To the right is the graphof an angle 0 on the unit circle, whereT < 0 <2Find simplified formulas(meaning: if your answer has trig func-tions in it, it's not simplified enough) forcos-1(cos 0) and sin-(sin 0) in terms of 0(justify your answers), and draw both an-gles on the unit circle.

To Find: i. cos-1cosθii. sin-1sinθ, whereπ<θ<3π2 Concept: a. sin-1sin x=x, ∀x∈-π2,π2b.…

Answered: To the right is the graph of an angle 0…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter6: The Trigonometric Functions

Section6.3: Trigonometric Functions Of Real Numbers

Problem 30E

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Related questions

Question

To the right is the graph
of an angle 0 on the unit circle, where
T < 0 <
2
Find simplified formulas
(meaning: if your answer has trig func-
tions in it, it's not simplified enough) for
cos-1(cos 0) and sin-(sin 0) in terms of 0
(justify your answers), and draw both an-
gles on the unit circle.

Expert Solution

Step 1

To Find:

$(i) . \cos^{- 1} (\cos θ) (i i) . \sin^{- 1} (\sin θ), where π < θ < \frac{3 π}{2}$

Concept:

$(a) . \sin^{- 1} (\sin x) = x, \forall x \in [- \frac{π}{2}, \frac{π}{2}] (b) . \cos^{- 1} (\cos x) = x, \forall x \in [0, π] (c) . \sin^{- 1} (\sin x) = \sin^{- 1} (\sin (π - x)) (d) . \cos^{- 1} (\cos x) = \cos^{- 1} (\cos (2 π - x)) = 2 π - x, \forall x \in [π, 2 π]$

Step 2

Explanation:

$We have π < θ < \frac{3 π}{2} (a) . we have to find \cos^{- 1} (\cos θ) as we know \cos^{- 1} (\cos θ) = θ, \forall θ \in [0, π] but here, (π, \frac{3 π}{2}) \notin [0, π] now, since π < θ < \frac{3 π}{2} \Rightarrow - π > - θ > - \frac{3 π}{2} or - \frac{3 π}{2} < - θ < - π \Rightarrow 2 π - \frac{3 π}{2} < 2 π - θ < 2 π - π \Rightarrow \frac{π}{2} < 2 π - θ < π \Rightarrow \cos^{- 1} \cos (\frac{π}{2}) < \cos^{- 1} \cos (2π-θ) < \cos^{- 1} \cos (π) \Rightarrow \frac{π}{2} < \cos^{- 1} (\cos θ) < π [∵ \cos^{- 1} (\cos θ) = \cos^{- 1} (\cos 2π-θ) = 2 π - θ, \forall θ \in [π, 2 π]]$

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