Express the following as the product of sines and cosines: \[ \sin 2x + \cos 4x \]
Solution
The expression
\[
\sin 2x + \cos 4x
\]
contains both sine and cosine functions, so first convert
\(\cos 4x\) into sine form using:
\[
\cos \theta = \sin \left(90^\circ – \theta\right)
\]
\[
\cos 4x = \sin (90^\circ – 4x)
\]
Therefore,
\[
\sin 2x + \cos 4x
=
\sin 2x + \sin (90^\circ – 4x)
\]
Now use the identity:
\[
\sin A + \sin B
=
2 \sin \frac{A+B}{2}
\cos \frac{A-B}{2}
\]
Here,
\[
A = 2x,\qquad B = 90^\circ – 4x
\]
Substituting:
\[
=
2 \sin \frac{2x + 90^\circ – 4x}{2}
\cos \frac{2x – (90^\circ – 4x)}{2}
\]
\[
=
2 \sin (45^\circ – x)
\cos (3x – 45^\circ)
\]
Hence,
\[
\boxed{
\sin 2x + \cos 4x
=
2 \sin (45^\circ – x)\cos(3x – 45^\circ)
}
\]