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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?

player3236
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user71207
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    mostly that there are Pythagorean triples $3,4,5$ and $7,24,25$ – Will Jagy Jan 05 '21 at 04:04
  • what? I dont understand. One pythagorean triple simplifies to another? – user71207 Jan 05 '21 at 04:08
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    Unless they're taking advantage of some identity I'm not aware of, I think they are just using the triangles like you are and skipping the steps inbetween. In any case if they didn't show steps I see no reason not to believe that. – Stephen Donovan Jan 05 '21 at 04:09
  • Ok yes I believe that was a stupid question. They probably did skip steps – user71207 Jan 05 '21 at 04:10
  • You don't understand how they made that simplification? –  Jan 05 '21 at 04:10
  • No, I understand a way they could've gotten that simplification but I don't understand that equation. However, I actually believe there is a way to do it without triangles as I've seen a similar question before – user71207 Jan 05 '21 at 04:14
  • I mean you can probably do it using the Pythagorean identities but that's basically equivalent to using the triangles. – Stephen Donovan Jan 05 '21 at 04:17
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    Not sure if this will help or not, but the first equation can be rewritten as $\theta_1+\theta_2=\theta_3$ where $\theta_3 = \frac{\pi}{2} - \theta_1$ thus $2\theta_1 + \theta_2 = \frac{\pi}{2}$. $2\theta_1$ is the opposite angle in the right triangle with sides $7,24,25$. $\tan\theta_2=\frac{7}{24}$ and $\cos(2\theta_1)=\frac{7}{25}=\cos^2(\theta_1)-\sin^2(\theta_1)=\left(\frac{4}{5}\right)^2-\left(\frac{3}{5}\right)^2$, indeed the equality must hold. You can always use triangles to guide along. – no lemon no melon Jan 05 '21 at 04:43

4 Answers4

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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: $$\cos \arcsin \left(\frac{3}{5}\right)\cos \arctan \left(\frac{7}{24}\right)-\sin \arcsin\left(\frac{3}{5}\right)\sin \arctan\left(\frac{7}{24}\right)=\frac{4}{5}\cdot \frac{24}{25}-\frac{3}{5}\cdot \frac{7}{25}=\frac{3}{5}$$

5201314
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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.

Transcending
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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 $-\frac{\pi}{2} < \arctan x < \frac{\pi}{2}$.

heropup
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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 $\sin u>0$

$\implies\sin u=\sqrt{1-\cos^2u}=\dfrac45$

$\implies\arcsin\dfrac45=u=\arccos\dfrac35$

lab bhattacharjee
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