Science:Math Exam Resources/Courses/MATH102/December 2013/Question A 02

MATH102 December 2013

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Question A 02
Which of the following gives a value of $\cos \left(\arcsin \left({\frac {\sqrt {9-x^{2}}}{3}}\right)\right)$ ? (a) ${\frac {3}{x}}$ (b) ${\frac {x}{3}}$ (c) ${\frac {\sqrt {9-x^{2}}}{x}}$ (d) ${\frac {x}{\sqrt {9-x^{2}}}}$ (e) ${\frac {3}{\sqrt {9-x^{2}}}}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Let $\theta =\arcsin \left({\frac {\sqrt {9-x^{2}}}{3}}\right)$ and try drawing a triangle with this information.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let $\theta =\arcsin \left({\frac {\sqrt {9-x^{2}}}{3}}\right)$. Then $\sin(\theta )={\frac {\sqrt {9-x^{2}}}{3}}$. This gives the following triangle: To solve for the last side, we simply use the Pythagorean theorem: $a^{2}=3^{2}-\left({\sqrt {9-x^{2}}}\right)^{2}=9-9+x^{2}=x^{2}$ and so $a=x$ (we only need to consider the positive root since $a$ has to be positive). Hence, taking the cosine of theta in our triangle above, we see that $\cos(\theta )=\cos \left(\arcsin \left({\frac {\sqrt {9-x^{2}}}{3}}\right)\right)={\frac {a}{3}}={\frac {x}{3}}$, which is answer ${\color {blue}{\text{(B)}}}$.