3) 0.96) 0.99) 3/412) 0.6 15) 2 cycles; a = 2; Period = 18) y = 4 sin (x) 21) y = 1.5 sin (120x)...
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3) 0.9 6) –0.9 9) –3/4 12) –0.6 15) 2 cycles; a = 2; Period = π 18) y = 4 sin (½x) 21) y = 1.5 sin (120x) 24) 27) 30) Period = π; y = 2.5 sin(2x) 33) Period = 4; y = 3 sin(90x) 13.4 Homework Answers
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Traces x – coordinate values of the unit circle Period of 360° or 2π Amplitude of 1 Begins at its maximum max, zero, min, zero, max 1 360° 2 π
Transcript of 3) 0.96) 0.99) 3/412) 0.6 15) 2 cycles; a = 2; Period = 18) y = 4 sin (x) 21) y = 1.5 sin (120x)...
3) 0.9 6) –0.9 9) –3/4 12) –0.615) 2 cycles; a = 2; Period = π18) y = 4 sin (½x) 21) y = 1.5 sin (120x)
24) 27)
30) Period = π; y = 2.5 sin(2x)33) Period = 4; y = 3 sin(90x)
13.4 Homework Answers
Section 13.5The Cosine Function
Traces x – coordinate values of the unit circle
Period of 360° or 2π Amplitude of 1 Begins at its maximum
max, zero, min, zero, max
Cosine Curve
1
360°
2 π
The Sine Curve The Cosine Curve
ComparingList the similarities and differences between these two curves.
Amplitude = |a|
Period = or
The Cosine Function: y = a cos (bθ)
360°b
2πb
Sketching a Graph
Sketch a cosine curve that has the following:
1)Amplitude of 2
2)Period of 720°
Assume a > 1
2
720°
Writing With given information we
can write equations to model a situation
The amplitude is half of a given wave height
The b value can be found by solving P = for the given period
Writing the WavesWrite a cosine function to model 10 in. waves that occur every 4 seconds.1) a = 10/2 = 52) P = 2π/b
4 = 2π/bb = 2π/4b = π/2
So, y = 5 cos ((π/2)θ)
Equations
2πb
We can use the intersect feature of the calculator to find certain values
Solve 3 = 5 cos ((π/2)x) for0 ≤ x ≤ 2π1) Graph y1 = 32) Graph y2 = 5 cos ((π/2)x)3) Set window [0, 2π] by [-4, 4]4) Use <CALC>, <5: intersect>
to find the intersections
Solving by Calculator
x ≈ 0.59x ≈ 3.41
x ≈ 4.59
For tomorrow, complete exercises 1 – 21 odd, starting on page 732
Homework
