Name: Multiple Angle Formula for Sine Inverse Function
Uploaded: 2019-07-19
Duration: 6 min 56 s
Description: What is twice of sine inverse function? Can we develop a formula for it? How to solve the problem using this function? Check out this video to know more.

Question

Find

\sin^{- 1} (\sin θ), \cos^{- 1} (\cos θ), \tan^{- 1} (\tan θ), \cot^{- 1} (\cot θ)

for

θ \in (\frac{5 π}{2}, 3 π)

Accepted Answer

Given, $\frac{5 π}{2} < θ < 3 π$

Now, $- \frac{5 π}{2} > - θ > - 3 π \Rightarrow \frac{π}{2} > 3 π - θ > 0$

$\Rightarrow 0 < 3 π - θ < \frac{π}{2}$

$\sin^{- 1} \sin θ = \sin^{- 1} (\sin (3 π - θ)) (∵ \sin (3 π - θ) = \sin θ)$

$= 3 π - θ (∵ \sin^{- 1} \sin x = x \forall x \in [- \frac{π}{2}, \frac{π}{2}])$

Also, $\frac{5 π}{2} < θ < 3 π \Rightarrow \frac{π}{2} < θ - 2 π < π$

$\cos^{- 1} \cos (θ) = \cos^{- 1} (\cos (θ - 2 π)) [∵ \cos (θ - 2 π) = \cos θ]$

$= θ - 2 π [∵ \cos^{- 1} \cos x = x \forall x \in [0, π]]$

Now, $\frac{5 π}{2} < θ < 3 π \Rightarrow - \frac{π}{2} < θ - 3 π < 0$

$\tan^{- 1} \tan θ = \tan^{- 1} (\tan (θ - 3 π)) [∵ \tan (θ - 3 π) = - \tan (3 π - θ) = \tan θ] = θ - 3 π [\tan^{- 1} \tan x = x \forall x \in (\frac{- π}{2}, \frac{π}{2})]$

$\cot^{- 1} \cot θ = \cot^{- 1} (\cot (θ - 2 π)) [∵ \cot (θ - 2 π) = - \cot (2 π - θ) = \cot θ]$

$= θ - 2 π [∵ \cot^{- 1} \cot x = x \forall x \in [0, π]]$

Find $\sin^{- 1} (\sin θ), \cos^{- 1} (\cos θ), \tan^{- 1} (\tan θ), \cot^{- 1} (\cot θ)$ for $θ \in (\frac{5 π}{2}, 3 π)$

Important Questions on Inverse Trigonometric Functions

Learn and Prepare for any exam you want !

Find sin-1 sin θ, cos-1 cos θ, tan-1 tan θ, cot-1 cot θ for θ∈5π2, 3π

Important Questions on Inverse Trigonometric Functions

Learn and Prepare for any exam you want !

Find $\sin^{- 1} (\sin θ), \cos^{- 1} (\cos θ), \tan^{- 1} (\tan θ), \cot^{- 1} (\cot θ)$ for $θ \in (\frac{5 π}{2}, 3 π)$