# Prove: $\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right) = \arcsin\left(\frac {16}{65}\right)$

This is not a homework question, its from sl loney I'm just practicing. To prove : $$\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right) = \arcsin\left(\frac {16}{65}\right)$$

So I changed all the angles to $\arctan$ which gives:

$$\arctan\left(\frac 34\right) - \arctan\left(\frac {12}{5}\right) = \arctan\left(\frac {16}{63}\right)$$

But the problem is after applying formula of $\arctan(X)-\arctan(Y)$ the lhs is negative and not equal to rhs? Is this because I have to add pi somewhere please help.

• Hint: simply take $\sin$ of both sides of the first equation. – lulu Apr 27 '16 at 11:38
• @lulu I just wanted to know why answer is not coming while taking in tan? – user3500780 Apr 27 '16 at 11:40
• – lab bhattacharjee Apr 27 '16 at 11:42
• If $\theta =\arccos\left(\frac {12}{13}\right)$ then $\theta =\arctan\left(\frac {5}{12}\right)$ – lulu Apr 27 '16 at 11:43

## 3 Answers

How exactly did you convert to arctan? Careful: $$\arccos\left(\frac {12}{13}\right) = \arctan\left(\frac {5}{12}\right) \ne \arctan\left(\frac {12}{5}\right)$$ Draw a right triangle with hypotenuse of length 13, adjacent side (from an angle $\alpha$) with length 12 and opposite side with length 5; then $\cos\alpha = 12/13$ and $\tan\alpha = 5/12$.

Perhaps easier: take the sine of both sides in the original equation: $$\sin\left(\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right)\right) = \sin\left( \arcsin\left(\frac {16}{65}\right)\right)$$ The RHS is $16/65$ and simplify the LHS with $\sin(a-b)=\sin a \cos b - \cos a \sin b$ to get: $$\frac{3}{5}\frac{12}{13}-\sqrt{1-\frac{9}{25}}\sqrt{1-\frac{144}{169} } = \frac{3}{5}\frac{12}{13}-\frac{4}{5}\frac{5}{13}= \cdots$$

• OH MY GOD i am so so stupid really thanks – user3500780 Apr 27 '16 at 11:45
• One thing though, how am I to look at the question and realize that I should take sin both sides? I mean under what circumstances it should click in my mind i gotta take sin both sides?? – user3500780 Apr 27 '16 at 11:47
• Other trigonometric functions could work as well, but will give you more work since you have two arcsines and $\sin\left(\arcsin\alpha\right) = \alpha$; so that simplifies easily. – StackTD Apr 27 '16 at 11:49

If $0<x<1$, then both $\arcsin x$ and $\arccos x$ are in $(0,\pi/2)$. I'll assume $0<x<1$ for the rest of the discussion.

If $\alpha=\arcsin x$, then $\sin\alpha=x$ and $\cos\alpha=\sqrt{1-x^2}$; therefore $$\tan\alpha=\frac{x}{\sqrt{1-x^2}}$$ and $$\arcsin x=\alpha=\arctan\frac{x}{\sqrt{1-x^2}}$$ Similarly, if $\beta=\arccos x$, then $x=\cos\beta$ and $$\arccos x=\arctan\frac{\sqrt{1-x^2}}{x}$$ For $x=3/5$ we have $\sqrt{1-x^2}=4/5$ and so $$\arcsin\frac{3}{5}=\arctan\frac{3}{4}$$ For $x=16/65$ we have $\sqrt{1-x^2}=63/65$, so $$\arcsin\frac{16}{65}=\arctan\frac{16}{63}$$ For $x=12/13$ we have $\sqrt{1-x^2}=5/13$, so $$\arccos\frac{12}{13}=\arctan\frac{5}{12}$$ Now $$\tan\left(\arctan\frac{3}{4}-\arctan\frac{5}{12}\right)= \frac{\dfrac{3}{4}-\dfrac{5}{12}}{1+\dfrac{3}{4}\dfrac{5}{12}}= \frac{\dfrac{1}{3}}{\;\dfrac{21}{16}\;}=\frac{16}{63}$$

To specifically use $arctan$ formula:

Change all angles to $tan$ as you outlined, using Pythagoras... $$\arctan(\frac34) - \arctan(\frac5{12}) = \arctan(\frac{16}{63})$$

Apply $arctan$ formula $$\arctan(x) - \arctan(y) = \arctan(\frac{x-y}{1+xy})$$

$$\arctan(\frac{\frac34 - \frac5{12}}{1+\frac34 \frac5{12}}) = \arctan(\frac{16}{63})$$ Simplify to give $$\arctan(\frac{16}{63}) = \arctan(\frac{16}{63})$$ $$LHS = RHS$$