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How to find the range of the following trigonometric function $\sin^2x-5\sin x-6$. Can some one help me out. Thank you

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2 Answers 2

up vote 4 down vote accepted

$$\sin^2x-5\sin x-6=\frac{(2\sin x)^2-2\cdot2\sin x\cdot5+5^2-24-25}4=\frac{(2\sin x-5)^2-49}4$$

Now, for real $\displaystyle x,-1\le\sin x\le1\iff -2\le\sin x\le2\iff-7\le2\sin x-5\le-3$

$\displaystyle2\sin x-5\le-3\implies(2\sin x-5)^2\ge9$

$\displaystyle2\sin x-5\ge-7\implies(2\sin x-5)^2\le49$

Can you take it home from here?

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Hint: if $\sin(x) = t$, this is $t^2 - 5 t - 6$, where $-1 \le t \le 1$. What are the maximum and minimum values of $t^2 - 5 t - 6$ on this interval?

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May I request for a little more elaboration of your method? –  lab bhattacharjee Apr 6 at 5:19

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