# Simplifying $\sin(4x)\cos(4x)$

Simplify $\sin(4x)\cos(4x)$ using double angle or compound trigonometry.

Can someone please show me how its done, Ive tried several times but no where near the answer.

• What do you mean by "solve"? What is there to solve? – Pedro Tamaroff Jul 10 '13 at 4:49
• As stated there is nothing to solve here. Do you mean $\sin(4x) \cos(4x) = 0$? – Joel Jul 10 '13 at 4:50
• Simplify, sorry. – Red Queen10101 Jul 10 '13 at 4:51
• @RedQueen10101 is it what i have answered or you mean different one? – dato datuashvili Jul 10 '13 at 4:56
• represent $8*x=2*(4*x)$ – dato datuashvili Jul 10 '13 at 4:59

The double angle formula is $\ 2 \sin\theta\cos\theta = \sin(2\theta) \iff \sin\theta\cos\theta = \frac{1}{2} \sin(2\theta)$.

By applying this formula with $\theta = 2x$, we obtain

$$\sin4x\cos4x=\frac{1}{2} \sin(8x).$$

$\sin4x\cos4x$ using identiity that $\sin(2\theta)=2\sin\theta\cos\theta$,we get

$\sin4x\cos4x=\frac{\sin(8x)}{2}$

• I'm reasonably sure the question is asking to simplify in terms of $\sin(x)$ and $\cos(x)$. – Zev Chonoles Jul 10 '13 at 4:54
• Also, named math operators should appear upright, and the common ones have their own code for this purpose (e.g. \sin, \log - see entry 11 in our MathJax guide). – Zev Chonoles Jul 10 '13 at 4:54
• ok i was thinking that he wanted like this,then it is his rights to ask it more clearly,what does mean simplify in terms of sin and cosine?by which way?i did not understand – dato datuashvili Jul 10 '13 at 4:55
• In textbooks, sometimes “simplify” means “complicate”. I like this answer fine. – Lubin Jul 10 '13 at 4:59
• I feel like Dato has the right of it. This is set up as a clear application of the double angle formula as you would find it in the exercises of a textbook. – Joel Jul 10 '13 at 5:01