1. ## 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

2. Hello, John!

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

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

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

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

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

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

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

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

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

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