Is there any formula to find $\sin^{-1}a +\sin^{-1}b$

Question

We known that$$\tan^{-1} a +\ tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right).$$

Now we derive the above formula. Let$$ \tan^{-1}a=\theta _1 \implies \tan\theta_1=a,\\ \tan^{-1}b=\theta _2 \implies \tan\theta_2=b,\\ \theta _1+\theta _2 = \tan^{-1}a+ \tan^{-1}b,\\ \tan(\theta _1+\theta _2) = \tan(\tan^{-1}a+\tan^{-1}b)· \tan(\theta _1+\theta _2)\\ =\frac{ \tan(\tan^{-1}a)+\tan( \tan^{-1}b)}{1- \tan(\tan^{-1}a)· \tan(\tan^{-1}b)},\\ (\theta _1+\theta _2) =\tan^{-1}\left(\frac {a+b}{1-ab}\right). $$ Therefore, $\tan^{-1} a +\ tan^{-1}b=\tan^{-1}\left(\dfrac{a+b}{1-ab}\right)$.

For all $\theta _1$ and $\theta _2$ such that $0 \le \theta _1,\theta _2 \lt 90$, in the similar way I try to find $\sin^{-1}a +\sin^{-1}b$. Let$$ \sin^{-1}a=\theta_1, \quad \sin^{-1}b=\theta_2,\\ \sin(\theta_1+\theta_2)=\sin(\sin^{-1}a +\sin^{-1}b),\\ \sin(\theta_1+\theta_2)=\sin(\sin^{-1}a)·\cos(sin^{-1}b) +\sin(\sin^{-1}b)·\cos(sin^{-1}a),\\ \theta_1+\theta_2=\sin(\sin^{-1}a)·\cos(sin^{-1}b) +\sin(\sin^{-1}b)·\cos(sin^{-1}a),\\ \sin^{-1}a+\sin^{-1}b=\sin(\sin^{-1}a).cos(sin^{-1}b) +\sin(\sin^{-1}b).cos(sin^{-1}a),\\ \sin^{-1}a+\sin^{-1}b=\sin(a)·\cos(\sin^{-1}b) +\sin(b)·\cos(\sin^{-1}a).$$ It cannot be write in the form only using $a$ and $b$. $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A·\tan B}.$$ It contains only $\tan$ terms. But $\sin(A+B)=\sin A·\cos B+\cos A·\sin B$, which contains both $\sin$ and $\cos$ terms.

Is there any formula for $\sin^{-1}a+\sin^{-1}b$?

https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y — lab bhattacharjee, May 30 '18 at 13:04

José Carlos Santos · Accepted Answer · 2018-05-30T13:10:40.647

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Assume that $\arcsin(a)+\arcsin(b)\in\left[-\frac\pi2,\frac\pi2\right]$. Then, since$$\sin\left(\arcsin(a)+\arcsin(b)\right)=a\sqrt{1-b^2}+b\sqrt{1-a^2},$$we have$$\arcsin(a)+\arcsin(b)=\arcsin\left(a\sqrt{1-b^2}+b\sqrt{1-a^2}\right).$$

edited May 30 '18 at 13:10

answered May 30 '18 at 13:02

José Carlos Santos

427,504

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What if $a=b=\dfrac{\sqrt3}2$? – lab bhattacharjee May 30 '18 at 13:05
@labbhattacharjee I've edited my answer. Thank you. – José Carlos Santos May 30 '18 at 13:10
@ALPHA24 When $a=b=\frac12$, both $\arcsin(a)+\arcsin(b)$ and $\arcsin\left(a\sqrt{1-b^2}+b\sqrt{1-a^2}\right)$ are equal to $\frac\pi3$. – José Carlos Santos May 30 '18 at 15:56
@ALPHA24 If $a=b=\frac12$, then $\arcsin(a)+\arcsin(b)=\frac\pi6+\frac\pi6=\frac\pi3$ and$$\arcsin\left(a\sqrt{1-b^2}+b\sqrt{1-a^2}\right)=\arcsin\left(\frac{\sqrt3}2\right)=\frac\pi3.$$Do you disagree? – José Carlos Santos May 30 '18 at 16:22
No, I agree this – May 31 '18 at 15:10

Is there any formula to find $\sin^{-1}a +\sin^{-1}b$

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