Can someone please show me step-by-step how to prove this trigonometric identity?
$$\frac{\cos(3A)-\sin(3A)}{1-2\sin(2A)}=\cos(A)+\sin(A)$$
This is what I've done so far:
Using the LHS,
$$\frac{cos(2A+A)-sin(2A+A)}{1-4sin(A)cos(A)}$$
$$=\frac{cos(2A)cos(A)-sin(2A)sin(A)-sin(2A)cos(A)-cos(2A)sin(A)}{1-4sin(A)cos(A)}$$
$$=\frac{cos(A)(cos(2A)-sin(2A))-sin(A)(sin(2A)+cos(2A))}{1-4sin(A)cos(A)}$$
$$=\frac{cos^3(A)-2cos^2(A)sin(A)-sin^2(A)cos(A)-cos^2(A)sin(A)-2sin^2(A)cos(A+sin^3(A))}{1-4sin(A)cos(A)}$$
$$=\frac{cos^3(A)+sin^3(A)-3cos^2(A)sin(A)-3sin^2(A)cos(A)}{1-4sin(A)cos(A)}$$
After this I have no idea how to continue to prove the identity.
If anyone could help I would be extremely grateful.
Best Answer
Let $c=\cos(A)$, $s=\sin(A)$, $C=\cos(2A)$, and $S=\sin(2A)$. Then
$$\cos(3A)=Cc-Ss\quad\text{and}\quad\sin(3A)=Sc+sC$$
so the numerator on the LHS is
$$(Cc-Ss)-(Sc+sC)=C(c-s)-S(s+c)$$
and, on clearing out the denominator, the RHS is
$$(1-2S)(c+s)$$
Moving the $2S$ term from right to left, the identity to verify becomes
$$C(c-s)+S(c+s)=c+s$$
Now invoke the identities $C=c^2-s^2$ and $S=2cs$ to see that the
$$\begin{align} C(c-s)+S(c+s) &=(c^2-s^2)(c-s)+2cs(c+s)\\ &=c^3+c^2s+cs^2+s^3\\ &=(c+s)(c^2+s^2)\\ &=(c+s) \end{align}$$
since $c^2+s^2=1$. This verifies the identity.
Remark: The use of $c$, $s$, $C$, and $S$ is mainly to reduce the amount of stuff that would otherwise have to be written out over and over again.