# Trig inequality?

• Nov 20th 2010, 04:22 AM
atreyyu
Trig inequality?
How would I go about solving $\displaystyle |\cos x|^{1+\sin x+\cos x} \geq 1$? All help is welcome.
• Nov 20th 2010, 04:47 AM
HallsofIvy
By THINKING, not using "formulas"!

You know that $\displaystyle |cos(x)|$ is always less than or equal to 1. You also know that powers of a number between 0 and 1 are always smaller than the number itself: if $\displaystyle 0< y< 1$ then $\displaystyle y^n< y$.

Saying that an absolute value of a power of cos(x) is "greater than or equal to 1" is exactly the same as saying it is equal to 1. And so the power itself is equal to either 1 or -1. That can only happen when the base is 1 or -1. For what x is cos(x)= 1 or cos(x)= -1?
• Nov 20th 2010, 04:56 AM
Also sprach Zarathustra
Quote:

Originally Posted by atreyyu
How would I go about solving $\displaystyle |\cos x|^{1+\sin x+\cos x} \geq 1$? All help is welcome.

You also can write it in the following form:

$\displaystyle cos^2(x)^{\frac{1}{2}( sin(x)+cos(x)+1)} \geq1$
• Nov 20th 2010, 04:59 AM
Plato
Actually there are more values of x than Prof Halls gave.
If $\displaystyle 1+\sin(x)+\cos(x)<0$ then the given inequality holds.