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- Feb 14, 2012

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- Thread starter anemone
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- Feb 14, 2012

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- Mar 10, 2012

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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 $a$.

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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.