Trig Identities Applications Question

[Auburn]

Hello! I've been tackling the question 'Express sin3x+sinx as a product and hence solve 1/2(sin3x+sinx)=sin2x ; x∈R' but I'm stumped - I'm not sure whether I've even approached it correctly. This is what I did:

sin(3x+x)=sin3x.cosx+sinx.cos3x

inserting this into the second equation,
1/2(sin3x.cosx+sinx.cos3x)=2sinx.cosx

Moving cosx over
4sinx=sin3x+3sinx

sinx=sin3x

sin(2x+x)-sinx=0

sin2xcosx+cos2x.sinx-sinx=0

2sinxcos^2x+(cos^2 x - sin^2 x) sinx - sinx = 0

2sinxcos^2 x + cos^2xsinx - sin^3 x -sinx=0

sinx(2cos^2 x + cos^2 x - sin^2 x - 1) = 0

sinx(3cos^2x - sin^2x -1) =0

BRACKETS:

3cos^2x - sin^2x - 1 = 0

3cos^2x - (1- cos^2x) - 1 = 0

3cos^2x - 1 + cos^2x - 1 = 0

4cos^2x - 2 = 0

cos^2x = 1/2
and
sinx = 0

But at this point I'm stumped. The answer says x = n∏/2 , and I would've gone about the second equation by doing:
cosx = √(1/2)
x = α = cos^-1(√(1/2)) but this is obviously wrong... :S Argh, how am I supposed to get that answer?

Thanks in advance.

 

[jing2178]

1. Is the equation 1/2(sin3x + sinx) =sin2x ie do you have the 2 and 1 the wrong way round in the first line of your question?

2. sin(3x) + sin(x) is not the same as sin(3x+x)

This page might help you.
http://www.sosmath.com/trig/prodform/prodform.html

 

[Auburn]

1. Is the equation 1/2(sin3x + sinx) =sin2x ie do you have the 2 and 1 the wrong way round in the first line of your question?

2. sin(3x) + sin(x) is not the same as sin(3x+x)

This page might help you.
http://www.sosmath.com/trig/prodform/prodform.html

Ah, yes it is, sorry. And thanks, I'll see what else I can do.

 

[Curious3141]

Hint: use Factor formulae.

Study this carefully: http://www.trigonometry-help.net/trig-formulas.php

Look at how they expressed sin P + sin Q as a product.

 

[Auburn]

My working now:
sin3x+sinx = 2sin(3x+x/2)cos(3x+x/2)=0,

2sin2x.cos2x=0

adding into parenthesis
1/2(2sin2x.cos2x)=sin2x

2sin2x.cos2x=2sin2x

2sin2x.cos2x - 2sin2x=0

2sin2x(cos2x-1)=0

2sin2x=0, cos2x=0

But using the general equation formula for cos2x=0 gives me x=∏n, when the answer should be x=∏n/2? 2sin2x isn't even close.

I apologise for my high school level ignorance in advance!

 

[e^(i Pi)+1=0]

My working now:
sin3x+sinx = 2sin(3x+x/2)cos(3x+x/2)=0,

It looks like you made a mistake in the cosine part, look at the formula again. Also, why did you set the right side equal to zero? It looks like you made some mistakes further down, but you need to restart from an earlier point anyway.

1/2(2sin2x.cos2x)=sin2x

does not become

2sin2x.cos2x=2sin2x

The 1/2 and the 2 merely cancel.

 

[jing2178]

My working now:
sin3x+sinx = 2sin(3x+x/2)cos(3x+x/2)=0,

Note that 3x+x/2 = 7x/2 not 2x what you should write is (3x +x)/2

sin3x+sinx=2sin((3x+x)/2)cos(...) I have left the cos blank as e^(i Pi)+1=0 said you need to check the formula to obtain the correct expression.