Question

How do you apply the double angle formula for $\sin 8x\cos 8x$?

Answer 1

This problem deals with solving the given equation with trigonometric identities and compound sum angles of trigonometric functions. A compound angle formula or addition formula is a trigonometric identity which expresses a trigonometric function of $\left( {A + B} \right)$ or $\left( {A - B} \right)$in terms of trigonometric functions of $A$ and $B$.

Formula Used:
$\Rightarrow \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
$\Rightarrow \cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
Here when $A = B,$ then $\sin \left( {A + B} \right) = \sin 2A$, then it is called as the double angle, its formula is given by:
$\Rightarrow \sin \left( {A + A} \right) = \sin A\cos A + \cos A\sin A$
$\therefore \sin \left( {2A} \right) = 2\sin A\cos A$
Where the double angle for cosine is given by $\cos \left( {A + B} \right) = \cos 2A$, which is expressed below:
$\Rightarrow \cos \left( {A + A} \right) = \cos A\cos A - \sin A\sin A$
$\therefore \cos \left( {2A} \right) = {\cos ^2}A - {\sin ^2}A$

Complete step-by-step answer:
The given expression is $\sin 8x\cos 8x$, consider this as given below:
$\Rightarrow \sin 8x\cos 8x$
Consider $\sin \left( {8x} \right)$, we know that from the double angle formula, it can be written as given below:
$\Rightarrow \sin \left( {8x} \right) = \sin \left( {4x + 4x} \right)$
Now applying the double angle formula to the given expression, as shown:
The expression for $\sin \left( {8x} \right)$ can be written as $\sin \left( {2\left( {4x} \right)} \right)$.
$\Rightarrow \sin \left( {2\left( {4x} \right)} \right) = 2\sin 4x\cos 4x$
Now consider $\cos \left( {8x} \right)$, we know that from the double angle formula, it can be written as given below:
$\Rightarrow \cos \left( {8x} \right) = \cos \left( {4x + 4x} \right)$
Now applying the double angle formula to the given expression, as shown:
The expression for $\cos \left( {8x} \right)$ can be written as $\cos \left( {2\left( {4x} \right)} \right)$.
$\Rightarrow \cos \left( {2\left( {4x} \right)} \right) = {\cos ^2}4x - {\sin ^2}4x$
Now substituting the above expressions, in the given expression $\sin 8x\cos 8x$ as shown below:
$\Rightarrow \sin 8x\cos 8x = \left( {2\sin 4x\cos 4x} \right)\left( {{{\cos }^2}4x - {{\sin }^2}4x} \right)$
Now simplifying the above expression, as given below:
$\Rightarrow \sin 8x\cos 8x = 2\left[ {\left( {\sin 4x\cos 4x} \right)\left( {{{\cos }^2}\left( {4x} \right)} \right) - \left( {\sin 4x\cos 4x} \right)\left( {{{\sin }^2}\left( {4x} \right)} \right)} \right]$
$\therefore \sin 8x\cos 8x = 2\left[ {\sin 4x{{\cos }^3}4x - {{\sin }^3}4x\cos 4x} \right]$

Final Answer: The expression $\sin 8x\cos 8x$ is equal to $2\left[ {\sin 4x{{\cos }^3}4x - {{\sin }^3}4x\cos 4x} \right]$.

Note:
Please note that the formula of cosine compound angles formula is used to solve this problem. But there are a few other trigonometric compound angle formulas of sine, cosine and tangent, here the double angle formulas for sine, cosine and tangent are also given below:
$\Rightarrow \sin \left( {2A} \right) = 2\sin A\cos A$
$\Rightarrow \cos \left( {2A} \right) = {\cos ^2}A - {\sin ^2}A$
$\Rightarrow \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
$\therefore \tan \left( {2A} \right) = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$