Question

Question #1e265

Answer 1

Please see below.

Explanation:

.
`y_1=20sin(3x+theta)`

`y_2=20sin(3x-theta)`

`y_1+y_2=20sin(3x+theta)+20sin(3x-theta)`

We have a Sum to Product formula that says:

`sinalpha+sinbeta=2sin((alpha+beta)/2)cos((alpha-beta)/2)`

Therefore:

`y_1+y_2=20[sin(3x+theta)+sin(3x-theta)]`

`y_1+y_2=20[2sin((3x+theta+3x-theta)/2)cos((3x+theta-3x+theta)/2)]`

`y_1+y_2=40sin3xcostheta`

Answer 2

See below

Explanation:

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

`sin(alpha+beta)=sinalphacosbeta+sinbetacosalpha`
`sin(alpha-beta)=sinalphacosbeta-sinbetacosalpha`

Using these formulas for y1 and y2 we get:

`y1=20sin(3x+theta)=20(sin3xcostheta+sinthetacos3x)`

Distribute the 20:
`y1=20sin3xcostheta+20sinthetacos3x`

`y2=20sin(3x-theta)=20(sin3xcostheta-sinthetacos3x)`

Distribute the 20
`y2=20sin3xcostheta-20sinthetacos3x`

Therefore:

`y1+y2=20sin3xcostheta+20sinthetacos3x+20sin3xcostheta-20sinthetacos3x`
`y1+y2=20sin3xcostheta+20sin3xcostheta=40sin3xcostheta`

It's all about the addition/subtraction formulas!