1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Hint: simply take $\sin$ of both sides of the first equation. $\endgroup$ – lulu Apr 27 '16 at 11:38
  • 1
    $\begingroup$ @lulu I just wanted to know why answer is not coming while taking in tan? $\endgroup$ – user3500780 Apr 27 '16 at 11:40
  • $\begingroup$ Use math.stackexchange.com/questions/672575/… $\endgroup$ – lab bhattacharjee Apr 27 '16 at 11:42
  • 2
    $\begingroup$ If $\theta =\arccos\left(\frac {12}{13}\right)$ then $\theta =\arctan\left(\frac {5}{12}\right)$ $\endgroup$ – lulu Apr 27 '16 at 11:43
0
$\begingroup$

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: $$\frac{3}{5}\frac{12}{13}-\sqrt{1-\frac{9}{25}}\sqrt{1-\frac{144}{169} } = \frac{3}{5}\frac{12}{13}-\frac{4}{5}\frac{5}{13}= \cdots$$

$\endgroup$
  • $\begingroup$ OH MY GOD i am so so stupid really thanks $\endgroup$ – user3500780 Apr 27 '16 at 11:45
  • $\begingroup$ 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?? $\endgroup$ – user3500780 Apr 27 '16 at 11:47
  • $\begingroup$ 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. $\endgroup$ – StackTD Apr 27 '16 at 11:49
1
$\begingroup$

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 $$ \tan\left(\arctan\frac{3}{4}-\arctan\frac{5}{12}\right)= \frac{\dfrac{3}{4}-\dfrac{5}{12}}{1+\dfrac{3}{4}\dfrac{5}{12}}= \frac{\dfrac{1}{3}}{\;\dfrac{21}{16}\;}=\frac{16}{63} $$

$\endgroup$
0
$\begingroup$

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 $$\arctan(\frac{16}{63}) = \arctan(\frac{16}{63})$$ $$LHS = RHS$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.