# Trigonometrical inequation problem

Solve the inequation: $\sin^4x+\cos^4x \geq 1/2$.

I did this: $(1-\cos^2x)^2+\cos^4x \geq 1/2$

$-2\cos^2x+2\cos^4x \geq -1/2$

$-2(\cos^2x-\cos^4x) \geq -1/2$

$\cos^2x(1-\cos^2x) \leq 1/4$

$\cos^2x\sin^2x \leq 1/4$

$|\cos x\sin x| \leq 1/2$

Now what? :S I feel it should be easy from here on out.

Where you've left correctly off,

$$4\sin^2x\cos^2x\le1$$

$$\iff-1\le2\sin x\cos x\le1\iff-1\sin2x\le1$$ which holds true for all real $x$

Alternatively, as $(a-b)^2\ge0$ for real $a,b$

$2(a^2+b^2)\ge(a+b)^2$

Set $a=\cos^2x,b=\sin^2x$ to get $2(\cos^4x+\sin^4x)\ge1^2$

• Thank you, so at least I did something right, bringing the inequation up to a point... I understand :) Jun 28, 2015 at 16:01

$$s^4+c^4=1-2\,s^2\,c^2=1- s^2_{2x} /2 \geq\frac12,$$ since maximum of $s^2_{2x}$ can be 1.

EDIT 1

so it is always $\ge 1/2$, for all values of x.

Hint: $\sin x \cos x =\dfrac{1}{2} \sin 2x$

• Thanks for the hint, I didn't have this formula written down. I swear trigonometry has so many formulas :( Jun 28, 2015 at 16:02

Using $\cos2A=1-2\sin^2A=2\cos^2A-1,$

$$(2\sin^2x)^2+(2\cos^2x)^2\ge2$$

$$\iff(1-\cos2x)^2+(1-\cos2x)^2\ge2$$

$$\iff1+\cos^22x\ge1\iff\cos^22x\ge0$$ which is true for all real $x$

$$\sin^4x+\cos^4x=(\sin^2+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac{\sin^2(2x)}{2}\geq\frac{1}{2}$$

The identity $$2\sin x\cos x=\sin 2x$$ should help, but maybe you are not supposed to know it. But note that $\sin^2 x\cos^2x\le\frac14$ implies $|\sin x\cos x|\le\frac12$. You can't "remove" the absolute value.

Alternatively, you can solve the inequality $$\cos^4x-\cos^2x+\frac14\ge 0$$ by substitution: $z=\cos^2x$.

• Yup, I fixed it now. I can't believe I forgot about the absolute value. Jun 28, 2015 at 16:03