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This is for my homework and I don't know how to approach it to get an exact value.

Given that $- \frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2}$, find the exact of $\cos[\arcsin(-\frac{1}{3})]$.

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use the fact that cosx = sqrt(1-(sinx)^2) and substitute x=arcsin(-1/3). –  Shu Xiao Li Feb 16 '13 at 21:29
    
So the answer is sqrt(1-[arcsin(-1/3)]^2) ? –  Wk_of_Angmar Feb 16 '13 at 21:46

3 Answers 3

up vote 2 down vote accepted

Yes, use $\cos^2\theta+\sin^2\theta$ and also $\sin(\arcsin(y))=y$ to find that $$ \cos^2(\arcsin(-1/3))=1-\left(\frac{-1}{3}\right)^2=\frac{8}{9}. $$

Now observe that $-\pi/2\leq \arcsin(-1/3)\leq \pi/2$, so $$ \cos(\arcsin(-1/3))\geq 0. $$

So it is the positive root of the above quadratic equation: $$ \cos(\arcsin(-1/3))=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}. $$

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@Wk_of_Angwar My initial answer was correct, editing again. It is the positive solution. –  1015 Feb 16 '13 at 22:35

Hint: Use the trigonometrical Pythagorean theorem: $(\cos\phi)^2+(\sin\phi)^2=1$ for $\phi:=\arcsin(-1/3)$.

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Draw a unit circle, mark the angle $v$ in the right halfplane such that $\sin v = -\dfrac13$. Use the Pythagorean theorem to compute $\cos v$.

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Thanks for this answer, I didn't really think about working this out with some sort of visual format. –  Wk_of_Angmar Feb 16 '13 at 22:03

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