# 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

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \cos\pars{\arcsin\pars{12 \over 13} + \arcsin\pars{4 \over 5}}} \\[5mm]&=\cos\pars{\arcsin\pars{12 \over 13}}\cos\pars{\arcsin\pars{4 \over 5}} -\ \overbrace{\sin\pars{\arcsin\pars{12 \over 13}}}^{\dsc{12 \over 13}}\ \overbrace{\sin\pars{\arcsin\pars{4 \over 5}}}^{\dsc{4 \over 5}} \\[5mm]&=\root{1 - \sin^{2}\pars{\arcsin\pars{12 \over 13}}} \root{1 - \sin^{2}\pars{\arcsin\pars{4 \over 5}}}\ -\ {12 \over 13}\,{4 \over 5} \\[5mm]&=\root{1 - \pars{12 \over 13}^{2}} \root{1 - \pars{4 \over 5}^{2}} - {48 \over 65} ={3 \over 13} - {48 \over 65}=\color{#66f}{\large -\,{33 \over 65}} \end{align}