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I've the proof of the following formulae.I've someother problem,i've spent a great amount time in remebering,but every time i commit mistakes while doing the problems of inverse trigonometry.I wanted to know if there is some quick approach of learning the following formulae especially $sin^{-1}x-sin^{-1}y$.

($I$)$$tan^{-1}x+tan^{-1}y = \begin{cases}tan^{-1} (\frac{x+y}{1-xy}), & \text{if $xy<1$ } \\[2ex] \pi +tan^{-1} (\frac{x+y}{1-xy}), & \text{if $x>0,y<0$ and $xy>1$ } \\[3ex] -\pi +tan^{-1} (\frac{x+y}{1-xy}), & \text{if $x<0,y<0$ and $xy<1$ } \end{cases}$$

($II$)$$tan^{-1}x-tan^{-1}y = \begin{cases}tan^{-1} (\frac{x-y}{1+xy}), & \text{if $xy>-1$ } \\[2ex] \pi +tan^{-1} (\frac{x-y}{1+xy}), & \text{if $x>0,y<0$ and $xy<-1$ } \\[3ex] -\pi +tan^{-1} (\frac{x-y}{1+xy}), & \text{if $x<0,y<0$ and $xy<-1$ } \end{cases}$$

($III$)$$sin^{-1}x-sin^{-1}y = \begin{cases}sin^{-1}(x\sqrt {1-y^2}-y\sqrt {1-x^2}), & \text{if $-1\leq x,y \leq1$ and $x^2+y^2\leq$1 or if $xy>0$ and $x^2+y^2>1$} \\[2ex] \pi +sin^{-1}(x\sqrt {1-y^2}-y\sqrt {1-x^2}), & \text{if $0< x \leq1$ and $-1\leq y\leq$0 and $x^2+y^2>1$}\\[3ex] -\pi +sin^{-1}(x\sqrt {1-y^2}-y\sqrt {1-x^2}), & \text{if $-1\leq x<0$, $0< y\leq$1 and $x^2+y^2>1$}\end{cases}$$

($IV$)$$sin^{-1}x+sin^{-1}y = \begin{cases}sin^{-1}(x\sqrt {1-y^2}+y\sqrt {1-x^2}), & \text{if $-1\leq x,y \leq1$ and $x^2+y^2\leq$1 or if $xy<0$ and $x^2+y^2>1$} \\[2ex] \pi +sin^{-1}(x\sqrt {1-y^2}+y\sqrt {1-x^2}), & \text{if $0< x,y \leq1$ and $x^2+y^2>1$}\\[3ex] -\pi +sin^{-1}(x\sqrt {1-y^2}+y\sqrt {1-x^2}), & \text{if $-1\leq x,y<0$ and $x^2+y^2>1$}\end{cases}$$

($V$) $$2sin^{-1}x = \begin{cases}sin^{-1} (2x\sqrt{1-x^2}), & \text{if $\frac{-1}{\sqrt 2}\leq x \leq \frac{1}{\sqrt 2}$ } \\[2ex] \pi -sin^{-1} (2x\sqrt{1-x^2}), & \text{if $\frac{1}{\sqrt 2}\leq x \leq1$ } \\[3ex] -\pi -sin^{-1} (2x\sqrt{1-x^2}), & \text{if $-1\leq x \leq \frac{-1}{\sqrt 2}$ } \end{cases}$$

($VI$)$$3sin^{-1}x = \begin{cases}sin^{-1} (3x-4x^3), & \text{if $\frac{-1}{ 2}\leq x \leq \frac{1}{ 2}$ } \\[2ex] \pi -sin^{-1} (3x-4x^3), & \text{if $\frac{1}{ 2}< x \leq1$ } \\[3ex] -\pi -sin^{-1} (3x-4x^3), & \text{if $-1\leq x < \frac{-1}{ 2}$ } \end{cases}$$

($VII$)$$2cos^{-1}x = \begin{cases}cos^{-1} (2x^2-1), & \text{if $0\leq x \leq1$ } \\[2ex] 2\pi -cos^{-1} (2x^2-1), & \text{if $-1\leq x\leq 0$ } \\[3ex] \end{cases}$$

($VIII$)$$3cos^{-1}x = \begin{cases}cos^{-1} (4x^3-3x), & \text{if $\frac{1}{ 2}\leq x \leq 1$ } \\[2ex] 2\pi -cos^{-1} (4x^3-3x), & \text{if $\frac{-1}{ 2}< x \leq \frac{1}{2}$ } \\[3ex] 2\pi +cos^{-1} (4x^3-3x), & \text{if $-1\leq x < \frac{-1}{ 2}$ } \end{cases}$$

($IX$)$$3tan^{-1}x = \begin{cases}tan^{-1}(\frac{3x-x^3}{1-3x^2}) , & \text{if $\frac{-1}{ \sqrt 3}\leq x \leq \frac{1}{\sqrt 3}$ } \\[2ex] \pi +tan^{-1}(\frac{3x-x^3}{1-3x^2}), & \text{if $ x >\frac{1}{\sqrt 3}$ } \\[3ex] -\pi +tan^{-1}(\frac{3x-x^3}{1-3x^2}), & \text{if $x < \frac{-1}{ \sqrt 3}$ } \end{cases}$$

Picaso
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    Remember $$\sin(A\pm B),\cos(A\pm B),\tan(A\pm B)$$ and https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values Then we can derive them like https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y and https://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan – lab bhattacharjee Aug 12 '17 at 09:43
  • @labbhattacharjee:I wanted to know the method of learning the formulae,not their derivations.Thank you!! – Picaso Aug 12 '17 at 14:06
  • @Picaso - lab bhattacharjee has told you what you really need. Now it is up to you to recognize that and stop chasing chimeras. Practically nobody memorizes such formulas as these. It is far more trouble than it could possibly ever be worth. Look them up when you need them. Learn how to derive them when you can't look them up. This is not only easier than memorizing tons of formulas, but will also serve you much, much, MUCH better than rote memorization. – Paul Sinclair Aug 12 '17 at 15:55

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