Prove that: \( \sin 5x = 5\sin x – 20\sin^3 x + 16\sin^5 x \)
Solution
We know that
\[
\sin 5x = \sin(2x+3x)
\]
Using the formula
\[
\sin(A+B)=\sin A \cos B + \cos A \sin B
\]
we get
\[
\sin 5x = \sin 2x \cos 3x + \cos 2x \sin 3x
\]
Now use
\[
\sin 2x = 2\sin x \cos x
\]
\[
\cos 2x = 1-2\sin^2 x
\]
\[
\sin 3x = 3\sin x – 4\sin^3 x
\]
\[
\cos 3x = 4\cos^3 x – 3\cos x
\]
Substituting these values,
\[
\sin 5x
= (2\sin x \cos x)(4\cos^3 x-3\cos x)
+(1-2\sin^2 x)(3\sin x-4\sin^3 x)
\]
Simplifying,
\[
= 8\sin x \cos^4 x – 6\sin x \cos^2 x
+3\sin x -4\sin^3 x
-6\sin^3 x +8\sin^5 x
\]
Using
\[
\cos^2 x = 1-\sin^2 x
\]
\[
\cos^4 x = (1-\sin^2 x)^2
=1-2\sin^2 x+\sin^4 x
\]
Substitute these values:
\[
\sin 5x
=8\sin x(1-2\sin^2 x+\sin^4 x)
-6\sin x(1-\sin^2 x)
+3\sin x
-10\sin^3 x
+8\sin^5 x
\]
Expanding,
\[
=8\sin x-16\sin^3 x+8\sin^5 x
-6\sin x+6\sin^3 x
+3\sin x
-10\sin^3 x
+8\sin^5 x
\]
\[
=(8-6+3)\sin x
+(-16+6-10)\sin^3 x
+(8+8)\sin^5 x
\]
\[
=5\sin x-20\sin^3 x+16\sin^5 x
\]
Hence proved.