Welcome to our community

Be a part of something great, join today!

Find the Range of Values for Negative a

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,690
If the inequality $\sin^2 x+a\cos x+a^2 \ge 1+\cos x$ holds for any $x\in R$, determine the range of values for negative $a$.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
If the inequality $\sin^2 x+a\cos x+a^2 \ge 1+\cos x$ holds for any $x\in R$, determine the range of values for negative $a$.
HINT:

Using $\sin^2x=1-\cos^2x$ the question essentially says that $x^2-(a-1)x-a^2\geq 0$ for all $x\in[-1,1]$. The quadratic $f(x)=x^2-(a-1)x-a^2$ achieves its minimum at $x_0=(a-1)/2$. We need to consider $3$ cases.

Case 1: $x_0<-1$.
Here its necessary and sufficient that $f(-1)\geq 0$, giving the range of $a$.

Case 2: $x_0>1$.
Here its necessary and sufficient that $f(1)\geq 0$, giving the range of $a$.

Case 3: $x_0\in [-1,1]$
Here its necessary and sufficient that $f(x_0)\geq 0$, giving the range of $a$.
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,690
We're told that the inequality $\sin^2 x+a\cos x+a^2 \ge 1+\cos x$ holds for any $x \in R$, so if $x=0$, then we see that $a+a^2 \ge 2$ which implies $a \le -2$.

And since $-1 \le \cos x \le 1$, when we multiplied it with $a$ and bear in mind that $a<0$, we have

$-a \ge a\cos x \ge a$

$a^2-a \ge a^2+\cos x \ge a^2+a$

$\therefore a^2+\cos x \ge a^2+a \ge 2$

And we also know that $\cos^2 x+\cos x \le 2$, thus $a^2+\cos x \ge \cos^2 x+\cos x \ge 1-\sin^2 x+cos x$ and this just gives us back the original inequality expression and this confirms that $a \le -2$ is the range of values of negative $a$ that we're after.