# Thread: [SOLVED] Solve cos54cos@ + sin54sin@ = sin2@

1. ## [SOLVED] Solve cos54cos@ + sin54sin@ = sin2@

Find the general solution of
cos54cos@ + sin54sin@ = sin2@

2. Originally Posted by differentiate
Find the general solution of
cos54cos@ + sin54sin@ = sin2@
see below

3. Originally Posted by differentiate
Find the general solution of
cos54cos@ + sin54sin@ = sin2@
On your LHS employ the identity $\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$ to simplify.

4. Hello differentiate
Originally Posted by differentiate
Find the general solution of
cos54cos@ + sin54sin@ = sin2@
My best guess is that you want the general solution of
$\cos54^o\cos\alpha+\sin54^o\sin\alpha = \sin2\alpha$
in which case, as pickslides has suggested:
$\cos(54^o-\alpha) =\sin2\alpha$
$=\cos(90^o-2\alpha)$
Now the general solution of
$\cos A = \cos B$
is (in degrees):
$A = 360n^o \pm B, n =0, \pm1, \pm2, ...$
So here we get:
$54^o-\alpha = 360n^o\pm (90^o-2\alpha)$
Taking the $+$ sign:
$54^o-\alpha = 360n^o+ (90^o-2\alpha)$
$=360n^o+90-2\alpha$
$\Rightarrow \alpha =360n^o+36^o$
Taking the $-$ sign:
$54^o-\alpha = 360n^o- (90^o-2\alpha)$
$=360n^o-90+2\alpha$
$\Rightarrow -3\alpha =360n^o-144^o$

$\Rightarrow \alpha = 120n^o+48^o$ (noting that this is now a 'new' $n$)
So there's the general solution: $\alpha = 360n^o+36^o$ or $120n^o+48^o, n = 0,\pm1,\pm2, ...$