# Question #1e265

Feb 13, 2018

#### Explanation:

.
${y}_{1} = 20 \sin \left(3 x + \theta\right)$

${y}_{2} = 20 \sin \left(3 x - \theta\right)$

${y}_{1} + {y}_{2} = 20 \sin \left(3 x + \theta\right) + 20 \sin \left(3 x - \theta\right)$

We have a Sum to Product formula that says:

$\sin \alpha + \sin \beta = 2 \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha - \beta}{2}\right)$

Therefore:

${y}_{1} + {y}_{2} = 20 \left[\sin \left(3 x + \theta\right) + \sin \left(3 x - \theta\right)\right]$

${y}_{1} + {y}_{2} = 20 \left[2 \sin \left(\frac{3 x + \theta + 3 x - \theta}{2}\right) \cos \left(\frac{3 x + \theta - 3 x + \theta}{2}\right)\right]$

${y}_{1} + {y}_{2} = 40 \sin 3 x \cos \theta$

Feb 13, 2018

See below

#### Explanation:

In order to complete this proof, you need to know the angle addition/subtraction formulas for sine:

$\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \sin \beta \cos \alpha$
$\sin \left(\alpha - \beta\right) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$

Using these formulas for y1 and y2 we get:

$y 1 = 20 \sin \left(3 x + \theta\right) = 20 \left(\sin 3 x \cos \theta + \sin \theta \cos 3 x\right)$

Distribute the 20:
$y 1 = 20 \sin 3 x \cos \theta + 20 \sin \theta \cos 3 x$

$y 2 = 20 \sin \left(3 x - \theta\right) = 20 \left(\sin 3 x \cos \theta - \sin \theta \cos 3 x\right)$

Distribute the 20
$y 2 = 20 \sin 3 x \cos \theta - 20 \sin \theta \cos 3 x$

Therefore:

$y 1 + y 2 = 20 \sin 3 x \cos \theta + 20 \sin \theta \cos 3 x + 20 \sin 3 x \cos \theta - 20 \sin \theta \cos 3 x$
$y 1 + y 2 = 20 \sin 3 x \cos \theta + 20 \sin 3 x \cos \theta = 40 \sin 3 x \cos \theta$