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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!

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  • $\begingroup$ $\arccos\frac{3}{4}=\arctan\frac{\sqrt{7}}{3}$, $\arctan\frac{\sqrt{7}}{3}-\arctan\frac{1}{3} = \arctan(\ldots)$ and $\sin\arctan(s)=\frac{s}{\sqrt{s^2+1}}$, for instance. $\endgroup$ – Jack D'Aurizio Apr 12 '18 at 2:44
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    $\begingroup$ Why the three downvotes and vote to close? I'm fully prepared to edit or add to the problem if needed, however no one made any sort of recommendations or criticisms. $\endgroup$ – Clclstdnt Apr 12 '18 at 2:50
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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 $$\frac{3(-1+\sqrt{7})}{4\sqrt{10}}$$

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  • $\begingroup$ Why does this answer have a downvote? Is it wrong? I don't see anything technically wrong with it. However feels the need to downvote could instead leave comments that point out possible errors. $\endgroup$ – Clclstdnt Apr 12 '18 at 2:52
  • $\begingroup$ I do not know why someone would possibly downvote me... For I think this solution to this answer is fairly detailed and I am quite certain it's a correct approach. $\endgroup$ – Jesse Meng Apr 12 '18 at 2:53
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    $\begingroup$ I upvoted..... I'm going through the solutions more carefully right now. But it looked good at first. Seems like someone is just being a bully.... they don't like me asking a "stupid easy question" and I have shown "no work" so they down vote then they down vote anyone who tries to answer. I've already shown some work, namely that it is likely that I should use the addition formula, and indeed this is what you have used. $\endgroup$ – Clclstdnt Apr 12 '18 at 2:55
  • $\begingroup$ @Clclstdnt yeah as long as my solution is helpful. Although this problem is not hard and you could have struggled a bit more with it :) $\endgroup$ – Jesse Meng Apr 12 '18 at 2:57
  • $\begingroup$ I could have and should have..... It's also late in the semester, I'm stressed and tired. I don't feel like this is a place I need to justify myself. I feel if people want to answer, let them answer.... if they don't want to, then don't. But don't be sanctimonious about it. $\endgroup$ – Clclstdnt Apr 12 '18 at 3:18
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Let $a=\cos^{-1}(\frac{3}{4})$, $\cos a= \frac{3}{4}$

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

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

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

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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

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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.

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