# Finding exact value of trigonometric functions

I was wondering, how do I get the exact fraction (the value) of this trigonometric function:

$$\cos\left(\sin^{-1}(12/13)+\sin^{-1}(4/5)\right)$$

Usually, I would evaluate the inverse sin in degree mode and multiply (by hand) by $\pi/180$.

But in this case, I don't get exact values of angles

For example :

$$\begin{eqnarray} \sin^{-1}(12/13)=67,38...^\circ\\ \sin^{-1}(4/5)=53,13...^\circ \end{eqnarray}$$ Is there any way of doing it ?

Thank you!

HINT:

Let $A=\arcsin\dfrac{12}{13}$

Using the definition of Principal value of $\arcsin,0<A<\dfrac\pi2$

$\implies \cos A=+\sqrt{1-\sin^2A}=\dfrac5{13}$

• ok, I found a way, not sure if you were referring to this : cos(arcsin(12/13)+arcsin(4/5))=cos(arcsin(12/13)cos(arcsin(4/5)-sin(arcsin(12/13))+sin(arcsin(4/5)) – user108343 Dec 9 '14 at 18:16
• @Astroman, Please share your method – lab bhattacharjee Dec 9 '14 at 18:17
• After, you just have to evaluate the two cos ( cos(arcsin(12/13) and cos(arcsin(4/5)) with your explanation – user108343 Dec 9 '14 at 18:19
• @Astroman, What's $\cos(A+B)$ – lab bhattacharjee Dec 9 '14 at 18:20
• well, A=arcsin(12/13 and B=arcsin(4/5 – user108343 Dec 9 '14 at 18:21

We have: $$\arcsin\frac{4}{5} = \arg(3+4i),\qquad \arcsin\frac{12}{13}=\arg(5+12i),$$ hence: $$\arcsin\frac{4}{5}+\arcsin\frac{12}{13}=\arg((3+4i)(5+12i))=\arg(-33+56i)$$ and: $$\cos\left(\arcsin\frac{4}{5}+\arcsin\frac{12}{13}\right)=\frac{-33}{\sqrt{33^2+56^2}}=-\frac{33}{65}.$$

Avoiding complex numbers. $$\arcsin\frac{4}{5} = \arctan\frac{4}{3},\qquad \arcsin\frac{12}{13}=\arctan\frac{12}{5},$$ so: $$\arcsin\frac{4}{5}+\arcsin\frac{12}{13}=\arctan\frac{\frac{4}{3}+\frac{12}{5}}{1-\frac{4}{3}\cdot\frac{12}{5}}=\arctan\left(-\frac{56}{33}\right)$$ and: $$\cos\arctan\left(-\frac{56}{33}\right)=\frac{-33}{\sqrt{33^2+56^2}}=-\frac{33}{56}.$$

• I don't even know what arg() means... And we're only working with real numbers. Thank you again for you answer) – user108343 Dec 9 '14 at 18:26
• @Astroman: I just added another proof using only the sum formula for the tangent function. – Jack D'Aurizio Dec 9 '14 at 19:02
