Verifying trig identity!

asterisk_man · Mar 13, 2007, 12:05 PM

This one was pretty easy. We're just deriving the "quad angle formula for sin", similar to the double angle formula. I just looked at a table of trig identities:
Table of Trigonometric Identities
There are 2 important ones there:
$\sin\left(2u\right)=2\sin\left(u\right)\cos\left(u \right) \\ \cos\left(2u\right)=2\cos^2\left(u\right)-1 $

So we just use these two to expand the sin(4x)
First let u=2x and expand with the double angle sin formula, then substitute the 2x for the you and use both double angle formulas to expand
$ U=2x \\ \sin(4x) \\ \text{substitute you for 2x} \\ \sin(2u) \\ \text{apply double angle \sin formula} \\ 2\sin(u)\cos(u) \\ \text{substitute 2x for u} 2\sin(2x)\cos(2x) \\ \text{apply double angle \sin and \cos formulas} \\ 2(2\sin(x)\cos(x))(2\cos^2(x)-1) \\ \text{multiply out the terms} \\ 8\sin(x)\cos^3(x)-4\sin(x)\cos(x) $

This is the same as what you wrote in your question.

If you looked at your table of trig identities you should have seen that cos^2(x)-1 was cos(2x) and maybe that would have given you an idea.

Ask questions if you have any left.
Thanks!

minnieirl55 · Dec 10, 2008, 07:46 PM

How to solve 1+ sin 2x = (sin x + cos x)^2

snowflake317 · May 20, 2010, 07:54 AM

If we take the Right Side of this equation (sinx + cosx)^2 then we can expand it according to the identity (a+b)^2= a^2 + b^2 +2ab
=> sin^2x + cos^2x +2sinxcosx
According to the basic trigonometry identities , sin^2x + cos^2x=1 and 2sinxcosx=sin2x
=> 1+sin2x which is equal to the left side of your equation.

If you want to solve it using the left side, you just have to go from bottom to top of the solution written above.

Glad to help

Unknown008 · May 20, 2010, 08:21 AM

Yes, except that you need to check the date... For the last question, it was 11 Dec 2008... Two years ago

Next time, check the date

snowflake317 · May 20, 2010, 08:41 AM

Thanks. . I'll keep that in mind next time :P

Verifying trig identity!

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