trig problem

[vanalm]

I'm taking trig online and the book does not cover some of the problems we are supposed to do. I am to verify the identity.

sin 4x = 4 sinxcosxcos2x

I'm not quite sure what to do with this. First of all, are you allowed to look at sin4x as two double angle identities? sin2x plus sin2x? second of all, on the right side of the equation, is that to be read 4(sinxcosxcos2x), or does the 4 only apply to the sinx?

My book does not go over any of this, and I'm not quite sure how to do it.

 

[royhaas]

Start with the double angle formula for \(\displaystyle sin(2y)\), then substitute \(\displaystyle y=2x\) and continue.

 

[wjm11]

I'm taking trig online and the book does not cover some of the problems we are supposed to do. I am to verify the identity.

sin 4x = 4 sinxcosxcos2x

I'm not quite sure what to do with this. First of all, are you allowed to look at sin4x as two double angle identities? sin2x plus sin2x? second of all, on the right side of the equation, is that to be read 4(sinxcosxcos2x), or does the 4 only apply to the sinx?

First of all, no, sin4x does NOT equal sin2x + sin2x. Treat it as

sin4x = sin (2x + 2x)

You can then apply either a double angle formula or a sum of two angles formula. It’s easier to see if you substitute another variable for 2x. Let y = 2x. Then the expression becomes sin(2y), to which you can apply the double angle formula. You can then substitute 2x back into the expression when you’re done.

Re your second question, the 4 is not tied to the sinx term. The right hand side is simply the multiplication of four separate terms: 4, sinx, cosx, and cos2x.

 

[vanalm]

You can then apply either a double angle formula or a sum of two angles formula. It’s easier to see if you substitute another variable for 2x. Let y = 2x. Then the expression becomes sin(2y), to which you can apply the double angle formula. You can then substitute 2x back into the expression when you’re done.

Re your second question, the 4 is not tied to the sinx term. The right hand side is simply the multiplication of four separate terms: 4, sinx, cosx, and cos2x.

This really helps ALOT, thank you so much. But I am running into one more problem. When I use the sum to product identity, i get:

2 sin [(2x + 2x)/2] cos[(2x - 2x)/2]

=2 sin 2x cos 0

=2*2 sinx cosx cos0 (double angle identity)

=4 sinx cosx cos0

But the answer is supposed to be 4sinx cosx cos2x

 

[wjm11]

When I use the sum to product identity, i get:

2 sin [(2x + 2x)/2] cos[(2x - 2x)/2]

NO, sin4x does NOT equal sin2x + sin2x. You cannot use the sum to product identity. Use the sine double-angle formula:

sin2x = 2sinxcosx

sin(4x) = sin(2(2x)) = 2sin(2x)cos(2x) = 2[2sinx(cosx)]cos2x = 4(sinx)(cosx)(cos2x)

 

[vanalm]

ok, the only reason i did it that way is because you said:

First of all, no, sin4x does NOT equal sin2x + sin2x. Treat it as

sin4x = sin (2x + 2x)

which is what i did. But anyway, thank you for clearing that up for me. That makes sense now.