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I'm sorry, if this question seems too simple..

Is it possible to reduce sin 8x/cos 4x it like this:

4 (sin 2x)/ 4(cos x)

sin 2x/cos x 


Or this trig. expression cannot be simplified any further?

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up vote 1 down vote accepted

We have $\sin(8x)=2\sin(4x)\cos(4x)$. This is a particular case of the double-angle identity $\sin 2u=2\sin u\cos u$ (we just take $u=4x$).

So our numerator can be written as $2\sin(4x)\cos(4x)$, and our denominator is $\cos(4x)$. There is cancellation, and we obtain $2\sin(4x)$.

Remark: Your proposed simplification was not correct. It seems to assume, for example, that $\cos 4x=4\cos x$. That is not true in general. Take almost any specific value of $x$, like $x=\pi/2$ ($90$ degrees). We have $\cos 4x=1$ and $4\cos x=0$. There was a similar incorrect assertion about $\sin 8x$.

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I have written out a simplified form $2\sin(4x)$. There is no reason to fool around with the denominator, since it cancels anyway. – André Nicolas May 19 '12 at 18:37
At first time, i didnt notice.. :( Thanks a lot!!) – DrStrangeLove May 19 '12 at 18:39

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