# Calculate trigonometry expression: $\frac{\cos 3\alpha - \sin 3\alpha}{\cos \alpha + \sin \alpha}$

I think that this expression very easy, but i don't know how resolve it. Please, help me, guys. So, there is: $$\dfrac{\cos 3\alpha - \sin 3\alpha}{\cos \alpha + \sin \alpha}, \;\;\; \mbox{if} \;\;\; \sin \left(\dfrac{\pi}{4} - \alpha\right) = 0,1.$$ I make following transformation: \begin{gather} \dfrac{\cos 3\alpha - \sin 3\alpha}{\cos \alpha + \sin \alpha} = \dfrac{4 \cos^3 \alpha - 3\cos \alpha -3\sin \alpha + 4\sin^3 \alpha}{\cos \alpha + \sin \alpha} =\\ = \dfrac{4\left(\cos^3 \alpha + \sin^3 \alpha\right) - 3\left(\cos \alpha + \sin \alpha\right)}{\cos \alpha + \sin \alpha} =\\ = \dfrac{4\left(\cos \alpha + \sin \alpha\right)\left(\cos^2 \alpha - \cos \alpha \cdot \sin \alpha + \sin^2 \alpha\right) - 3 \left(\cos \alpha + \sin \alpha\right)}{\cos \alpha + \sin \alpha} =\\ \dfrac{\left(\cos \alpha + \sin \alpha\right)\left[\;4\;(1 - \sin \alpha \cos \alpha) - 3\;\right]}{\cos \alpha + \sin \alpha} = 4 - 4 \sin \alpha \cos \alpha - 3 = 1 - 2\sin 2\alpha. \end{gather} Look very nice, but what next? How i can use substitution? Thank's all.

Notice that $$1-2\sin 2\alpha= 1-2\cos\left(2\left(\frac \pi 4 -\alpha\right)\right)=1-2\sqrt{1-\sin^2\left(2\left(\frac \pi 4 -\alpha\right)\right)}$$

• Beautiful observation. (+1). – Aditya Agarwal Sep 25 '15 at 9:16
• Thanks for idea. In my case i got: $1-2\sin2\alpha=1-2\cos\left(2\left(\frac{\pi}{4}-\alpha\right)\right)=1-2\left[1-\sin^2\left(\frac{\pi}{4}-\alpha\right)\;\right]$. – Yura Sep 25 '15 at 9:50

Expand $\sin \left(\frac {\pi}4-\alpha\right)=\sin \frac {\pi}4 \cos\alpha -\cos \frac {\pi}4 \sin \alpha$ and you can compute $\sin \alpha-\cos \alpha=d$ because $\sin \frac {\pi}4=\cos \frac {\pi}4$

Then $d^2=\sin^2 \alpha -2\sin\alpha \cos \alpha +\cos^2\alpha=1-2\sin \alpha\cos \alpha$

So you can substitute $2\sin \alpha \cos \alpha =1-d^2$

Notice,

1. If $$\sin\left(\frac{\pi}{4}-\alpha\right)=0$$$$\sin\alpha\cos\frac{\pi}{4}-\cos\alpha\sin\frac{\pi}{4}=0\implies \tan\alpha=1$$ Hence, $$\color{red}{1-2\sin 2\alpha}=1-2\frac{2\tan \alpha}{1+\tan^2\alpha}=1-2\frac{2(1)}{1+(1)^2}=\color{red}{-1}$$
2. If $$\sin\left(\frac{\pi}{4}-\alpha\right)=1\implies \alpha=-\left(2n\pi+\frac{\pi}{4}\right)$$
Hence, $$\color{red}{1-2\sin 2\alpha}=1-2\sin 2\left(-\left(2n\pi+\frac{\pi}{4}\right)\right)=1+2\sin \left(4n\pi+\frac{\pi}{2}\right)$$$$=1+2\sin\frac{\pi}{2}=1+2(1)=\color{red}{3}$$
• $\sin\left(\frac{\pi}{4}-\alpha\right)=\color{red}{0.1}$. – Yura Sep 25 '15 at 11:28