Advertisements
Advertisements
प्रश्न
Write the principal value of `sin^-1(-1/2)`
Advertisements
उत्तर
Let `y=sin^-1(-1/2)`
Then,
\[\sin{y} = - \frac{1}{2} = \sin\left( - \frac{\pi}{6} \right)\]
\[y = - \frac{\pi}{6} \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\]
Here,
\[\left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\] is the range of the principal value branch of the inverse sine function.
∴ \[\sin^{- 1} \left( - \frac{1}{2} \right) = - \frac{\pi}{6}\]
APPEARS IN
संबंधित प्रश्न
Show that:
`2 sin^-1 (3/5)-tan^-1 (17/31)=pi/4`
If `(sin^-1x)^2 + (sin^-1y)^2+(sin^-1z)^2=3/4pi^2,` find the value of x2 + y2 + z2
Find the domain of `f(x) =2cos^-1 2x+sin^-1x.`
Find the principal values of the following:
`cos^-1(tan (3pi)/4)`
`sin^-1(sin (7pi)/6)`
`sin^-1(sin4)`
`sin^-1(sin2)`
Evaluate the following:
`tan^-1(tan12)`
Evaluate the following:
`cot^-1(cot (9pi)/4)`
Write the following in the simplest form:
`tan^-1{(sqrt(1+x^2)+1)/x},x !=0`
Write the following in the simplest form:
`sin^-1{(sqrt(1+x)+sqrt(1-x))/2},0<x<1`
Write the following in the simplest form:
`sin{2tan^-1sqrt((1-x)/(1+x))}`
Solve: `cos(sin^-1x)=1/6`
`sin(sin^-1 1/5+cos^-1x)=1`
Solve the following equation for x:
tan−1(x −1) + tan−1x tan−1(x + 1) = tan−13x
Solve the following equation for x:
cot−1x − cot−1(x + 2) =`pi/12`, x > 0
Sum the following series:
`tan^-1 1/3+tan^-1 2/9+tan^-1 4/33+...+tan^-1 (2^(n-1))/(1+2^(2n-1))`
`sin^-1 63/65=sin^-1 5/13+cos^-1 3/5`
Evaluate the following:
`sin(1/2cos^-1 4/5)`
Solve the following equation for x:
`tan^-1((x-2)/(x-1))+tan^-1((x+2)/(x+1))=pi/4`
What is the value of cos−1 `(cos (2x)/3)+sin^-1(sin (2x)/3)?`
Write the value of cos−1 (cos 1540°).
Write the value of sin−1 (sin 1550°).
If tan−1 x + tan−1 y = `pi/4`, then write the value of x + y + xy.
What is the principal value of `sin^-1(-sqrt3/2)?`
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 \[\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha, then\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = \]
Let f (x) = `e^(cos^-1){sin(x+pi/3}.`
Then, f (8π/9) =
If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is
\[\cot\left( \frac{\pi}{4} - 2 \cot^{- 1} 3 \right) =\]
If \[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right),\text{ where }a, x \in \left( 0, 1 \right)\] , then, the value of x is
The value of \[\tan\left( \cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4} \right)\]
Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ; 1 < x < 1\].
If 2 tan−1 (cos θ) = tan−1 (2 cosec θ), (θ ≠ 0), then find the value of θ.
The period of the function f(x) = tan3x is ____________.
