# Trigonometry QUESTION?

• November 24th 2010, 01:02 AM
JohnRamey
Trigonometry QUESTION?
PROVE that sin^4 x + cos^4 x= 1/4(3+cos4x). hence find the largest possible value and smallest possible value of sin^4 x+cos^4 x

Brainquicken
• November 24th 2010, 09:22 AM
Soroban
Hello, John!

I'm delighted that you solved it.
Here's my approach . . .

Quote:

$\text{Prove that: }\:\sin^4\!x + \cos^4\!x \:=\: \frac{1}{4}(3+\cos4x)$

We have: . $\left(\sin^2\!x\right)^2 + \left(\cos^2\!x\right)^2 \;=\;\left(\dfrac{1-\cos2x}{2}\right)^2 + \left(\dfrac{1+\cos2x}{2}\right)^2$

. . . . . . $\displaystyle =\;\frac{1 - 2\cos2x + \cos^2\!2x}{4} + \frac{1 + 2\cos2x + \cos^2\!2x}{4}$

. . . . . . $=\;\displaystyle \frac{2 + 2\cos^2\!2x}{4} \;=\;\frac{1+\cos^2\!2x}{2} \;=\;\frac{1}{2}(1 + \cos^2\!2x)$

. . . . . . $\displaystyle=\;\frac{1}{2}\left(1 + \frac{1+\cos4x}{2}\right) \;=\;\frac{1}{2}\left(\frac{3 + \cos4x}{2}\right) \;=\;\dfrac{1}{4}(3 + \cos 4x)$

Quote:

$\text{Hence, find the largest and smallest possible value of }\sin^4\!x+\cos^4\!x$

The maximum value of $\cos4x$ is 1.
The minimum value of $\cos4x$ is -1.

Therefore: . $\frac{1}{4}(3 - 1)\;\leq \;\sin^4\!x + \cos^4\!x \;\leq \;\frac{1}{4}(3+1)$

. . . . . . . . . . . . . $\frac{1}{2}\;\le \;\sin^4\!x+\cos^4\!x \;\le \; 1$