How to simplify $\frac {\sin 3A - \cos 3A}{\sin A + \cos A} + 1$? So I started by using $\sin 3A$ and $\cos 3A$ identities and then I added the lone $1$ to the trigonometric term. (Done in the picture below)
But after this I don't have any clue on how to proceed.
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-\left(4 \cos ^{3} \theta-\sin \theta\right)}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4\sin^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta+\sin \theta+\cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+4 \cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta+\cos \theta-\sin ^{3} \theta-\cos ^{3} \theta}
{\sin \theta+\cos \theta}
$$
Original image

$$\sin3A-\cos3A$$
$$=\sqrt2\sin\left(3A-\dfrac\pi4\right)$$
$$=\sqrt2\sin\left(3A+\dfrac{3\pi}4-\pi\right)$$
$$=-\sqrt2\sin3\left(A+\dfrac\pi4\right)$$
Similarly, $\sin A+\cos A=\sqrt2\sin(?)$
Finally for $\sin B\ne0,$ $$\dfrac{\sin3B}{\sin B}=\dfrac{\sin B(3-4\sin^2B)}{\sin B}=3-2(1-\cos2B)$$
