Trigonometric and Inverse Trigonometric Functions

Let us now establish some elementary but useful results involving trigonometric and inverse trigonometric functions.


a. For a given angle $\theta$,

1. $\theta=\sin^{-1}\sin\theta$

Let $\sin\theta=x$, then, \[\theta=\sin^{-1}x\] \[\therefore\theta=\sin^{-1}\sin\theta\]

2. $\theta=\sin\sin^{-1}\theta$

Let $y=\sin^{-1}\theta$, then, \[\theta=\sin y\] \[\theta=\sin\sin^{-1}y\] The rest of the trigonometric functions; cosine, tangent, etc. follows similarly.


b. For a given numerical value $x$, \[\operatorname{cosec}^{-1}x=\sin^{-1}\frac{1}{x}\]

Let $\theta=\operatorname{cosec}^{-1}x$, then, \[\operatorname{cosec}\theta=x\] \[\sin\theta=\frac{1}{x}\] \[\theta=\sin^{-1}\frac{1}{x}\] \[\therefore\operatorname{cosec}^{-1}x=\sin^{-1}\frac{1}{x}\]

Similarly, we can show, \[\sin^{-1}x=\operatorname{cosec}^{-1}\frac{1}{x}\] \[\sec^{-1}x=\cos^{-1}\frac{1}{x}\] \[\cot^{-1}x=\tan^{-1}\frac{1}{x},\;\;\text{etc.}\]


c. Expressions of a given inverse trigonometric function in terms of the remaining inverse trigonometric functions

1. $\sin^{-1}x=\cos^{-1}\sqrt{1-x^2}$

Let $\theta=\sin^{-1}x$, then, $\sin\theta=x$. Now, \[\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-x^2}\] \[\theta=\cos^{-1}\sqrt{1-x^2}\] \[\therefore\sin^{-1}x=\cos^{-1}\sqrt{1-x^2}\]

2. $\sin^{-1}x=\tan^{-1}\frac{x}{\sqrt{1-x^2}}$

Let $\theta=\sin^{-1}x$, then, $\sin\theta=x$. Now, \[\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{x}{\sqrt{1-x^2}}\] \[\theta=\tan^{-1}\frac{x}{\sqrt{1-x^2}}\] \[\therefore\sin^{-1}x=\tan^{-1}\frac{x}{\sqrt{1-x^2}}\]

Similarly, we can prove, \[\sin^{-1}x=\cot^{-1}\frac{\sqrt{1-x^2}}{x}\] \[\sin^{-1}x=\sec^{-1}\frac{1}{\sqrt{1-x^2}}\]

Expressions of other inverse trigonometric functions can also be deduced in the same way.


d. For a given numerical value $x$, \[\sin^{-1}x+\cos^{-1}x=\frac{π}{2}\]

Let $\theta=\sin^{-1}x$, then, $x=\sin\theta$. Hence, \[x=\cos\left(\frac{π}{2}-\theta\right)\] \[\cos^{-1}x=\frac{π}{2}-\theta\] \[\therefore\sin^{-1}x+\cos^{-1}x=\frac{π}{2}\]

Similarly, we can prove,

\[\tan^{-1}x+\cot^{-1}x=\frac{π}{2}\] \[\operatorname{cosec}^{-1}x+\sec^{-1}x=\frac{π}{2}\]


e. For a given numerical value $x$,

1. $\sin^{-1}(-x)=-\sin^{-1}x$

Let $\sin^{-1}x=\theta$, then, $x=\sin\theta$. Now, \[-x=-\sin\theta\] \[-x=\sin(-\theta)\] \[\sin^{-1}(-x)=-\theta\] \[\therefore\sin^{-1}(-x)=-\sin^{-1}x\]

2. $\cos^{-1}(-x)=π-\cos^{-1}x$

Let $\cos^{-1}x=\theta$, then, $x=\cos\theta$. Now, \[-x=-\cos\theta\] \[-x=\cos(π-\theta)\] \[\cos^{-1}(-x)=π-\theta\] \[\therefore\cos^{-1}(-x)=π-\cos^{-1}x\]

Similarly, we can prove, \[\tan^{-1}(-x)=-\tan^{-1}x\] \[\cot^{-1}(-x)=π-\cot^{-1}x\]


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