Simplify $\sin[\cos^{-1}(\frac{3}{4}) - \tan^{-1}(\frac{1}{3})]$

Question

I am trying to solve the following:

$$\sin\left[\cos^{-1}\left(\frac{3}{4}\right) - \tan^{-1}\left(\frac{1}{3}\right)\right]$$

but I have no clue..... I used wolfram alpha and it says the solution is

$$\frac{3(-1+\sqrt{7})}{4\sqrt{10}}$$

and it says the method is by the use of the addition method of trig functions; I don't see how though. Any hints would be greatly appreciated!

Answer 1

You can do this:

Apply the formula $\sin(A-B)=\sin A \cos B-\cos A \sin B$: $$\sin\left[\cos^{-1}\left(\frac{3}{4}\right) - \tan^{-1}\left(\frac{1}{3}\right)\right]=\sin\left[\cos^{-1}\left(\frac{3}{4}\right)\right]\cos\left[\tan^{-1}\left(\frac{3}{4}\right)\right]-\frac{3}{4}\sin\left[\tan^{-1}\left(\frac{3}{4}\right)\right]$$

For the calculation of each of the terms: for example, $\sin\left[\cos^{-1}\left(\frac{3}{4}\right)\right]$, let $\theta=\cos^{-1}\left(\frac{3}{4}\right)$, this means $\cos\theta=\left(\frac{3}{4}\right)$, from this, you can draw a right triangle and arrive at $\sin\theta=\left(\frac{\sqrt{7}}{4}\right)$. This means $\sin\left[\cos^{-1}\left(\frac{3}{4}\right)\right]$=$\sin\theta=\left(\frac{\sqrt{7}}{4}\right)$. Similarly, you can calculate $$\cos\left[\tan^{-1}\left(\frac{3}{4}\right)\right],\sin\left[\tan^{-1}\left(\frac{3}{4}\right)\right]$$ Then you substitute them back and indeed you have your answer $$\frac{3(-1+\sqrt{7})}{4\sqrt{10}}$$

Answer 2

You need some ingredients here. First, a formula for $\sin(\alpha+\beta)$. And then, you'll need to compute $\sin\cos^{-1} (3/4)$ and other expressions like that.

In order to so, draw a Pythagoras triangle in which an angle is such that $\cos\theta = 3/4$. Now compute $\sin\theta$. Good luck and feel free to come at the comments for further guidance.

Answer 3

Let $a=\cos^{-1}(\frac{3}{4})$, $\cos a= \frac{3}{4}$

Let $b=\tan^{-1}(\frac{1}{3})$, $\tan b= \frac{1}{3}$

$\sin\left[\cos^{-1}\left(\frac{3}{4}\right) - \tan^{-1}\left(\frac{1}{3}\right)\right]=\sin(a-b)$

$\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)=\frac{\sqrt{7}}{4}\times\frac{3}{\sqrt{10}}-\frac{3}{4}\times\frac{1}{\sqrt{10}}=\frac{3\sqrt{7}-3}{4\sqrt{10}}=\frac{3(-1+\sqrt{7})}{4\sqrt{10}}$

Answer 4

Always sketch right angled triangle in matters of inverse trig evaluation $$ \cos^{-1} \frac34 - \tan^{-1} \frac13 $$ $$ \tan^{-1}\frac{ \sqrt7/3 -1/3}{1+\sqrt7/9} $$ $$ \tan^{-1}\frac{ \sqrt7 -1}{3+\sqrt7/3} $$ Find Hypotenuse by Pythagoras thm $$ 7+1- 2\sqrt7 +7/9 +9 + 2\sqrt 7 = \frac{160}{9}$$ whose square root is $$ \frac{\sqrt10 \cdot 4}{3}$$

Now find sine.. opposite side/ hypotenuse,

the WA result follows