Question

If $\sin x+\cos x=t,$ then find the value of $\sin 3x-\cos 3x.$

Answer 1

We have to simplify/modify
$\sin 3x-\cos 3x$ to express in terms of $\sin x+\cos x.$

We will express them in terms of $\sin x\text{and}\cos x$

$\sin 3x-\cos 3x=3\sin x-4{\sin }^{3}x-4{\cos }^{3}x+3\cos x=3(\sin x+\cos x)-4({\sin }^{3}x+{\cos }^{3}x)=3(\sin x+\cos x)-4[(\sin x+\cos x{)}^3-3\sin x\cos x(\sin x+\cos x)]$

We know,
$t^2=(\sin x+\cos x{)}^2=1+2\sin x\cos x$
$\Rightarrow \sin x\cos x=\frac{t^2-1}{2}$

$\Rightarrow \sin 3x-\cos 3x=3t-4[t^3-3\frac{(t^2-1)}{2}t]$
$=3t-4[t^3-\frac{3t^3}{2}+\frac{3t}{2}]$
$=3t-4[-\frac{t^3}{2}+\frac{3t}{2}]$
$=3t+2t^3-6t$
$=2t^3-3t$
$=t(2t^2-3)$