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I am trying to evaluate and simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$.

I am getting $\frac{11}{10}$ but the answer is $\frac{3-4\sqrt{3}}{10}$

My Process:

$\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$

$\cos[\cos^{-1}(\frac{3}{5})] + \cos(\frac{\pi}{3})$

$(\frac{2}{2}) \cdot \frac{3}{5} + \frac{1}{2} \cdot (\frac{5}{5})$

$\frac{6}{10} + \frac{5}{10}$

$\frac{11}{10}$

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    $\begingroup$ Your $\frac{11}{10}$ can't possibly be the cosine of anything! $\endgroup$
    – TonyK
    Commented Nov 22, 2018 at 18:05
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    $\begingroup$ @TonyK Not for real ;^) $\endgroup$ Commented Nov 22, 2018 at 19:28

5 Answers 5

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You cannot separate out the $\cos$ function as you have done in step two.

You can remember this identity.

$$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$

Here $\arccos(\frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)

Using thus result you should get the desired answer.

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    $\begingroup$ (+1) ... for discussing the error in the OP $\endgroup$
    – Mark Viola
    Commented Nov 22, 2018 at 18:05
  • $\begingroup$ @MarkViola thanks for your upvote. $\endgroup$
    – Akash Roy
    Commented Nov 22, 2018 at 18:06
  • $\begingroup$ Thanks once again for your valuable edits @Mark, will bear these in mind. $\endgroup$
    – Akash Roy
    Commented Nov 22, 2018 at 18:08
  • $\begingroup$ Thank you! I get it now. I just have to simplify $\cos[\cos^{-1}(\frac{3}{5})] \cdot \cos(\frac{3}{5}) - \sin[\cos^{-1}(\frac{3}{5})] \cdot \sin(\frac{\pi}{3})$ $\endgroup$ Commented Nov 22, 2018 at 18:09
  • $\begingroup$ @Luminous you can accept my answer if you understood. $\endgroup$
    – Akash Roy
    Commented Nov 22, 2018 at 18:10
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The $\cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$.

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    $\begingroup$ (+1) for mentioning the error in the OP $\endgroup$
    – Mark Viola
    Commented Nov 22, 2018 at 18:10
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$$ \cos(a+b)=\cos a\cos b-\sin a\sin b $$ and hence $$ \cos\left[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}\right]=\cos\left(\cos^{-1}\left(\frac{3}{5}\right)\right)\cos\left(\frac{\pi}{3}\right)-\sin\left(\cos^{-1}\left(\frac{3}{5}\right)\right)\sin\left(\frac{\pi}{3}\right)\\=\frac{3}{5}\cdot\frac{1}{2}-\sqrt{1-\frac{3^2}{5^2}}\cdot\frac{\sqrt{3}}{2}=\cdots $$

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By compound-angle formula, \begin{eqnarray*} \cos(\cos^{-1}(\frac{3}{5})+\frac{\pi}{3}) & = & \cos\left(\cos^{-1}(\frac{3}{5})\right)\cos\frac{\pi}{3}-\sin\left(\cos^{-1}(\frac{3}{5})\right)\sin(\frac{\pi}{3})\\ & = & \frac{3}{5}\cdot\frac{1}{2}-\sin\left(\cos^{-1}(\frac{3}{5})\right)\cdot\frac{\sqrt{3}}{2}. \end{eqnarray*}

To evaluate $\sin\left(\cos^{-1}(\frac{3}{5})\right)$, we let $\theta=\cos^{-1}(\frac{3}{5})$. Then $\cos\theta=\frac{3}{5}$. Recall that $\sin^{2}\theta+\cos^{2}\theta=1$ and observe that $0<\theta<\frac{\pi}{2}$. Therefore $\sin\theta>0$ and $\sin\theta$ is given by $\sin\theta=\sqrt{1-\cos^{2}\theta}=\frac{4}{5}$. It follows that \begin{eqnarray*} \cos(\cos^{-1}(\frac{3}{5})+\frac{\pi}{3}) & = & \frac{3}{5}\cdot\frac{1}{2}-\frac{4}{5}\cdot\frac{\sqrt{3}}{2}\\ & = & \frac{3}{10}-\frac{2\sqrt{3}}{5} \end{eqnarray*}

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Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.

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  • $\begingroup$ The first sentence in this answer is false: $\arccos\frac 35$ is not $37^\circ$. $\endgroup$
    – user562983
    Commented Nov 24, 2018 at 10:09
  • $\begingroup$ ... and $\cos 97^\circ$ is not $-0.39$. $\endgroup$
    – user562983
    Commented Nov 24, 2018 at 11:17

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