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I am needing help learning how to simplify the following equation: $$\sin\left(\arccos\left(\frac{x}{x+1}\right)\right)$$ Also any steps on how to get the answer would be greatly appreciated!

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    $\begingroup$ Hint: Let $\arccos(\frac x{x+1})=\theta$. Then $\cos\theta=\frac x{x+1}$. You want $\sin\theta$. $\endgroup$ – Rahul Jun 10 '16 at 0:50
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Use $\sin^2\theta+\cos^2\theta = 1$

$$\begin{align} &~\sin\arccos\dfrac{x}{x+1}\\[1ex] =&~ \pm\sqrt{~1-\cos^2\arccos\dfrac x{x+1}~} \\[1ex] =&~ \pm\sqrt{~1-\dfrac{x^2}{(x+1)^2}~} \\[1ex] =&~ \pm\dfrac{\sqrt{~2x+1~}}{x+1}\end{align}$$

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as Rahul said... let $\arccos(\frac{x}{x+1})=\theta$ so $\cos\theta=\frac{x}{x+1}$ but you want $\sin\theta$ which is $\sqrt{1-\cos^2\theta}$.

  • Remember $\arccos t$ is between $0$ and $\pi$, so $\sin(\arccos t)$ is positive.

so the answer would be $$\sqrt{1-(\frac{x}{x+1})^2}=\sqrt{1-\frac{x^2}{(x+1)^2}}=\sqrt{\frac{(x+1)^2-x^2}{(x+1)^2}}=\sqrt{\frac{2x+1}{(x+1)^2}}=\frac{\sqrt{2x+1}}{x+1}.$$

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HINT

Put $\cos^{-1} \frac{x}{x+1} =t$

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Draw a picture. Let $ABC$ be a right triangle with right angle $\angle ABC$, and $\theta = \angle CAB$. Assume $AB = x, AC = x+1$. You can see that $$\cos \theta = \frac{AB}{AC} = \frac{x}{x+1}$$ By the Pythagorean theorem, $$BC = \sqrt{(x+1)^2 - x^2} = \sqrt{2x+1}$$ Now what's $\sin \theta$?

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There is a subtle point: you first need to ensure that $x/(x+1)$ belongs to the domain of $\arccos$, that is, $$ -1\le\frac{x}{x+1}\le 1 $$ With standard manipulations you get that your expression is defined for $x\ge-1/2$ (so, in particular, $x+1>0$).

Then, set $\alpha=\arccos(x/(x+1))$, so your expression is $\sin\alpha$, with $0\le\alpha\le\pi$. Hence, $$ \sin\left(\arccos\frac{x}{x+1}\right)= \sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\frac{x^2}{(x+1)^2}} =\sqrt{\frac{2x+1}{(x+1)^2}}=\frac{\sqrt{2x+1}}{x+1} $$ (the last step is justified by the fact that $x+1>0$).

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