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### Pythagorean Identity

Main formulas:

• The Pythagorean theorem: The side lengths of a right triangle satisfy a2 + b2 = c2.
• The Pythagorean identity: For any angle x, we have sin2 x + cos2 x = 1.

Example 1:

Use the Pythagorean theorem to prove the Pythagorean identity.

Example 2:

If cosθ = 0.47 and θ is in the fourth quadrant, find sinθ .

Example 3:

Verify the following trigonometric identity :
 1+cosθ sinθ = sinθ 1 − cosθ

Example 4:

Use the Pythagorean identity to prove the Pythagorean theorem.

Example 5:

If sinθ = − [5/13] and θ is in the third quadrant, find cosθ .
rac{1}{2}cos(3x + rac{\pi }{6}) - 3

## Identify the amplitude, period, phase shift, and vertical shift of the following function: [1/2]cos(3x + [(π)/6]) − 3

• Recall: Acos(Bx + C) + D where |A| is the amplitude; [(2π)/B] is the period; − [C/B] is the phase shift; D is the vertical shift
• A = [1/2], B = 3, C = [(π)/6], D = - 3

## Identify the amplitude, period, phase shift, and vertical shift of the following function: − 4sin(2x − [(π)/4]) + 1

• Recall: Asin(Bx + C) + D where |A| is the amplitude; [(2π)/B] is the period; − [C/B] is the phase shift; D is the vertical shift
• A = - 4, B = 2, C = − [(π)/4], D = 1

## Find a cosine wave with amplitude - 2, period [(π)/3], phase shift p, and vertical shift of 4

• Recall: Acos(Bx + C) + D where |A| is the amplitude; [(2π)/B] is the period; − [C/B] is the phase shift; D is the vertical shift
• A = - 2, D = 4, We need to find B and C by working backwards
• First find B by using the period. We know Period = [(2π)/B] so,
• [(π)/3] = [(2π)/B] ⇒ Bπ = 6π ⇒ B = 6
• Now find C by using the phase shift and the value of B we just calculated
• π = − [C/6] ⇒ 6π = - C ⇒ C = - 6π

## Find a sine wave with amplitude [1/2], period 4π, phase shift [(π)/4], and vertical shift of - 2

• Recall: Asin(Bx + C) + D where |A| is the amplitude; [(2π)/B] is the period; − [C/B] is the phase shift; D is the vertical shift
• A = [1/2], D = - 2, We need to find B and C by working backwards
• First find B by using the period. We know Period = [(2π)/B] so,
• 4π = [(2π)/B] ⇒ B4π = 2π ⇒ B = [1/2]
• Now find C by using the phase shift and the value of B we just calculated
• [(π)/4] = − [C/([1/2])] ⇒ [1/2]π = − 4C ⇒ C = − [(π)/8]

## Identify the amplitude, period, phase shift, and vertical shift of the following function: [1/3]sin([1/2]x − [(π)/3]) - 5

• Recall: Asin(Bx + C) + D where |A| is the amplitude; [(2π)/B] is the period; − [C/B] is the phase shift; D is the vertical shift
• A = [1/3], B = [1/2], C = − [(π)/3], D = - 5

## Identify the amplitude, period, phase shift, and vertical shift of the following function. Graph the function. 3cos(x + π) - 3

• Recall: Acos(Bx + C) + D where |A| is the amplitude; [(2π)/B] is the period; − [C/B] is the phase shift; D is the vertical shift
• Amplitude = |A| = 3, Period = [(2π)/B] = [(2π)/1] = 2π, Phase Shift = − [C/B] = − [(π)/1] = − π, Vertival Shift = D = - 3
• First graph f(x) = cos(x)
• Now graph the amplitude f(x) = 3cos(x)
• Now graph the period. Since the period is 2π, the graph will stay the same
• Now graph the phase shift f(x) = 3cos(x + π). The graph will shift p units to the left
• Now graph the vertical shift f(x) = 3cos(x + π) - 3. The graph will shift down from 3 to 0 and - 3 to - 6

## Identify the amplitude, period, phase shift, and vertical shift of the following function. Graph the function. 2sin(2x - [(π)/3]) + 2

• Recall: Asin(Bx + C) + D where |A| is the amplitude; [(2π)/B] is the period; − [C/B] is the phase shift; D is the vertical shift
• Amplitude = |A| = 2, Period = [(2π)/2] = [(2π)/2] = π, Phase Shift = − [C/B] = − [( − [(π)/3])/2] = [(π)/6], Vertival Shift = D = 2
• First graph f(x) = sin(x)
• Now graph the amplitude f(x) = 2sin(x)
• Now graph the period. The period is p, so the graph will cycle every p for f(x) = 2sin(2x)
• Now graph the phase shift f(x) = 2sin(2x - [(π)/3]). The graph will shift [(π)/6] units to the right
• Now graph the vertical shift f(x) = 2sin(2x - [(π)/3]) + 2. The graph will shift up from 2 to 4 and - 2 to 0

## Identify the amplitude, period, phase shift, and vertical shift of the following function. Graph the function. - cos([1/2]x − [(π)/4]) + 2

• Recall: Acos(Bx + C) + D where |A| is the amplitude; [(2π)/B] is the period; − [C/B] is the phase shift; D is the vertical shift
• Amplitude = |A| = 1, Period = [(2π)/2] = [(2π)/([1/2])] = 4π, Phase Shift = − [C/B] = − [( − [(π)/4])/([1/2])] = [(π)/2], Vertival Shift = D = 2
• First graph f(x) = cos(x)
• Now graph the amplitude f(x) = - cos(x)
• Now graph the period. The period is 4π, so the graph will cycle every 4π for f(x) = - cos([1/2]x)
• Now graph the phase shift f(x) = - cos([1/2]x - [(π)/4]). The graph will shift [(π)/2] units to the right
• Now graph the vertical shift f(x) = - cos([1/2]x - [(π)/4]) + 2. The graph will shift up from - 1 to 1 and 1 to 3

## Identify the amplitude, period, phase shift, and vertical shift of the following function. Graph the function. - 3sin(4x + 4π) - 1

• Recall: Asin(Bx + C) + D where |A| is the amplitude; [(2π)/B] is the period; − [C/B] is the phase shift; D is the vertical shift
• Amplitude = |A| = 3 ,Period = [(2π)/B] = [(2π)/4] = [(π)/2], Phase Shift = − [C/B] = − [(4π)/4] = − π, Vertival Shift = D = - 1
• First graph f(x) = sin(x)
• Now graph the amplitude f(x) = - 3sin(x)
• Now graph the period. The period is [(π)/2], so the graph will cycle every [(π)/2] for f(x) = - 3sin(4x)
• Now graph the phase shift f(x) = - 3sin(4x + 4π). The graph will shift π units to the left
• Now graph the vertical shift f(x) = - 3sin(4x + 4π) - 1. The graph will shift down from 3 to 2 and - 3 to - 4

## Find a cosine wave with amplitude 4, period [(π)/4], phase shift [(π)/3], and vertical shift of 7

• Recall: Acos(Bx + C) + D where |A| is the amplitude; [(2π)/B] is the period; − [C/B] is the phase shift; D is the vertical shift
• A = 4, D = 7, We need to find B and C by working backwards
• First find B by using the period. We know Period = [(2π)/B] so,
• [(π)/4] = [(2π)/B] ⇒ Bπ = 8π ⇒ B = 8
• Now find C by using the phase shift and the value of B we just calculated
• [(π)/3] = − [C/8] ⇒ 8π = - 3C ⇒ C = − [(8π)/3]

### 4cos(8x - [(8π)/3]) + 7

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.