Prove that: \[ \sin 5x = 5\cos^4 x \sin x -10\cos^2 x \sin^3 x +\sin^5 x \]
Solution
We know that
\[
\sin 5x
=
\sin(4x+x)
\]
Using the formula
\[
\sin(A+B)
=
\sin A\cos B+\cos A\sin B
\]
we get
\[
\sin 5x
=
\sin 4x\cos x+\cos 4x\sin x
\]
Now use
\[
\sin 4x
=
4\sin x\cos x(\cos^2 x-\sin^2 x)
\]
\[
\cos 4x
=
\cos^4 x-6\sin^2 x\cos^2 x+\sin^4 x
\]
Substituting these values,
\[
\sin 5x
=
4\sin x\cos x(\cos^2 x-\sin^2 x)\cos x
\]
\[
\qquad
+
(\cos^4 x-6\sin^2 x\cos^2 x+\sin^4 x)\sin x
\]
Simplifying,
\[
=
4\sin x\cos^2 x(\cos^2 x-\sin^2 x)
+\sin x\cos^4 x
-6\sin^3 x\cos^2 x
+\sin^5 x
\]
\[
=
4\sin x\cos^4 x
-4\sin^3 x\cos^2 x
+\sin x\cos^4 x
-6\sin^3 x\cos^2 x
+\sin^5 x
\]
Combine like terms:
\[
=
(4+1)\sin x\cos^4 x
+(-4-6)\sin^3 x\cos^2 x
+\sin^5 x
\]
\[
=
5\cos^4 x\sin x
-10\cos^2 x\sin^3 x
+\sin^5 x
\]
Hence proved.