Question

%5C%20" id="TexFormula1" title=" 2 \sin( \alpha ) - \cos( \alpha ) = 2 \\ " alt=" 2 \sin( \alpha ) - \cos( \alpha ) = 2 \\ " align="absmiddle" class="latex-formula">
find the value of:
$\sin( \alpha ) + 2 \cos( \alpha )$
​

Answer 1

$2\sin\alpha-\cos\alpha=2$

Consider the substitution $\tan\dfrac\alpha2=\beta$. Then by the double angle identities we get

$\sin\alpha=2\sin\dfrac\alpha2\cos\dfrac\alpha2$

$\cos\alpha=\cos^2\dfrac\alpha2-\sin^2\dfrac\alpha2$

We also have

$\tan\dfrac\alpha2=\beta\implies\begin{cases}\sin\dfrac\alpha2=\dfrac\beta{\sqrt{1+\beta^2}}\\\\\cos\dfrac\alpha2=\dfrac1{\sqrt{1+\beta^2}}\end{cases}$

so that

$\sin\alpha=\dfrac{2\beta^2}{1+\beta^2}$

$\cos\alpha=\dfrac{1-\beta^2}{1+\beta^2}$

and the original equation has been transformed to

$\dfrac{4\beta^2-(1-\beta^2)}{1+\beta^2}=2$

Solve for $\beta$:

$5\beta^2-1=2+2\beta^2$

$3\beta^2=3$

$\beta^2=1$

$\beta=\pm1$

Solving for $\alpha$ gives

$\tan\dfrac\alpha2=-1\implies\dfrac\alpha2=-\dfrac\pi4+n\pi\implies\alpha=-\dfrac\pi2+2n\pi$

$\tan\dfrac\alpha2=1\implies\dfrac\alpha2=\dfrac\pi4+n\pi\implies\alpha=\dfrac\pi2+2n\pi$

where $n$ is any integer. Both $\sin$ and $\cos$ are $2\pi$-periodic, which is to say

$\cos(x+2n\pi)=\cos x$

$\sin(x+2n\pi)=\sin x$

so that

$\sin\alpha=\sin\left(\pm\dfrac\pi2+2n\pi\right)=\sin\left(\pm\dfrac\pi2\right)=\pm1$

$\cos\alpha=\cos\left(\pm\dfrac\pi2+2n\pi\right)=\cos\left(\pm\dfrac\pi2\right)=0$

and we find that

$\sin\alpha+2\cos\alpha=\pm1$