I can't solve this for the life of me and I am sure I will feel like an idiot after seeing the solution. Any help would be greatly appreciated before I pull my hair out over this little problem! Argh!

verify the identity:

sin4x = 8cos"cubed"xsinx - 4cosxsinx

if any help, I tried factoring out 4cosxsinx leaving 2cos"squared"x - 1, however I didn't get too far into this before getting stuck.. :)

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:


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 u and use both double angle formulas to expand


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!

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

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 :D

Yes, except that you need to check the date... for the last question, it was 11 Dec 2008... two years ago :rolleyes:

Next time, check the date :)

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