Advertisements
Advertisements
Question
`sin^-1(sin12)`
Solution
We know
`sin(sin^-1theta)=theta if - pi/2<=theta<=pi/2`
We have
= `sin^-1sin36°`
= `- sin^-1sin(4pi-12)`
= `-(4pi-12)`
= `-4pi+12`
= `12-4pi`
APPEARS IN
RELATED QUESTIONS
Prove that
`tan^(-1) [(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]=pi/4-1/2 cos^(-1)x,-1/sqrt2<=x<=1`
`sin^-1(sin3)`
Evaluate the following:
`cos^-1{cos (5pi)/4}`
Evaluate the following:
`cos^-1(cos12)`
Evaluate the following:
`tan^-1(tan (6pi)/7)`
Evaluate the following:
`sec^-1{sec (-(7pi)/3)}`
Evaluate the following:
`\text(cosec)^-1(\text{cosec} pi/4)`
Evaluate the following:
`cosec^-1(cosec (13pi)/6)`
Evaluate the following:
`cosec^-1{cosec (-(9pi)/4)}`
Evaluate the following:
`cot^-1{cot ((21pi)/4)}`
Write the following in the simplest form:
`tan^-1{x+sqrt(1+x^2)},x in R `
Write the following in the simplest form:
`tan^-1{(sqrt(1+x^2)-1)/x},x !=0`
Evaluate the following:
`sin(tan^-1 24/7)`
Evaluate the following:
`tan(cos^-1 8/17)`
Prove the following result
`sin(cos^-1 3/5+sin^-1 5/13)=63/65`
Evaluate: `sin{cos^-1(-3/5)+cot^-1(-5/12)}`
If `cos^-1x + cos^-1y =pi/4,` find the value of `sin^-1x+sin^-1y`
If `cot(cos^-1 3/5+sin^-1x)=0`, find the values of x.
Solve the following:
`cos^-1x+sin^-1 x/2=π/6`
Solve `cos^-1sqrt3x+cos^-1x=pi/2`
`tan^-1 1/4+tan^-1 2/9=1/2cos^-1 3/2=1/2sin^-1(4/5)`
`tan^-1 1/7+2tan^-1 1/3=pi/4`
Solve the following equation for x:
`tan^-1 1/4+2tan^-1 1/5+tan^-1 1/6+tan^-1 1/x=pi/4`
Write the value of cos−1 (cos 6).
Evaluate: \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]
If \[\sin^{- 1} \left( \frac{1}{3} \right) + \cos^{- 1} x = \frac{\pi}{2},\] then find x.
What is the principal value of `sin^-1(-sqrt3/2)?`
Write the principal value of \[\cos^{- 1} \left( \cos680^\circ \right)\]
If \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right)\] = α, then x2 =
If α = \[\tan^{- 1} \left( \tan\frac{5\pi}{4} \right) \text{ and }\beta = \tan^{- 1} \left( - \tan\frac{2\pi}{3} \right)\] , then
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
If tan−1 3 + tan−1 x = tan−1 8, then x =
If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is
In a ∆ ABC, if C is a right angle, then
\[\tan^{- 1} \left( \frac{a}{b + c} \right) + \tan^{- 1} \left( \frac{b}{c + a} \right) =\]
If y = sin (sin x), prove that \[\frac{d^2 y}{d x^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0 .\]
If 2 tan−1 (cos θ) = tan−1 (2 cosec θ), (θ ≠ 0), then find the value of θ.
Find the simplified form of `cos^-1 (3/5 cosx + 4/5 sin x)`, where x ∈ `[(-3pi)/4, pi/4]`
The period of the function f(x) = tan3x is ____________.
Solve for x : {xcos(cot-1 x) + sin(cot-1 x)}2 = `51/50`
