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Wize University Calculus 1 Textbook > Pre-Calculus (Review)

Inverse Trigonometric Functions

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Inverse Trigonometric Functions

Inverse Sine Function

f(x)=sin⁡x   ↔  f−1(x)=arcsin⁡x   or   sin⁡−1xf\left(x\right)=\sin x\ \ \ \leftrightarrow\ \ f^{-1}\left(x\right)=\arcsin x\ \ \ \text{or}\ \ \ \sin^{-1}x


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Inverse Cosine Function

f(x)=cos⁡x   ↔  f−1(x)=arccos⁡x   or   cos⁡−1xf\left(x\right)=\cos x\ \ \ \leftrightarrow\ \ f^{-1}\left(x\right)=\arccos x\ \ \ \text{or}\ \ \ \cos^{-1}x


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Inverse Tangent Function

f(x)tan⁡x   ↔  f−1(x)=arctan⁡x   or   tan⁡−1xf\left(x\right)\tan x\ \ \ \leftrightarrow\ \ f^{-1}\left(x\right)=\arctan x\ \ \ \text{or}\ \ \ \tan^{-1}x


Inverse Tangent has horizontal asymptotes at y=π2  and  y=−π2\displaystyle y=\frac{\pi}{2} \ \text{ and } \ y=\frac{-\pi}{2}


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Dealing with Inverse Trig Functions

If you are given a question with different trig functions like tan⁡(arcsin⁡x)=?\tan\left(\arcsin x\right)=?
  • let θ=trig−1(x)\theta=\text{trig}^{-1}\left(x\right)
  • then we have trig(θ)=x\text{trig} (\theta)=x
  • Draw a triangle based on this θ\theta
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Example: Trig & Inverse Trig Values

Determine the values of the following:

a.cos⁡(arccos⁡ (12))\cos\left(\arccos\ \left(\frac{1}{2}\right)\right)
The domain of arccos⁡\arccos is [−1,1]\left[-1,1\right].
You can use the cancellation rule directly since the argument is in [−1,1]\left[-1,1\right]
Therefore, this evaluates to 12\frac{1}{2}.


b. arcsin⁡(sin⁡(π6))\arcsin\left(\sin\left(\frac{\pi}{6}\right)\right)
The range of arcsin⁡\arcsin is [−π2, π2]\left[-\frac{\pi}{2},\ \frac{\pi}{2}\right].
You can use the cancellation rule directly since the argument is in [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
Therefore, this evaluates to π6\frac{\pi}{6}


c. arcsin⁡(sin⁡3π4)\arcsin\left(\sin\frac{3\pi}{4}\right)
The range of arcsin⁡\arcsin is [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
Cancellation rule does not work in this case because the argument is not in [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
We need to find an angle xx in [−π2, π2]\left[-\frac{\pi}{2},\ \frac{\pi}{2}\right]such that sin⁡x=sin⁡3π4\sin x=\sin\frac{3\pi}{4}.
Notice that sin⁡3π4\sin\frac{3\pi}{4} (quadrant 2) is positive, so the angle we're looking for is in quadrant 1.
Therefore, the angle xx we need is π4\frac{\pi}{4}

d. sin⁡(cos⁡−10)\sin\left(\cos^{-1}0\right).
*Recall that cos⁡−1=arccos⁡\cos^{-1}=\arccos
The domain of arccos⁡\arccos is [−1,1]\left[-1,1\right]. Notice that 0 is in that domain.
Let θ=arccos⁡0\theta=\arccos0,
Then cos⁡θ=0\cos\theta=0.
Since the range of arccos⁡\arccos is [0,π]\left[0,\pi\right], we get that θ=π2\theta=\frac{\pi}{2}.

Putting this back into the 𝑠𝑖𝑛 function, sin⁡(arccos⁡0)=sin⁡(θ)=sin⁡(π2)=1\sin\left(\arccos0\right)=\sin\left(\theta\right)=\sin\left(\frac{\pi}{2}\right)=1.
Find the exact value of tan⁡[arcsin⁡(−35)]\tan\left[\arcsin\left(-\frac{3}{5}\right)\right]




Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right)




Extra Practice
Inverse Trigonometric Functions
Compute the exact value of
arccos⁡(cos⁡(3π4))\arccos \left ( \cos \left ( \frac{3 \pi}{4} \right ) \right )
Find the exact value of arcsin⁡(sin⁡(−17π6))\arcsin(\sin(-\frac{17\pi}{6}))
Trig & Inverse Trig Values
Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right).
Trig & Inverse Trig Values
Find the exact value of tan⁡[arcsin⁡(−35)]\tan\left[\arcsin\left(-\frac{3}{5}\right)\right].
Practice: Trig & Inverse Trig Values (level 2)
Find the exact value of tan⁡[arcsin⁡(−35)]\tan\left[\arcsin\left(-\frac{3}{5}\right)\right].
Trig & Inverse Trig Values
Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right).
Trig & Inverse Trig Values
Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right).
Trig & Inverse Trig Values
Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right).
Inverse Trigonometric Functions
Which of the following results in a value that is positive:
Domain & Range with Inverse Trig
Find the domain and range of f(x)=1arcsin⁡xf\left(x\right)=\frac{1}{\arcsin x}.
Trig & Inverse Trig Values
Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right).
Trig & Inverse Trig Values
Find the exact value of tan⁡[arcsin⁡(−35)]\tan\left[\arcsin\left(-\frac{3}{5}\right)\right].
Inverse Trigonometric Functions
Evaluate tan⁡(tan⁡−1π4)+arcsin⁡(sin⁡5π4)\tan\left(\tan^{-1}\frac{\pi}{4}\right)+\arcsin\left(\sin\frac{5\pi}{4}\right).
Inverse Trigonometric Functions
Find the exact value of cos⁡[tan⁡−12+tan⁡−13]\cos[\tan^{-1}2+\tan^{-1}3].
Inverse Trig with Product
Find the exact value of sin⁡[2cos⁡−1(15)]\sin\bigg[2\cos^{-1}\bigg(\dfrac{1}{5}\bigg)\bigg].
Inverse Trigonometric Functions: Restrict the Domain
Find the exact value of arccos⁡(cos⁡23π3)\arccos\bigg(\cos\dfrac{23\pi}{3}\bigg) .
Inverse Trigonometric Functions
Simplify cos⁡(arctan⁡(4))\cos(\arctan(4)).
Exponential, Logarithmic and Trigonometric Functions
Which of the following numbers is the largest?
Inverse Trigonometric Functions
Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right).
Practice: Trig & Inverse Trig Values (level 3)
Practice: Trig & Inverse Trig Values
Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right).
Practice: Trig & Inverse Trig Values (level 2)
Find the exact value of tan⁡[arcsin⁡(−35)]\tan\left[\arcsin\left(-\frac{3}{5}\right)\right].
Practice: Trig & Inverse Trig Values (level 3)
Practice: Trig & Inverse Trig Values
Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right).
Practice: Trig & Inverse Trig Values (level 3)
Practice: Trig & Inverse Trig Values
Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right).
Inverse Trigonometric Functions
Without the using a calculator, compute the exact value of
arctan⁡(tan⁡(13π11))\arctan \left ( \tan \left ( \frac{13\pi}{11} \right ) \right )
*The π\pi symbol is obtained by typing in \pi
Inverse Trigonometric Functions
Answer the following Trig and Inverse Trig questions.
Inverse Trigonometric Functions
Without the use a calculator, compute the exact value of
arccos⁡(cos⁡(3π4))\arccos \left ( \cos \left ( \frac{3 \pi}{4} \right ) \right )
Additional Functions Practice Problems
Additional Practice Problems--Functions
1. Solve the following equations for x:
a. 9x−2=(127)2x+19^{x-2}=\left(\frac{1}{27}\right)^{2x+1}
Inverse Trigonometric Functions
Find the exact value of sin⁡(2cos⁡−1(x))\sin\left(2\cos^{-1}\left(x\right)\right).
Inverse Trigonometric Functions
Find the value of sin (arccos (− √3/2 )).
Inverse Trigonometric Functions
Find the exact value of tan⁡[arcsin⁡(−35)]\tan\left[\arcsin\left(-\frac{3}{5}\right)\right].
Inverse Trigonometric Functions
Simplify the following expression:
sec⁡2(arctan⁡(1/x2))\sec^2(\arctan(1/x^2))
Inverse Trigonometric Functions
Simplify
sin⁡(2arcsin⁡x)\sin(2\arcsin x)
Inverse Trigonometric Functions
Find the exact value of tan⁡[arcsin⁡(−3/5)]\tan[\arcsin(-3/5)]
Inverse Trigonometric Functions
Find the exact value of sin⁡(2cos⁡−1x)\sin(2\cos^{-1}x)
Inverse Trigonometric Functions
Find the domain and range of f(x)=1arcsin⁡xf(x) = \frac{1}{\arcsin x}
Additional Functions Practice Problems
Additional Practice Problems--Functions
1. Solve the following equations for x:
a. 9x−2=(127)2x+19^{x-2}=\left(\frac{1}{27}\right)^{2x+1}
Simplify the following expression: sin⁡[−tan⁡−1(x2−1)]\sin[-\tan^{-1}(\sqrt{x^2-1})]
Practice: Trigonometric Modeling
Q:\bf{Q:} The temperature of the body oscillates throughout the day and can be modeled by the function: T(t)=37 + 0.4cos⁡(π12 t−π3)T\left(t\right)=37\ +\ 0.4\cos\left(\dfrac{\pi}{12} \ t-\dfrac{\pi}{3}\right) where TT is the temperature in degrees Celsius, and tt is the time in hours after 12pm.
a) What is the maximum and minimum temperatures of the body throughout the day?
b) What is the period of this function? Does this make sense given the context of the problem?
Inverse Trigonometric Functions: Restrict the Domain
Find the exact value of arccos⁡(cos⁡23π3)\arccos\bigg(\cos\dfrac{23\pi}{3}\bigg) .
Inverse Trigonometric Functions
Find the exact value of cos⁡[tan⁡−12+tan⁡−13]\cos[\tan^{-1}2+\tan^{-1}3].
Inverse Trig with Product
Find the exact value of sin⁡[2cos⁡−1(15)]\sin\bigg[2\cos^{-1}\bigg(\dfrac{1}{5}\bigg)\bigg].
Inverse Trigonometric Functions
Simplify the expression cos⁡(tan⁡−1(x2))\cos\left(\tan^{-1}\left(\frac{x}{2}\right)\right)
Inverse Trigonometric Functions
Simplify the following expression;
sec⁡−1(2)=\sec^{-1}(2)=
Inverse Trigonometric Functions
Calculate the value of the expression sin⁡−1(sin⁡(5π3))\sin^{-1}\left(\sin\left(\frac{5\pi}{3}\right)\right)
Find the exact value of arcsin⁡(sin⁡(−17π6))\arcsin(\sin(-\frac{17\pi}{6}))
Inverse Trigonometric Functions
Find sin⁡(arctan⁡(x))\sin\left(\arctan\left(x\right)\right).
Practice: Trig & Inverse Trig Values (level 2)
Practice: Trig & Inverse Trig Values
Find the exact value of tan⁡[arcsin⁡(−35)]\tan\left[\arcsin\left(-\frac{3}{5}\right)\right].
Practice: Trig & Inverse Trig Values (level 3)
Practice: Trig & Inverse Trig Values
Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right).
Practice: Domain & Range with Inverse Trig
Practice: Domain & Range with Inverse Trig
Find the domain and range of f(x)=1arcsin⁡xf\left(x\right)=\frac{1}{\arcsin x}.
Find the value of sin⁡[arccos⁡(−32)]\sin\left[\arccos\left(-\frac{\sqrt{3}}{2}\right)\right] .
Inverse Trigonometric Functions
Find sin (arctan 𝑥).
Trig & Inverse Trig Values
Find the exact value of tan⁡[arcsin⁡(−35)]\tan\left[\arcsin\left(-\frac{3}{5}\right)\right].
Trig & Inverse Trig Values
Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right).
Domain & Range with Inverse Trig
Find the domain and range of f(x)=1arcsin⁡xf\left(x\right)=\frac{1}{\arcsin x}.
Inverse trigonometric functions
Evaluate the expression arcsin⁡(−12)×cos⁡(7π6)+2ln⁡(3e)\arcsin\left(-\frac{1}{\sqrt{2}}\right)\times\cos\left(\frac{7\pi}{6}\right)+2\ln\left(3\sqrt{e}\right)
Inverse Trigonometric Functions
Which of the following results in a value that is positive:
Inverse Trigonometric Functions
Evaluate tan⁡(tan⁡−1π4)+arcsin⁡(sin⁡5π4)\tan\left(\tan^{-1}\frac{\pi}{4}\right)+\arcsin\left(\sin\frac{5\pi}{4}\right).
Practice: Trig & Inverse Trig Values (level 2)
Find the exact value of tan⁡[arcsin⁡(−35)]\tan\left[\arcsin\left(-\frac{3}{5}\right)\right].