# trigo sum/different~do not do~!

• Sep 11th 2008, 04:52 AM
sanikui
trigo sum/different~do not do~!
Express product as a sum/different containing only sines or cosines.

sin4Asin2A=????

I stuck at beginning.

THX WHO HELP~!
• Sep 11th 2008, 05:07 AM
thelostchild
remember that
$\displaystyle \sin 2A=2\sin A \cos A$

(so obviously $\displaystyle \sin 4A = 2\sin2A \cos 2A$)

and $\displaystyle \cos 2A = \cos^2A - \sin^2 A$

make substitutions and simplify :)
• Sep 11th 2008, 05:08 AM
mr fantastic
Quote:

Originally Posted by sanikui
Express product as a sum/different containing only sines or cosines.

sin4Asin2A=????

I stuck at beginning.

THX WHO HELP~!

See my reply in this thread for the formulae you need: http://www.mathhelpforum.com/math-he...dentities.html
• Sep 11th 2008, 05:20 AM
Mr.Green
sin4Asin2A=sin2(2A)sin2A
=2sin2Acos2Asin2A
=2(2sinAcosA)(1-2sin^2A)(2sinAcosA)
=8sin^2Acos^2A(1-2sin^2A)
=8sin^2A(1-sin^2A)(1-2sin^2A)
• Sep 11th 2008, 05:34 AM
mr fantastic
Quote:

Originally Posted by Mr.Green
sin4Asin2A=sin2(2A)sin2A
=2sin2Acos2Asin2A
=2(2sinAcosA)(1-2sin^2A)(2sinAcosA)
=8sin^2Acos^2A(1-2sin^2A)
=8sin^2A(1-sin^2A)(1-2sin^2A)

No. The answer has to be a sum.

From the compound angle formulae: cos(x - y) - cos(x + y) = 2 sin(x) sin(y).

Substitute x = 4A and y = 2A:

$\displaystyle \sin (4A) \sin (2A) = \frac{1}{2} (\cos (2A) - \cos(6A) )$.
• Sep 11th 2008, 06:43 AM
sanikui
I got it from Mr.fantastic.
THx Mr.Green ans Mr.fantastic~!!
I'm continues do my homework~!!(Happy)