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How can I evaluate the sine of an inverse cosine?

for example: sin(arccos((x)^1/2))

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This is the full solution. I apologize that I do not know how to make/insert I diagram. Draw a right triangle with one leg of length $\sqrt{x}$ and hypotenuse $1$. By the Pythagorean theorem, we can conclude that the other leg has length $1-x^2$. Let $\theta$ be the angle so that the $\sqrt{x}$ leg is adjacent to it. Thus, $\cos(\theta) =\sqrt{x}$. Thus, $\theta = \arccos(\sqrt{x})$. Now consider $\sin(\theta) = \sqrt{1^2-x^2}/1=\sqrt{1-x^2}$ And that $\sin(\theta)=\sin(\arccos(\sqrt{x}))$ Thus, we have the final answer we wanted, $\sqrt{1-x^2}$.

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    $\begingroup$ That makes sense, I was reading too much into it thinking there was some identity relating the two. Drawing the triangle does help you see the solution much clearer! $\endgroup$ – user8028 Jul 13 '14 at 2:00
  • $\begingroup$ I also usually look for the identity because I want to feel superior to the problem and not have to messy my hands with diagrams, but all of the identities come from such figures (in that they can be derived) and using them is very powerful. $\endgroup$ – Thoth19 Jul 13 '14 at 2:03
  • $\begingroup$ There might be something using the pythaogrean identity, but it isn't working out to be as nice as the triangle method. $\endgroup$ – Thoth19 Jul 13 '14 at 2:05
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    $\begingroup$ @Thoth19 I've edited your LaTeX. Putting a \ in front of trig functions (and many common functions) will cause them to display more beautifully. $\endgroup$ – Austin Mohr Jul 13 '14 at 2:08
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    $\begingroup$ What I meant was the type of arrogance one develops from the SAT, i.e. I should be smart enough to figure it out without drawing because it will save me time. Indeed for deeper meaning, being able to derive it is much superior. Also based on the identities, I have also that $sin(1-x)$ is another solution. $\endgroup$ – Thoth19 Jul 13 '14 at 2:09

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