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
 A: Recall the cube sum identity: $x^3+y^3 = (x+y)(x^2-xy+y^2)$
Continuing from where you left off:
\begin{align}&\quad4\frac {\sin\theta+\cos\theta-\sin^3\theta-\cos^3\theta}{\sin\theta+\cos\theta}\\
&=4-4\frac {\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta}\\
&=4-4(\sin^2\theta - \sin\theta\cos\theta+\cos^2\theta)\\
&=4-4(1-\frac12\sin2\theta)\\
&=2\sin2\theta
\end{align}
A: $$\frac{\sin3x-\cos3x}{\sin x+\cos x}+1=\frac{\sin3x+\sin x+\cos x-\cos3x}{\sin x+\cos x}=\frac{\sin(2x+x)+\sin (2x-x)+\cos (2x-x)-\cos(2x+x)}{\sin x+\cos x}=\frac{\sin2x\cos x+\sin x\cos2x+\sin2x\cos x-\sin x\cos 2x+\cos 2x\cos x+ \sin 2x\sin x-(\cos 2x\cos x- \sin 2x\sin x)}{\sin x+\cos x}=\frac{2 \sin2x\cos x +2 \sin 2x\sin x}{\sin x+\cos x}=\frac{2 \sin2x(\cos x +\sin x)}{\sin x+\cos x}=2\sin2x$$
A: $$\sin3A-\cos3A$$
$$=\sqrt2\sin\left(3A-\dfrac\pi4\right)$$
$$=\sqrt2\sin\left(3A+\dfrac{3\pi}4-\pi\right)$$
$$=-\sqrt2\sin3\left(A+\dfrac\pi4\right)$$
Similarly, $\sin A+\cos A=\sqrt2\sin(?)$
Finally for $\sin B\ne0,$ $$\dfrac{\sin3B}{\sin B}=\dfrac{\sin B(3-4\sin^2B)}{\sin B}=3-2(1-\cos2B)$$
