0
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Beautiful observation. (+1). $\endgroup$ – Aditya Agarwal Sep 25 '15 at 9:16
  • 1
    $\begingroup$ Thanks for idea. In my case i got: $1-2\sin2\alpha=1-2\cos\left(2\left(\frac{\pi}{4}-\alpha\right)\right)=1-2\left[1-\sin^2\left(\frac{\pi}{4}-\alpha\right)\;\right]$. $\endgroup$ – Yura Sep 25 '15 at 9:50
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.