Question #1e265

2 Answers
Feb 13, 2018

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

Feb 13, 2018

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!