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

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:

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

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

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

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

$$\frac{11}{10}$$

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

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.

• (+1) ... for discussing the error in the OP Commented Nov 22, 2018 at 18:05
• @MarkViola thanks for your upvote. Commented Nov 22, 2018 at 18:06
• Thanks once again for your valuable edits @Mark, will bear these in mind. Commented Nov 22, 2018 at 18:08
• Thank you! I get it now. I just have to simplify $\cos[\cos^{-1}(\frac{3}{5})] \cdot \cos(\frac{3}{5}) - \sin[\cos^{-1}(\frac{3}{5})] \cdot \sin(\frac{\pi}{3})$ Commented Nov 22, 2018 at 18:09
• @Luminous you can accept my answer if you understood. Commented Nov 22, 2018 at 18:10

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: $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$.

• (+1) for mentioning the error in the OP Commented Nov 22, 2018 at 18:10

$$\cos(a+b)=\cos a\cos b-\sin a\sin b$$ and hence $$\cos\left[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}\right]=\cos\left(\cos^{-1}\left(\frac{3}{5}\right)\right)\cos\left(\frac{\pi}{3}\right)-\sin\left(\cos^{-1}\left(\frac{3}{5}\right)\right)\sin\left(\frac{\pi}{3}\right)\\=\frac{3}{5}\cdot\frac{1}{2}-\sqrt{1-\frac{3^2}{5^2}}\cdot\frac{\sqrt{3}}{2}=\cdots$$

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 $$\begin{eqnarray*} \cos(\cos^{-1}(\frac{3}{5})+\frac{\pi}{3}) & = & \frac{3}{5}\cdot\frac{1}{2}-\frac{4}{5}\cdot\frac{\sqrt{3}}{2}\\ & = & \frac{3}{10}-\frac{2\sqrt{3}}{5} \end{eqnarray*}$$

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.

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