Math  /  Trigonometry

QuestionProduct-to-Sum Formulas:
1. sin(u)cos(v)=12[sin(u+v))+sin(uv)]\left.\sin (u) \cos (v)=\frac{1}{2}[\sin (u+v))+\sin (u-v)\right]
2. cos(u)sin(v)=12[sin(u+v))sin(uv)]\left.\cos (u) \sin (v)=\frac{1}{2}[\sin (u+v))-\sin (u-v)\right]
3. cos(u)cos(v)=12[cos(u+v))+cos(uv)]\left.\cos (u) \cos (v)=\frac{1}{2}[\cos (u+v))+\cos (u-v)\right]
4. sin(u)sin(v)=12[cos(uv))cos(u+v)]\left.\sin (u) \sin (v)=\frac{1}{2}[\cos (u-v))-\cos (u+v)\right]

Rewrite the expression below using one of the given formulas. sin(3x)sin(5x)\sin (3 x) \sin (5 x) Using formula number one: sin(3x)sin(5x)=12[cos(3x5x)cos(3x+5x)]\sin (3 x) \sin (5 x)=\frac{1}{2}[\cos (3 x-5 x)-\cos (3 x+5 x)] Using formula four: sin(3x)sin(5x)=12[cos(3x+5x)cos(3x5x)]\sin (3 x) \sin (5 x)=\frac{1}{2}[\cos (3 x+5 x)-\cos (3 x-5 x)] Using formula one: sin(3x)sin(5x)=12[cos(3x+5x)cos(3x5x)]\sin (3 x) \sin (5 x)=\frac{1}{2}[\cos (3 x+5 x)-\cos (3 x-5 x)] Using formula number four: sin(3x)sin(5x)=12[cos(3x5x)cos(3x+5x)]\sin (3 x) \sin (5 x)=\frac{1}{2}[\cos (3 x-5 x)-\cos (3 x+5 x)]

Studdy Solution

STEP 1

1. We are given a trigonometric expression sin(3x)sin(5x)\sin(3x) \sin(5x) to rewrite using product-to-sum formulas.
2. We have four product-to-sum formulas available for use.
3. We need to identify which formula correctly applies to the given expression.

STEP 2

1. Identify the correct product-to-sum formula for sin(u)sin(v)\sin(u) \sin(v).
2. Apply the identified formula to rewrite sin(3x)sin(5x)\sin(3x) \sin(5x).

STEP 3

Identify the correct product-to-sum formula for the expression sin(3x)sin(5x)\sin(3x) \sin(5x).
The relevant formula for sin(u)sin(v)\sin(u) \sin(v) is:
sin(u)sin(v)=12[cos(uv)cos(u+v)] \sin(u) \sin(v) = \frac{1}{2}[\cos(u-v) - \cos(u+v)]
This is formula number four from the provided list.

STEP 4

Apply the identified formula to rewrite the expression sin(3x)sin(5x)\sin(3x) \sin(5x).
Using the formula:
sin(3x)sin(5x)=12[cos(3x5x)cos(3x+5x)] \sin(3x) \sin(5x) = \frac{1}{2}[\cos(3x - 5x) - \cos(3x + 5x)]
Simplify the terms inside the cosine functions:
cos(3x5x)=cos(2x)=cos(2x) \cos(3x - 5x) = \cos(-2x) = \cos(2x)
cos(3x+5x)=cos(8x) \cos(3x + 5x) = \cos(8x)
Thus, the expression becomes:
sin(3x)sin(5x)=12[cos(2x)cos(8x)] \sin(3x) \sin(5x) = \frac{1}{2}[\cos(2x) - \cos(8x)]

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