How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$? The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both side and drawing triangles. However, the worked solutions does a simplification from LHS to RHS instead, namely $$\cos\arcsin\left(\frac{3}{5}\right)\cos\arctan\left(\frac{7}{24}\right)-\sin\arcsin\left(\frac{3}{5}\right)\sin\arctan\left(\frac{7}{24}\right)=\frac{3}{5}$$which I don't understand. Can someone please explain?
 A: By drawing triangles it should be easy to see that \begin{align*}
\cos \arcsin \left(\frac{3}{5}\right)&=\frac{4}{5}\\
\cos \arctan \left(\frac{7}{24}\right)&=\frac{24}{25}\\
\sin \arcsin\left(\frac{3}{5}\right)&=\frac{3}{5}\\ 
\sin \arctan\left(\frac{7}{24}\right)&=\frac{7}{25}
\end{align*}
(note the Pythagorean triples (3,4,5), and (24,7,25).) Putting this all together: $$\cos \arcsin \left(\frac{3}{5}\right)\cos \arctan \left(\frac{7}{24}\right)-\sin \arcsin\left(\frac{3}{5}\right)\sin \arctan\left(\frac{7}{24}\right)=\frac{4}{5}\cdot \frac{24}{25}-\frac{3}{5}\cdot \frac{7}{25}=\frac{3}{5}$$
A: The solution makes use of the Pythagorean identities.
I. $\arcsin(3/5)=\theta\implies \sin(\theta)=3/5$. Using the identity: $\cos^2(\theta)=1-\sin^2(\theta)\iff \cos(\theta)=\sqrt{1-\sin^2(\theta)}$. Hence, $\cos(\arcsin(3/5))=\sqrt{1-(\frac{3}{5})^2}=\sqrt{\frac{16}{25}}=4/5$.
II. $\arctan(7/24)=\theta\implies \tan(\theta)=7/24$. Using the identity $1+\tan^2(\theta)=\sec^2(\theta) \iff \cos(\theta)=\frac{1}{\sqrt{1+\tan^2(\theta)}}$. Hence, $\cos(\arctan(7/24))=\frac{1}{\sqrt{1+(\frac{7}{24})^2}}=24/25$.
III. $\sin(\arcsin(3/5))=3/5$ is trivial.
IV. $\arctan(7/24)=\theta \implies \tan(\theta)=7/24$ Using the identity $1+\cot^2(\theta)=\csc^2(\theta) \iff \sin(\theta)=\sqrt{\frac{\tan^2(\theta)}{1+\tan^2(\theta)}}$. Hence, $\sin(\arctan(7/24))=\sqrt{\frac{(\frac{7}{24})^2}{1+(\frac{7}{24})^2}}=7/25$.
Combining the 4 results we have: $I*II-III*IV=4/5*24/25-3/5*7/25=3/5$ yields the desired result.
A: The provided solution first employs the angle addition identity $$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta,$$ with the choices $$\alpha = \arcsin \frac{3}{5}, \quad \beta = \arctan \frac{7}{24}.$$  That is to say, $$\sin \alpha = \frac{3}{5}, \quad \tan \beta = \frac{7}{24}.$$  In order to proceed with the evaluation, we now recall $$\sin^2 \theta + \cos^2 \theta = 1, \\ \tan^2 \theta + 1 = \sec^2 \theta.$$  So $$\frac{9}{25} = \sin^2 \alpha = 1 - \cos^2 \alpha,$$ hence $$\cos \alpha = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}.$$  Similarly, $$\frac{49}{24^2} = \tan^2 \beta = \sec^2 \beta - 1,$$ or $$\cos \beta = \frac{24}{25}.$$  It follows that $\sin \beta = \cos \beta \tan \beta = \frac{7}{25}$ and $$\cos(\alpha + \beta) = \frac{4}{5} \frac{24}{25} - \frac{3}{5} \frac{7}{25} = \frac{3}{5}.$$  This solution assumes that $-\frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2}$ and  $-\frac{\pi}{2} < \arctan x < \frac{\pi}{2}$.
A: As the inverse trigonometric functions are multi-valued with a respective principle value
or
as sine/cosine/tangent functions are periodic, it is incorrect to apply cosines in both sides.
For example,

*

*$x$ in general $\ne -x$  but $\cos(-x)=\cos(x)$

*$x$ in general $\ne \pi-x$  but $\sin(\pi-x)=\sin(x)$

*$x\ne\pi+x$  but $\tan(\pi+x)=\tan(x)$
Using the ranges of principle values of $\arctan,$ if $\arctan\dfrac7{24}=y$
$\implies0<y<\dfrac\pi2$
and $\tan y=\dfrac7{24}$
and $\sin y,\cos y>0$
Consequently, $\dfrac{\sin y}7=\dfrac{\cos y}{24}=+\sqrt{\dfrac{\sin^2y+\cos^2y}{7^2+24^2}}$
$\implies\arctan\dfrac7{24}=\arcsin\dfrac7{25}=\arccos\dfrac{24}{25}$
Now use Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ in the left hand side
For the right hand side, if $\arccos\dfrac35=u$
$\implies0<u<\dfrac\pi2$ and $\cos u=\dfrac35$ and $\sin u>0$
$\implies\sin u=\sqrt{1-\cos^2u}=\dfrac45$
$\implies\arcsin\dfrac45=u=\arccos\dfrac35$
