# Thread: trigo problem no. 14

1. ## trigo problem no. 14

p216 q14 re.ex10

2. Hello, afeasfaerw23231233!@

Consider the equation: .$\displaystyle \sin^8\!x + \cos^8\!x \;=\;\frac{17}{16}\cos^2\!2x\;\hdots\;(*)$

(a) Let $\displaystyle c \:=\:\cos2x$

Show that (*) can be written as: .$\displaystyle \left(\frac{1-c}{2}\right)^4 + \left(\frac{1+c}{2}\right)^4 \:=\:\frac{17}{16}c^2$

We have: .$\displaystyle \left(\sin^2\!x\right)^4 +\left(\cos^2\!x\right)^4 \;=\;\frac{17}{16}\cos^2\!2x$

Then: .$\displaystyle \left(\frac{1-\cos2x}{2}\right)^4 + \left(\frac{1+\cos2x}{2}\right)^4 \;=\;\frac{17}{16}\cos^2\!2x$

Let $\displaystyle c = \cos2x$ and we have: .$\displaystyle \left(\frac{1-c}{2}\right)^4 + \left(\frac{1+c}{2}\right)^2 \;=\;\frac{17}{16}c^2$

(b) Hence find the general solution of (*).

We have: .$\displaystyle \left(\frac{1-c}{2}\right)^4 + \left(\frac{1+c}{2}\right)^2 \;=\;\frac{17}{16}c^2$

Expand: .$\displaystyle \frac{1 - 4c + 6c^2 - 4c^3 + c^4}{16} + \frac{1+4x + 6c^2 - 4c^3 + c^4}{16}\;=\;\frac{17}{16}c^2$

Simplify: .$\displaystyle \frac{2 + 12c^2 + 2c^4}{16}\;=\;\frac{17}{16}c^2\quad\Rightarrow\qua d 2 + 12c^2 + 2c^4 \;=\;17c^2$

And we have the quadratic: .$\displaystyle 2c^2 - 5c^2 + 2 \:=\:0$

. . which factors: .$\displaystyle (c^2-2)(2c^2-1) \:=\:0$

. . and has roots: .$\displaystyle \begin{array}{ccccccc}c^2-2&=&0 & \to & c &=& \pm\sqrt{2} \\ 2c^2-1 &=&0 & \to & c &=&\pm\frac{1}{\sqrt{2}} \end{array}$

$\displaystyle \cos2x \:=\:\pm\sqrt{2}$ has no real roots.

$\displaystyle \cos2x \:=\:\pm\frac{1}{\sqrt{2}}\quad\Rightarrow\quad 2x \:=\:\frac{\pi}{4} + \frac{\pi}{2}n$

. . Therefore: .$\displaystyle x \;=\;\frac{\pi}{8} + \frac{\pi}{4}n$

3. thanks. i made a calculating mistake and that's why my answer was wrong