An easy trigo question. Solve sin2xsin6x=sin^2(4x) for 0° ≦ x ≦ 360°
cos(6x - 2x) - cos(2x+6x) = 2 sin2xsin6x <-- this might work
sin2xsin6x=(1/2)(cos 8x - cos4x) sin2xsin6x=sin^2(4x) (1/2)(cos 8x - cos4x) = 1- cos^2 (4x) (1/2)(2cos^2 (4x) -1 -cos4x) = 1-cos^2 (4x) 2cos^2 (4x) -1 - cos4x = 2- 2cos^2 (4x) 4cos^2 (4x) - cos4x -3=0 (cos4x -1)(4 cos 4x+3)=0 cos 4x-1 =0 or cos4x=-3/4 Last few steps are left for you :P
Alright... my calculation mistakes again :S Thanks
I got it up to 2cos^2 (4x) -1 - cos4x = 2- 2cos^2 (4x :(
Is it me or did any body see anything wrong with \[\sin2x \sin6x= (1/2)(\cos 8x - \cos4x) \]
It should be cos 4x - cos 8x
well, I thought we will have quadratic equation on Cos 4x
Oh well, that's makes everything buggy.
Well.... That goes back to my problem again :S
\[\sin \left( 6x \right) \sin \left( 2x \right) = -\frac{1}{2}\left( \cos \left( 8x \right) - \cos \left( 4x \right) \right)\]\[-\frac{1}{2}\left( \cos \left( 8x \right) - \cos \left( 4x \right) \right) = \sin^{2}\left( 4x \right)\]\[\cos \left( 8x \right) - \cos \left( 4x \right) = -2\left( 1 - \cos^{2}\left( 4x \right) \right)\]\[(2 \cos^{2} \left( 4x \right) -1 - \cos \left( 4x) \right) = -2 + 2 \cos^{2} \left( 4x \right)\]\[\cos \left( 4x \right) = 1\]
looks like my guess didn't work at all ... :( it was much simpler
\[\cos \left( A + B \right) = \cos A \cos B - \sin A \sin B\]\[\cos (A - B) = \cos A \cos B + \sin A \sin B\]simply substract them and we get\[-2 \sin A \sin B = \cos \left( A + B \right) - \cos \left( A - B \right)\]\[\sin A \sin B = -\frac{1}{2}\left( \cos (A + B) - \cos (A - B) \right)\]
I know what's wrong with me... Thank you so much!
How about just changing the \(\sin \) angle? \[ \frac{1}{2}\left( \cos \left( 4x \right) - \cos \left( 8x \right) \right) = \sin^{2}\left( 4x \right)=\frac 12 (1-\cos 8x) \] \( \implies \cos 4x = 1 \) Faster ain't it?
(Hard to decide which is the best answer :S )
roll a dice ..!!! :D
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