# Trig Functions

• December 6th 2009, 03:20 PM
KingV15
Trig Functions
If sec 4x-sec2x = 2, find the value of cos^2x, 0<x<pi/2
• December 6th 2009, 07:50 PM
Soroban
Hello, KingV15!

We need two identities: . $\begin{array}{ccc}
\cos2\theta &=& 2\cos^2\!\theta - 1 \\ \cos4\theta &=& 8\cos^4\!\theta - 8\cos^2\!\theta + 1 \end{array}$

Quote:

If $\sec4x-\sec2x \:=\: 2$, find the value of $\cos^2\!x,\:\text{ for } 0 \leq x \leq \tfrac{\pi}{2}$

We have: . $\sec4x - \sec2x \:=\:2 \quad \Rightarrow\quad \frac{1}{\cos4x} - \frac{1}{\cos2x} \:=\:2 \quad \Rightarrow \quad \cos2x - \cos4x \:=\:2\cos2x\cos4x$

. . $(2\cos^2\!x-1) - (8\cos^4\!x - 8\cos^2\!x + 1) \;=\;2(2\cos^2\!x-1)(8\cos^4\!x - 8\cos^2\!x + 1)$

. . $32\cos^6\!x - 40\cos^4\!x + 10\cos^2\!x \:=\:0$

. . $2\cos^2\!x\left(16\cos^4\!x - 20\cos^2\!x + 5\right) \;=\;0$

Then:

. . $2\cos^2\!x \:=\:0 \quad\Rightarrow\quad \boxed{\cos^2\!x \:=\:0}\quad [1]$

. . $16\cos^4\!x - 20\cos^2\!x + 5 \:=\:0 \quad\Rightarrow\quad\boxed{ \cos^2\!x \:=\:\frac{5\pm\sqrt{5}}{8}} \quad [2]$

From [1]: . $x \:=\:\frac{\pi}{2}$

From [2]: . $x \:=\:\frac{\pi}{10},\:\frac{3\pi}{10}$