛𝐵
𝑏
𝑠𝑖𝑛𝐶
𝑐
If a SIDE is asked
𝑎
𝑠𝑖𝑛𝐴
𝑏
𝑠𝑖𝑛𝐵
𝑐
𝑠𝑖𝑛𝐶
Cosine rule
If an ANGLE is asked
𝑐𝑜𝑠𝐴 𝑏 𝑐 𝑎
𝑏𝑐
𝑐𝑜𝑠𝐵 𝑎 𝑐 𝑏
𝑎𝑐
𝑐𝑜𝑠𝐶 𝑎 𝑏 𝑐
𝑎𝑏
If a SIDE is asked
𝑎 𝑏 𝑐 𝑏𝑐 𝑐𝑜𝑠𝐴 𝑏 𝑎 𝑐 𝑎𝑐 𝑐𝑜𝑠𝐵 𝑐 𝑎 𝑏 𝑎𝑏 𝑐𝑜𝑠𝐶
Area rule 𝐀𝐫𝐞𝐚 𝐨𝐟 𝐀𝐁𝐂
𝑎𝑏 𝑠𝑖𝑛𝐶 or
𝑏𝑐 𝑠𝑖𝑛𝐴 or
𝑎𝑐 𝑠𝑖𝑛𝐵
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Tips for Solving 2D Problems
1. The diagram usually consists of 2 or more triangles with COMMON sides.
2. One of the triangles is sometimes right-angled, so you can use the trig ratios to solve it.
(In triangles without right angles, the Sine, Cosine and Area rules must be applied.)
You can also apply the Sine rule in a right-angled triangle.
3. Make use of basic Geometry to obtain additional information, such as vertical opposite
angles, interior angles of a triangle, etc.
4. Start in the triangle that contains the most information, then move along to the triangle
in which the required line/angle is.
5. In applications, we often use angles of DEPRESSION and ELEVATION. Both are
measured from the horisontal.
Using the Sine, Cosine and Area Rules
Worked Example 1
A soccer player (S) is 15 m from the back line of a soccer field (CH). She aims towards the
goal (GH). The angle from the left goal post (G) to the soccer player is 116°. The goal posts
are 7,32 metres apart. The diagram below represents the situation.
a) Calculate the size of SGC ˆ .
b) Calculate SG, the distance between the soccer player and the left goal post FG.
c) Calculate the size of HSG ˆ , the angle within which the soccer player could possibly score
a goal.
SOLUTION:
a) 64ˆSGC (angles on straight line CGH)
b) In CGS , right-angled at C
Angle of elevation
Angle of depression
HORISONTAL
H
15
G C 7,3
2 116°
S
𝐶𝐺𝑆 is right-angled
use the trig ratios
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= 16,689 02911 … metres
16,69 metres
c) In SGH we know the lengths of two sides (SG and GH and the size of the included
angle ) so we use the cosine rule
116cos69,1632,7269,1632,7222SH
√( ) ( ) ( ) ( ) SH = 20,957 389 36
20,96 metres
Enough information in SGH is known to use either the sine or cosine rule to calculate .
Using the sine rule:
(
)
= 18,293 926 57
18,3
Using the cosine rule:
( ) ( ) ( )
( )( )
(( ) ( ) ( )
( )( )
= 18,296 286 19
18,3
Worked Example 2
In the figure below ˆSPQ = , ˆPQS and PQ = h. PQ and SR are perpendicular on RQ.
Prove that
sin .cos
sin
hRS
Solution:
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In ΔPQS:
180ˆQSP
sin
sin.SQ
sinsin
SQ
180sinsin
SQ
SsinPsin
h
h
h
sp
In ΔSRQ:
90ˆRQS
sin
cos.sin.
cossin
sin.
cosSQRS
90sinSQ
RS
h
hRS
Do Exercise 16 on page 171
Do Exercise 17 on page 174
Do Exercise 18 on page 178
Do Exercise 19 on page 181
Do Exercise 20 on page 184
Do Revision Exercise on page 185
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Extra Exercise
1. Triangle PQS represents a certain area of a park. R is a point on line PS such that QR
divides the area of the park into two triangular parts, as shown below. PQ
, RS
and RQ x units.
1.1 Calculate the size of .
1.2 Determine the area of Δ QRS in terms of x.
2. In the diagram below, ABCD is a cyclic quadrilateral with DC = 6 units, AD = 10
units, and .
Calculate the following, correct to ONE decimal place:
2.1 The length of BC
2.2 The area of ∆ABC
P
Q
R
S
𝑥
𝑥
𝑥
𝑥
6
D
C
B
A
100°
10
40°