Advertisements
Advertisements
Question
Write the value of sin−1 \[\left( \cos\frac{\pi}{9} \right)\]
Advertisements
Solution
Consider,
\[\sin^{- 1} \left( \cos\frac{\pi}{9} \right) = \sin^{- 1} \left\{ \sin\left( \frac{\pi}{2} - \frac{\pi}{9} \right) \right\} \left[ \because \cos x = \sin\left( \frac{\pi}{2} - x \right) \right]\]
\[ = \sin^{- 1} \left\{ \sin\left( \frac{7\pi}{18} \right) \right\}\]
\[ = \frac{7\pi}{18} \left[ \because \sin^{- 1} \left( \sin{x} \right) = x \right]\]
∴ \[\sin^{- 1} \left( \cos\frac{\pi}{9} \right) = \frac{7\pi}{18}\]
APPEARS IN
RELATED QUESTIONS
If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
Find the principal values of the following:
`cos^-1(-1/sqrt2)`
Evaluate the following:
`tan^-1(tan4)`
Evaluate the following:
`sec^-1(sec pi/3)`
Write the following in the simplest form:
`tan^-1sqrt((a-x)/(a+x)),-a<x<a`
Write the following in the simplest form:
`sin{2tan^-1sqrt((1-x)/(1+x))}`
Evaluate:
`cot{sec^-1(-13/5)}`
`sin(sin^-1 1/5+cos^-1x)=1`
`sin^-1x=pi/6+cos^-1x`
`tan^-1x+2cot^-1x=(2x)/3`
`5tan^-1x+3cot^-1x=2x`
Solve the following equation for x:
tan−1(x −1) + tan−1x tan−1(x + 1) = tan−13x
Solve `cos^-1sqrt3x+cos^-1x=pi/2`
Evaluate the following:
`sin(1/2cos^-1 4/5)`
`2sin^-1 3/5-tan^-1 17/31=pi/4`
`2tan^-1 1/5+tan^-1 1/8=tan^-1 4/7`
Solve the following equation for x:
`2tan^-1(sinx)=tan^-1(2sinx),x!=pi/2`
Solve the following equation for x:
`tan^-1((x-2)/(x-1))+tan^-1((x+2)/(x+1))=pi/4`
If x < 0, then write the value of cos−1 `((1-x^2)/(1+x^2))` in terms of tan−1 x.
Write the value of tan−1x + tan−1 `(1/x)`for x > 0.
Write the value of cos \[\left( 2 \sin^{- 1} \frac{1}{2} \right)\]
Write the value of `cot^-1(-x)` for all `x in R` in terms of `cot^-1(x)`
Wnte the value of\[\cos\left( \frac{\tan^{- 1} x + \cot^{- 1} x}{3} \right), \text{ when } x = - \frac{1}{\sqrt{3}}\]
If \[\cos\left( \sin^{- 1} \frac{2}{5} + \cos^{- 1} x \right) = 0\], find the value of x.
The number of real solutions of the equation \[\sqrt{1 + \cos 2x} = \sqrt{2} \sin^{- 1} (\sin x), - \pi \leq x \leq \pi\]
If x < 0, y < 0 such that xy = 1, then tan−1 x + tan−1 y equals
sin \[\left\{ 2 \cos^{- 1} \left( \frac{- 3}{5} \right) \right\}\] is equal to
If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is
If \[3\sin^{- 1} \left( \frac{2x}{1 + x^2} \right) - 4 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) + 2 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) = \frac{\pi}{3}\] is equal to
If tan−1 (cot θ) = 2 θ, then θ =
Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ; 1 < x < 1\].
Find the domain of `sec^(-1)(3x-1)`.
The equation sin-1 x – cos-1 x = cos-1 `(sqrt3/2)` has ____________.
