# How to solve cos(pi/4-x)=sin(pi/6+x)?

Jun 11, 2018

$\implies x = k \pi + \frac{7 \pi}{24} , k \in \mathbb{Z}$

#### Explanation:

We know that,
color(red)((1)sinC-sinD=2cos((C+D)/2)sin((C-D)/2)
color(blue)((2)cos(pi/2-theta)=sintheta

Here,

$\cos \left(\frac{\pi}{4} - x\right) = \sin \left(\frac{\pi}{6} + x\right)$

$\implies \cos \left(\frac{\pi}{4} - x\right) - \sin \left(\frac{\pi}{6} + x\right) = 0$

$\implies \cos \left[\frac{\pi}{2} - \frac{\pi}{4} - x\right] - \sin \left(\frac{\pi}{6} + x\right)$=$0 \to \left[\because \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}\right]$

=>color(blue)(cos[pi/2-(pi/4+x)])-sin(pi/6+x)=0tocolor(blue)(Apply(2)

=>color(red)(sin(pi/4+x)-sin(pi/6+x)=0

Using color(red)((1)  we get

color(red)(2cos((pi/4+x+pi/6+x)/2)sin((pi/4+x-pi/6-x)/2)=0

$\implies 2 \cos \left(\frac{\frac{5 \pi}{12} + 2 x}{2}\right) \sin \left(\frac{\frac{\pi}{12}}{2}\right) = 0$

$\implies \cos \left(\frac{5 \pi}{24} + x\right) = 0 \ldots \to \left[\because 2 \sin \left(\frac{\pi}{24}\right) \ne 0\right]$

$\implies \cos \left(x + \frac{5 \pi}{24}\right) = 0$

=>color(brown)(x+(5pi)/24=(2k+1)pi/2,k inZZ

$\implies x = k \pi + \frac{\pi}{2} - \frac{5 \pi}{24} , k \in \mathbb{Z}$

$\implies x = k \pi + \frac{7 \pi}{24} , k \in \mathbb{Z}$

Note:

color(brown)(costheta=0=>theta=(2k+1)pi/2,k in ZZ

Jun 11, 2018

$x = \frac{7 \pi}{24} + k \pi$

#### Explanation:

$\cos \left(\frac{\pi}{4} - x\right) = \sin \left(\frac{\pi}{6} + x\right)$
Reminder:
$\sin \left(\frac{\pi}{6} - x\right) = \cos \left(\frac{\pi}{2} - \left(\frac{\pi}{6} + x\right)\right) = \cos \left(\frac{\pi}{3} - x\right)$
Therefor,
$\cos \left(\frac{\pi}{4} - x\right) = \cos \left(\frac{\pi}{3} - x\right)$
Trig table, unit circle and property of cos function -->
$\frac{\pi}{4} - x = \pm \left(\frac{\pi}{3} - x\right)$

a. $\frac{\pi}{4} - x = \frac{\pi}{3} - x$ (rejected),
b. $\frac{\pi}{4} - x = - \frac{\pi}{3} + x$
$2 x = \frac{\pi}{4} + \frac{\pi}{3} = \frac{7 \pi}{12} + 2 k \pi$
$x = \frac{7 \pi}{24} + k \pi$
Check
$x = \frac{7 \pi}{24} = {52}^{\circ} 5$ --> cos (45 - 52.5) = cos (7.5) = 0.99 -->
sin (30 + 52.5) = sin (82.5) = 0.99. Proved.

Jun 11, 2018

$\implies x = k \pi + \frac{7 \pi}{24} , k \in \mathbb{Z}$

#### Explanation:

$I {I}^{n d} m e t h o d :$

We know that,

color(brown)((1)sintheta=cos(pi/2-theta)

color(red)((2)cosC-cosD=-2sin((c+D)/2)sin((C-D)/2)

color(blue)((3)sintheta=0=>theta=kpi ,k inZZ

Here,

cos(pi/4-x)=color(brown)(sin(pi/6+x)...tocolor(brown)(Apply(1)

=>cos(pi/4-x)=color(brown)(cos[pi/2-(pi/6+x)]

$\implies \cos \left(\frac{\pi}{4} - x\right) = \cos \left(\frac{\pi}{2} - \frac{\pi}{6} - x\right)$

$\implies \cos \left(\frac{\pi}{4} - x\right) = \cos \left(\frac{\pi}{3} - x\right)$

=>color(red)(cos(pi/4-x)-cos(pi/3-x)=0

Using color(red)((1),we get

=>color(red)(-2sin((pi/4-x+pi/3-x)/2)sin((pi/4-x-pi/3+x)/2)=0

$\implies - 2 \sin \left(\frac{\frac{7 \pi}{12} - 2 x}{2}\right) \sin \left(\frac{- \frac{\pi}{12}}{2}\right) = 0$

$\implies \sin \left(\frac{7 \pi}{24} - x\right) = 0 \to \left[\because - 2 \sin \left(- \frac{\pi}{24}\right) \ne 0\right]$

=>color(blue)(sin(x-(7pi)/24)=0...tocolor(blue)(Apply(3)

=>color(blue)(x-(7pi)/24=kpi, k inZZ

$\implies x = k \pi + \frac{7 \pi}{24} , k \in \mathbb{Z}$