The value of cos -1 - 1 2 -sin -1 - 3 2 is-

The correct option is D π3 As for nbsp;cos ^ 1 x x∈[ 0,π nbsp;] and nbsp;sin ^ 1 x x∈[ π nbsp;/ 2 ,π nbsp;/ 2 nbsp;] Hence ...

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Standard XII

Mathematics

First Fundamental Theorem of Calculus

The value of ...

Question

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

\frac{π}{3}

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\frac{2 π}{3}

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Solution

The correct option is D $\frac{π}{3}$
As for ${cos}^{- 1} x x \in [0, π]$ and ${sin}^{- 1} x x \in [- \frac{π}{2}, \frac{π}{2}]$
Hence
${cos}^{- 1} (- \frac{1}{2}) + {sin}^{- 1} (- \frac{\sqrt{3}}{2}) = \frac{2 π}{3} + (- \frac{π}{3}) = \frac{π}{3}$

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Similar questions

{sin}^{- 1} x - {sin}^{- 1} y = \frac{2 π}{3}; {cos}^{- 1} x - {cos}^{- 1} = \frac{π}{3}

Q. Find the value of

cos [\frac{π}{2} - {sin}^{- 1} (\frac{1}{3})]

Q. Assertion :

3 s i n^{- 1} (\frac{1}{3}) + s i n^{- 1} (\frac{3}{5}) < 2 π / 3

and

t a n^{- 1} (2 \sqrt{2} - 1) > π / 3

Reason: If

3 s i n^{- 1} x = π / 6,

then

6 x - 8 x^{3} - 1 = 0

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 :
(a) $sin [\frac{π}{3} - {sin}^{- 1} (- \frac{1}{2})]$
(b) $sin [\frac{π}{2} - {sin}^{- 1} (- \frac{\sqrt{3}}{2})]$

Q. What is the value of cos⁻¹

(\cos \frac{2 π}{3}) + \sin^{- 1} (\sin \frac{2 π}{3})

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The value of cos−1(−12)+sin−1(−√32) is-

The correct option is D π3As for cos−1xx∈[0,π] and sin−1xx∈[−π2,π2]Hencecos−1(−12)+sin−1(−√32)=2π3+(−π3)=π3

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

The correct option is D $\frac{π}{3}$
As for ${cos}^{- 1} x x \in [0, π]$ and ${sin}^{- 1} x x \in [- \frac{π}{2}, \frac{π}{2}]$
Hence
${cos}^{- 1} (- \frac{1}{2}) + {sin}^{- 1} (- \frac{\sqrt{3}}{2}) = \frac{2 π}{3} + (- \frac{π}{3}) = \frac{π}{3}$