# What does \cos x \sin x equal?

What does $$\displaystyle{\cos{{x}}}{\sin{{x}}}$$ equal?

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Serita Dewitt
Explanation:
So we have
$$\displaystyle{\cos{{\left({x}\right)}}}{\sin{{\left({x}\right)}}}$$
If we multiply it by two we have
$$\displaystyle{2}{\cos{{\left({x}\right)}}}{\sin{{\left({x}\right)}}}$$
Which we can say it's a sum
$$\displaystyle{\cos{{\left({x}\right)}}}{\sin{{\left({x}\right)}}}+{\cos{{\left({x}\right)}}}{\sin{{\left({x}\right)}}}$$
Which is the double angle formula of the sine
$$\displaystyle{\cos{{\left({x}\right)}}}{\sin{{\left({x}\right)}}}+{\cos{{\left({x}\right)}}}{\sin{{\left({x}\right)}}}={\sin{{\left({2}{x}\right)}}}$$
But since we multiplied by 2 early on to get to that, we need to divide by two to make the equality, so
$$\displaystyle{\cos{{\left({x}\right)}}}{\sin{{\left({x}\right)}}}={\frac{{{\sin{{\left({2}{x}\right)}}}}}{{{2}}}}$$
###### Not exactly what youâ€™re looking for?
Corgnatiui

$$\displaystyle{\sin{{\left({A}+{B}\right)}}}={\sin{{\left({A}\right)}}}{\cos{{\left({B}\right)}}}+{\cos{{\left({A}\right)}}}{\sin{{\left({B}\right)}}}$$
Now let A = B = x. So we get:
$$\displaystyle{\sin{{\left({x}+{x}\right)}}}={\sin{{\left({2}{x}\right)}}}={\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}+{\cos{{\left({x}\right)}}}{\sin{{\left({x}\right)}}}={2}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}$$
Therefore,
$$\sin(x)\cos(x) = (1/2)\sin(2x)$$
Hope this helps!

Vasquez

Look into the following:
$$\begin{array}{}\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y) \\Now\ set\ x=y, \\Now, \\\sin(x+x)=\sin(x)\cos(x)+\cos(x)\sin(x) \\\Rightarrow \sin(2x)=2\sin(x)\cos(x) \\\Rightarrow \sin(x)\cos(x)=\frac{1}{2}\sin(2x) \end{array}$$