Question

Given that
$\sqrt{2}cos(x)$
$cos(x+\frac{\pi }{4} )$
$cos(x)$
are 3 consecutive terms of an arithmetic sequence, where x is an acute angle, find the value of x.

Answer 1

Answer:

Step-by-step explanation:

$cos(x+\frac{\pi }{4})$ - $\sqrt{2}cos(x)$ = $cos(x)$ - $cos(x+\frac{\pi }{4})$

2$cos(x+\frac{\pi }{4})$ =  $\sqrt{2}cos(x)$ + $cos(x)$

($cos(A+B) = cosAcosB - sinAsinB$)

$cos(x + \frac{\pi}{4}) = cos(\frac{\pi}{4})cosx - sinxsin(\frac{\pi}{4})\\\\cos(x + \frac{\pi}{4}) = \frac{cosx}{\sqrt{2}} - \frac{sinx}{\sqrt{2}}\\\\2cos(x + \frac{\pi}{4}) = \sqrt{2}cosx- \sqrt{2}sinx\\\\\sqrt{2}cosx- \sqrt{2}sinx = \sqrt{2}cosx + cosx\\\\- \sqrt{2}sinx = cosx\\\\tanx = \frac{-1}{\sqrt{2}}\\\\tanx \geq 0 (0\leq x\leq 90)$

This is there fore not possible!

Please check question again!

Thanks!