Weird trig identity (1 Viewer)

[MarsBarz]

Can someone tell me where this identity comes from and if it's in the 3U course + proof

(sinA)^2=1/2(1-cos2A)

 

[MarsBarz]

Ok I've found out that it's a double angle trig formula but it is not in my math book. Someone mind proving it?

 

[Slidey]

cos2A=cos^2(A)-sin^2(A)=1-sin^2(A)-sin^2(A)
1-cos2A=2sin^2(A)
sin^2(A)=(1-cos2A)/2

 

[MarsBarz]

Thanks.

What about this integration(I):

I (sin 2x)^2 dx

It was in my trial today and I didn't know how to do it because of the 2x. Had it been x I could have simply substituted the trig identity (sinA)^2=1/2(1-cos2A) but what should I have done in this particular case?

 

[withoutaface]

S sin²2x dx
S (2sinxcosx)² dx
S 2sin²x(1-sin²x) dx
S 2sin²x - 2sin⁴x dx
S 1-cos2x - (1-cos2x)²/2 dx
S 1-cos2x - (1 -2cos2x + cos²2x)/2 dx
1/2 S 1 -(cos²x - sin²x) dx
1/2 S (1-cos²x)+sin²x dx
1/2 S 2sin²x dx
1/2 S 1-cos2x dx
1/2 (x - 2sin2x) + C

There's probably a more elegant method.

 

[alphatango]

Of course, if you're used to doing Int(sin²x dx), then you can always make the simple substitution u=2x and then proceed as normal.

 

[shafqat]

Thanks.

What about this integration(I):

I (sin 2x)^2 dx

It was in my trial today and I didn't know how to do it because of the 2x. Had it been x I could have simply substituted the trig identity (sinA)^2=1/2(1-cos2A) but what should I have done in this particular case?

Using that identity, it becomes S 1/2 (1 - cos4x) = ....

 

[alphatango]

A better method would probably be to take the identity you know:

sin²x = (1/2)(1-cos2x)

and replace x with 2x, giving you:

sin²2x = (1/2)(1-cos4x)

which is trivial to integrate.

 

[MarsBarz]

Ok thanks. I'm a tool. I've got 6 hours of exams tomorrow but then nothing for a week so I'll use that week to brush up on my trig.