Advertisements
Advertisements
Question
Evaluate the following:
`cos^-1{cos (13pi)/6}`
Advertisements
Solution
We know
`cos^-1(costheta)=thetaif 0<=theta<=pi`
We have
`cos^-1{cos (13pi)/6}=cos^-1{cos(2pi+pi/6)}`
`= cos^-1{cos(pi/6)}`
`=pi/6`
APPEARS IN
RELATED QUESTIONS
Find the value of the following: `tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))],|x| <1,y>0 and xy <1`
Solve the following for x :
`tan^(-1)((x-2)/(x-3))+tan^(-1)((x+2)/(x+3))=pi/4,|x|<1`
If `tan^(-1)((x-2)/(x-4)) +tan^(-1)((x+2)/(x+4))=pi/4` ,find the value of x
`sin^-1(sin (7pi)/6)`
`sin^-1(sin (13pi)/7)`
`sin^-1{(sin - (17pi)/8)}`
`sin^-1(sin4)`
Evaluate the following:
`cos^-1(cos3)`
Evaluate the following:
`tan^-1(tan2)`
Evaluate the following:
`tan^-1(tan4)`
Evaluate the following:
`sec^-1(sec (25pi)/6)`
Evaluate the following:
`cosec^-1(cosec (3pi)/4)`
Evaluate the following:
`cot^-1(cot (19pi)/6)`
Write the following in the simplest form:
`tan^-1(x/(a+sqrt(a^2-x^2))),-a<x<a`
Evaluate the following:
`sin(tan^-1 24/7)`
Prove the following result-
`tan^-1 63/16 = sin^-1 5/13 + cos^-1 3/5`
Evaluate:
`cos(sec^-1x+\text(cosec)^-1x)`,|x|≥1
`tan^-1x+2cot^-1x=(2x)/3`
Prove the following result:
`sin^-1 12/13+cos^-1 4/5+tan^-1 63/16=pi`
Solve the following equation for x:
cot−1x − cot−1(x + 2) =`pi/12`, x > 0
If `sin^-1 (2a)/(1+a^2)-cos^-1 (1-b^2)/(1+b^2)=tan^-1 (2x)/(1-x^2)`, then prove that `x=(a-b)/(1+ab)`
Evaluate sin
\[\left( \frac{1}{2} \cos^{- 1} \frac{4}{5} \right)\]
Write the value of cos2 \[\left( \frac{1}{2} \cos^{- 1} \frac{3}{5} \right)\]
Write the value ofWrite the value of \[2 \sin^{- 1} \frac{1}{2} + \cos^{- 1} \left( - \frac{1}{2} \right)\]
Write the value of \[\tan^{- 1} \frac{a}{b} - \tan^{- 1} \left( \frac{a - b}{a + b} \right)\]
Write the value of \[\sin^{- 1} \left( \frac{1}{3} \right) - \cos^{- 1} \left( - \frac{1}{3} \right)\]
If \[\cos\left( \sin^{- 1} \frac{2}{5} + \cos^{- 1} x \right) = 0\], find the value of x.
The number of solutions of the equation \[\tan^{- 1} 2x + \tan^{- 1} 3x = \frac{\pi}{4}\] is
If 4 cos−1 x + sin−1 x = π, then the value of x is
In a ∆ ABC, if C is a right angle, then
\[\tan^{- 1} \left( \frac{a}{b + c} \right) + \tan^{- 1} \left( \frac{b}{c + a} \right) =\]
Prove that : \[\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \frac{x}{2}, 0 < x < \frac{\pi}{2}\] .
Find the domain of `sec^(-1)(3x-1)`.
Find the real solutions of the equation
`tan^-1 sqrt(x(x + 1)) + sin^-1 sqrt(x^2 + x + 1) = pi/2`
Find the simplified form of `cos^-1 (3/5 cosx + 4/5 sin x)`, where x ∈ `[(-3pi)/4, pi/4]`
Solve for x : {xcos(cot-1 x) + sin(cot-1 x)}2 = `51/50`
