Express product as a sum/different containing only sines or cosines. sin4Asin2A=???? I stuck at beginning. THX WHO HELP~!
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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
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
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)
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: $\displaystyle \sin (4A) \sin (2A) = \frac{1}{2} (\cos (2A) - \cos(6A) )$.
I got it from Mr.fantastic. THx Mr.Green ans Mr.fantastic~!! I'm continues do my homework~!!
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