Chapter 7, Problem 18RE

To determine

To prove: The identity $\frac{\sin 3x+\cos 3x}{\cos x-\sin x}=1+2\sin 2x$ .

Explanation of Solution

Given information:

The given identity is $\frac{\sin 3x+\cos 3x}{\cos x-\sin x}=1+2\sin 2x$ .

Formula used:

Identity:

  $\begin{array}{c}\sin \left(x+y\right)=\sin x\cos y+\cos x\sin y\\ \cos \left(x+y\right)=\cos x\cos y-\sin x\sin y\end{array}$

Proof:

To prove the identity, take left side of the identity and solve it.

  $\begin{array}{c}\frac{\sin 3x+\cos 3x}{\cos x-\sin x}=\frac{\sin \left(x+2x\right)+\cos \left(x+2x\right)}{\cos x-\sin x}\\ =\frac{\left(\sin x\cos 2x+\cos x\sin 2x\right)+\left(\cos x\cos 2x-\sin x\sin 2x\right)}{\cos x-\sin x}\\ =\left[\frac{\begin{array}{l}\left(\sin x\left(2{\cos }^{2}x-1\right)+2\sin x\cos x\cdot \cos x\right)+\\ \left(\cos x\left(1-2{\sin }^{2}x\right)-2\sin x\cos x\cdot \sin x\right)\end{array}}{\cos x-\sin x}\right]\\ =\left[\frac{2{\cos }^{2}x\sin x-\sin x+2{\cos }^{2}x\sin x+\cos x-2{\sin }^{2}x\cos x-2{\sin }^{2}x\cos x}{\cos x-\sin x}\right]\end{array}$

Further simplify the equation.

  $\begin{array}{c}\frac{\sin 3x+\cos 3x}{\cos x-\sin x}=\left[\frac{2{\cos }^{2}x\sin x-\sin x+2{\cos }^{2}x\sin x+\cos x-2{\sin }^{2}x\cos x-2{\sin }^{2}x\cos x}{\cos x-\sin x}\right]\\ =\left[\frac{\left(\cos x-\sin x\right)+\left(4{\cos }^{2}x\sin x-4{\sin }^{2}x\cos x\right)}{\cos x-\sin x}\right]\\ =\frac{\cos x-\sin x}{\cos x-\sin x}+\frac{4\cos x\sin x\left(\cos x-\sin x\right)}{\cos x-\sin x}\\ =1+4\cos x\sin x\\ =1+2\sin 2x\end{array}$

From the above, it is clear that the left hand side of the identity is equal to the right side of the identity.

Hence, proved.