The principal value of sin -1cos sin -1-32 is

The correct option is A π/6 The principle value of sin^ 1(cos(sin^ 1(√(3)2))) =π2 cos^ 1(cos(sin^ 1(√(3)2))) =π2 sin^ 1(√(3)2) =cos^ 1 ...

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Standard XII

Mathematics

Basic Inverse Trigonometric Functions

The principal...

Question

The principal value of ${sin}^{- 1} {cos ({sin}^{- 1} \frac{\sqrt{3}}{2})}$ is

\frac{π}{6}

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\frac{π}{3}

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- \frac{π}{3}

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none of these

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Solution

The correct option is A $\frac{π}{6}$
The principle value of ${sin}^{- 1} (cos ({sin}^{- 1} (\frac{\sqrt{3}}{2})))$

$= \frac{π}{2} - {cos}^{- 1} (cos ({sin}^{- 1} (\frac{\sqrt{3}}{2})))$

$= \frac{π}{2} - {sin}^{- 1} (\frac{\sqrt{3}}{2})$

$= {cos}^{- 1} (\frac{\sqrt{3}}{2})$

$= \frac{π}{6}$

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Similar questions

Q. The range of

f (x) = {sin}^{- 1} (\frac{x^{2} + 1}{x^{2} + 2})

Inverse circular functions,Principal values of ${sin}^{- 1} x, {c o s}^{- 1} x, {tan}^{- 1} x$ .
${tan}^{- 1} x + {tan}^{- 1} y = {tan}^{- 1} \frac{x + y}{1 - x y}$ , $x y < 1$
$π + {tan}^{- 1} \frac{x + y}{1 - x y}$ , $x y > 1$ .
Evaluate the following :
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(b) $sin [\frac{π}{2} - {sin}^{- 1} (- \frac{\sqrt{3}}{2})]$

Prove that:

sin [\frac{π}{6} + A] . cos [\frac{π}{3} - B] + sin [\frac{π}{3} - B] . cos [\frac{π}{6} + A] = cos (A - B)

Q. The value of

x

in the interval

[0, 2 π]

for which

4 {sin}^{2} x - 8 sin x + 3 \leq 0

Find the principal solution of $t a n^{2} x + (\sqrt{3} - 1) t a n x - \sqrt{3} = 0$

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The principal value of sin−1{cos(sin−1√32)} is

The correct option is A π6The principle value of sin−1(cos(sin−1(√32)))=π2−cos−1(cos(sin−1(√32)))=π2−sin−1(√32)=cos−1(√32)=π6

The principal value of ${sin}^{- 1} {cos ({sin}^{- 1} \frac{\sqrt{3}}{2})}$ is

The correct option is A $\frac{π}{6}$
The principle value of ${sin}^{- 1} (cos ({sin}^{- 1} (\frac{\sqrt{3}}{2})))$

$= \frac{π}{2} - {cos}^{- 1} (cos ({sin}^{- 1} (\frac{\sqrt{3}}{2})))$

$= \frac{π}{2} - {sin}^{- 1} (\frac{\sqrt{3}}{2})$

$= {cos}^{- 1} (\frac{\sqrt{3}}{2})$

$= \frac{π}{6}$