7.7. Principal Value of Inverse Trigonometric Functions

The least numeral value (positive or negative) of inverse trigonometric function is known as principal value of inverse trigonometric function.

Inverse Trigonometric Functions	Principal Value when `x\geq0`	Principal Value when `x<0`
`\sin^{-1}x`	`0\leq\sin^{-1}(x)\leq\frac{\pi}2`	`-\frac{\pi}2\leq\sin^{-1}(x)<0`
`\cos^{-1}x`	`0\leq\cos^{-1}(x)\leq\frac{\pi}2`	`\frac{\pi}2\leq\cos^{-1}(x)\leq\pi`
`\tan^{-1}x`	`0\leq\tan^{-1}(x)\leq\frac{\pi}2`	`-\frac{\pi}2\leq\tan^{-1}(x)<0`
`\cot^{-1}x`	`0\leq\cot^{-1}(x)\leq\frac{\pi}2`	`\frac{\pi}2\leq\cot^{-1}(x)\leq\pi`
`\sec^{-1}x`	`0\leq\sec^{-1}(x)\leq\frac{\pi}2`	`\frac{\pi}2\leq\sec^{-1}(x)\leq\pi`
`\cosec^{-1}x`	`0\leq\cosec^{-1}(x)\leq\frac{\pi}2`	`-\frac{\pi}2\leq\cosec^{-1}(x<0`

Example

1. Find the principal value of `\sin^{-1}\left(\frac1\2right)`.

Solution: We know that,

`\sin^{-1}(\frac(1)(2))=\sin^{-1}{\sin(\frac(pi)(6))}=\frac(pi)(6)`           and           `\sin^{-1}(\frac(1)(2))=\sin^{-1}{\sin(2\pi+\frac(pi)(6))}=\frac(13\pi)(6)`.

Since, `0\leq\sin^{-1}(x)\leq\frac(\pi)(2)` for `x\geq0` and `x=\frac(1)(2)` 

So, the principal value of `\sin^{-1}\left(\frac1\2right)` is `\frac\pi\6`. (Answer:)   

2. Find the principal value of `\cos^{-1}\left(\frac1{\sqrt2}\right)`.    

Solution: We know that,

`\cos^{-1}(\frac(1)(\sqrt2))=\cos^{-1}{(\cos(\frac(\pi)(4)}=\frac(\pi)(4)`           and           `\cos^{-1}(\frac(1)(\sqrt2))=\cos^{-1}{\cos(2\pi+\frac(pi)(4))}=\frac(9\pi)(4)`.

 

Since, `0\leq\cos^{-1}(x)\leq\frac(\pi)(2)` for `x\geq0` and `x=\frac(1)(\sqrt2)`

So, the principal value of `\cos^{-1}\left(\frac1{\sqrt2}\right)` is `\frac\pi\4`. Answer:

3. Find the principal value of `\sin^{-1}\left(-\frac1{\sqrt2}\right)`.

Solution: We know that,

`\sin^{-1}\left(-\frac1{\sqrt2}\right)` = `\sin^{-1}\left(-sin\frac\pi\4\right)` = `\sin^{-1}\{sin\left(-\frac\pi\4\right)\}` =  `-\frac\pi\4`

Since, `-\frac(\pi)(2)\leq\sin^{-1}(x)\leq0` for `x<0` and `x=-\frac(1)(\sqrt2)`

So, the principal value of `\sin^{-1}\left(-\frac1{\sqrt2}\right)` is `-\frac\pi\4`. Answer:

4. Find the principal value of `\sin^{-1}\left(-1right)`.

Try yourself.

5. Find the principal value of `\sin^{-1}\left(-\frac\sqrt3\2right)`.

Try yourself.

6. Find the principal value of `\sin^{-1}\left(\frac\sqrt3\2right)`.

Try yourself.

7. Find the principal value of `\cos^{-1}\left(\frac\sqrt3\2right)`.

Try yourself.

8. Find the principal value of `sin{\cos^{-1}\left(\frac\sqrt3\2right)}`.

Try yourself.

9. Find the principal value of `tan{\cos^{-1}\left(\frac1{\sqrt2}right)}`.

Try yourself.

10. Find the principal value of `\sin^{-1}\left(\frac1\2right)`.

Try yourself.

11. Find the principal value of `\tan^{-1}8+\tan^{-1}\left(\frac9\7right)`.

Solution: We know that,

So, the principal value of `\tan^{-1}8+\tan^{-1}\left(\frac9\7right)` is `\tan^{-1}\left(\frac(8+\frac9\7)\(1-8*\frac9\7)right)` = `\tan^{-1}\left(\frac(56+9)\(7-72)right)` = `\tan^{-1}(-1)` = `\tan^{-1}(-\tan\frac\pi\4)` = `\tan^{-1}{\tan(-\frac\pi\4)}`=  `\tan^{-1}{\tan(-\frac\pi\4)}`,  `\tan^{-1}{\tan(\pi-\frac\pi\4)}` = `-\frac\pi\4`, `\frac{3\pi}4` = `\frac{3\pi}4`. [Since sum of two positive angles cannot be negative]