This page covers inverse trigonometric functions specifically. Inverses of functions in general are discussed on the inverse functions page.
Topics You Need To Understand For This Page 

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Plot 1  \(\sin(x)\) 

As mentioned on the inverse functions page, the trig functions do not pass the horizontal line test and, therefore, do not, in general, have inverses. However, in order to get inverses, we restrict the domain. This allows us to talk about inverses when we have trig functions.
For example, if we have the trig function \(f(x)=\sin x\), we can restrict the domain and end up with a onetoone function. This is shown in plot 1. The blue line is \(y=\sin x\). If we restrict the domain to \(\pi/2 \leq x \leq \pi/2\), we end up with the red part of \(y=\sin x\). This part passes the horizontal line test and, therefore, has an inverse.
Here is a great video that goes into detail for several trig functions like we did above with \(y=\sin x\).
video by PatrickJMT 

Working with Inverse Trig Functions
Inverse trig functions are used so much in calculus that we have special notation to write them by putting arc in front.
function 
inverse  

\(\sin(x)\) 
\(\sin^{1}(x) = \) \(\arcsin(x)\)  
\(\cos(x)\) 
\(\cos^{1}(x) = \) \(\arccos(x)\) 
The other trig functions have similar notation. Of course, we assume that the domains of these functions are restricted so that they pass the horizontal line test, which allows the inverse trig function to make sense. We do not state it and most instructors and textbooks do not either. So you just need to remember it.
Since \(\arcsin(x)\) is the inverse, we know that \(\sin(\arcsin(x)) = \arcsin(\sin(x)) = x\). This works not just with x but with any expression. For example, \(\sin(\arcsin(72k)) = \arcsin(\sin(72k)) = 72k\) and \(\sin(\arcsin(17)) = \arcsin(\sin(17)) = 17\).
Here is a good video showing some examples working with inverse trig functions.
video by PatrickJMT 

One type of problem that you will probably be asked to solve is to evaluate an expression like \(\sin(\arccos(x))\). Notice that we have an inverse cosine inside of a sine, so you can't directly write x. To solve an expression like this, we use the technique of substitution. See the practice problems for explanation on how to do this.
Practice
Unless otherwise instructed, evaluate (simplify) the given function without using a calculator and give your answers in radians in exact, simplified form.
\( \sin( \arccos( 5 / 13 ) ) \)
Problem Statement 

Evaluate (simplify) \( \sin( \arccos( 5 / 13 ) ) \) without using a calculator and give your answer in radians in exact, simplified form.
Final Answer 

\( \sin( \arccos( 5 / 13 ) ) \) \( = 12 / 13\)
Problem Statement
Evaluate (simplify) \( \sin( \arccos( 5 / 13 ) ) \) without using a calculator and give your answer in radians in exact, simplified form.
Solution
We start on the inside by letting \( \theta = \arccos(5 / 13) \). Taking the cosine of each side, we get \( \cos(\theta) = 5 / 13\).
From this equation, we set up the triangle on the right, which means \( b = 5 \) and \( c = 13 \).
Using the Pythagorean Theorem, we know that \( c^2 = a^2 + b^2 \) or \( 13^2 = a^2 + 5^2 \). Solving for \(a\) we get \( a^2 = 13^2  5^2 = 169  25 = 144 \) or \( a = \pm 12 \). We choose \( a = +12 \).
Now that we have a triangle with the length of all sides known, we can write \( \sin( \arccos(5 / 13)) = \sin(\theta) \) and use the triangle to get \( \sin(\theta) = 12 / 13 \).
Final Answer
\( \sin( \arccos( 5 / 13 ) ) \) \( = 12 / 13\)
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\( \sin^{1}(1/2) \)
Problem Statement
Evaluate \( \sin^{1}(1/2) \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \sin^{1}(\sqrt{3}/2) \)
Problem Statement
Evaluate \( \sin^{1}(\sqrt{3}/2) \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \sin^{1}(1/2) \)
Problem Statement
Evaluate \( \sin^{1}(1/2) \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \sin^{1}(\sqrt{2}/2) \)
Problem Statement
Evaluate \( \sin^{1}(\sqrt{2}/2) \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \cos^{1}(1/2) \)
Problem Statement
Evaluate \( \cos^{1}(1/2) \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \cos^{1}(\sqrt{3}/2) \)
Problem Statement
Evaluate \( \cos^{1}(\sqrt{3}/2) \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \cos^{1}(\sqrt{2}/2) \)
Problem Statement
Evaluate \( \cos^{1}(\sqrt{2}/2) \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \tan^{1}(0) \)
Problem Statement
Evaluate \( \tan^{1}(0) \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \tan^{1}(1) \)
Problem Statement
Evaluate \( \tan^{1}(1) \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \tan^{1}(1) \)
Problem Statement
Evaluate \( \tan^{1}(1) \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \tan^{1}(\sqrt{3}) \)
Problem Statement
Evaluate \( \tan^{1}(\sqrt{3}) \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \tan^{1}(\sqrt{3}/3) \)
Problem Statement
Evaluate \( \tan^{1}(\sqrt{3}/3) \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \sin^{1}[\cos(\pi/3)] \)
Problem Statement
Evaluate \( \sin^{1}[\cos(\pi/3)] \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \tan^{1}[\sin(\pi/2)] \)
Problem Statement
Evaluate \( \tan^{1}[\sin(\pi/2)] \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \cos^{1}[\tan \pi] \)
Problem Statement
Evaluate \( \cos^{1}[\tan \pi] \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \sin[\cos^{1}(3/5)] \)
Problem Statement
Evaluate \( \sin[\cos^{1}(3/5)] \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \cos[\tan^{1}(8/15)] \)
Problem Statement
Evaluate \( \cos[\tan^{1}(8/15)] \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \tan[\sin^{1}(5/13)] \)
Problem Statement
Evaluate \( \tan[\sin^{1}(5/13)] \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \sec[\sin^{1}(1/4)] \)
Problem Statement
Evaluate \( \sec[\sin^{1}(1/4)] \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\( \csc[\cos^{1}(7/9)] \)
Problem Statement
Evaluate \( \csc[\cos^{1}(7/9)] \) without using a calculator, giving your answer in exact terms in radians.
Solution
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\(\cos(\arcsin(1/2))\)
Problem Statement
Evaluate (simplify) \(\cos(\arcsin(1/2))\) without using a calculator and give your answer in exact, simplified form.
Solution
video by PatrickJMT 

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\(\tan(\arcsin(2/3))\)
Problem Statement
Evaluate (simplify) \(\tan(\arcsin(2/3))\) without using a calculator and give your answer in exact, simplified form.
Solution
video by PatrickJMT 

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\(\sin(2\arctan(\sqrt{2}))\)
Problem Statement
Evaluate (simplify) \(\sin(2\arctan(\sqrt{2}))\) without using a calculator and give your answer in exact, simplified form.
Solution
video by PatrickJMT 

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\( \cos( \arcsin(x))\)
Problem Statement 

Evaluate (simplify) \( \cos( \arcsin(x))\) without using a calculator and give your answer in exact, simplified form.
Final Answer 

\( \cos( \arcsin(x))\) \( = \sqrt{1x^2} \)
Problem Statement
Evaluate (simplify) \( \cos( \arcsin(x))\) without using a calculator and give your answer in exact, simplified form.
Solution
Starting on the inside, we have \( \theta = \arcsin(x) \). Taking the sine of both sides yields \( \sin(\theta) = x \). We use this equation to set up the triangle to the right, \( a = x \) and \( c = 1 \).
Now we use the Pythagorean Theorem to solve for \(b\).
\(\begin{array}{rcl}
x^2 + b^2 & = & 1 \\
b^2 & = & 1  x^2 \\
b & = & \pm \sqrt{1x^2}
\end{array} \)
We choose the positive square root.
Now we can write \( \cos(\theta) = \sqrt{1x^2} \).
Final Answer
\( \cos( \arcsin(x))\) \( = \sqrt{1x^2} \)
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\( \sec( \arctan(x/3))\)
Problem Statement 

Evaluate (simplify) \( \sec( \arctan(x/3))\) without using a calculator and give your answer in exact, simplified form.
Final Answer 

\( \sec( \arctan(x/3))\) \(\displaystyle{ = \frac{\sqrt{x^2+9}}{3}}\)
Problem Statement
Evaluate (simplify) \( \sec( \arctan(x/3))\) without using a calculator and give your answer in exact, simplified form.
Solution
We start on the inside by letting \( \theta = \arctan(x/3) \). Take the tangent of both sides to get \( \tan(\theta) = x/3 \). We use this equation to set up the triangle to the right, \( a = x, ~~ b = 3 \). Using the Pythagorean Theorem, we can find \(c \), i.e. \( c^2 = x^2 + 3^2 ~~ \to ~~ c = \pm \sqrt{x^2 + 9} \).
We choose the positive square root.
Now we have \( \displaystyle{ \sec(\theta) = \frac{\sqrt{x^2+9}}{3} }\)
Final Answer
\( \sec( \arctan(x/3))\) \(\displaystyle{ = \frac{\sqrt{x^2+9}}{3}}\)
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\( \sin[\cos^{1}(x)] \)
Problem Statement
Evaluate \( \sin[\cos^{1}(x)] \) without using a calculator, giving your answer in terms of \(x\).
Solution
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\( \tan[\sin^{1}(2x)] \)
Problem Statement
Evaluate \( \tan[\sin^{1}(2x)] \) without using a calculator, giving your answer in terms of \(x\).
Solution
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\( \cos[\tan^{1}(1/x)] \)
Problem Statement
Evaluate \( \cos[\tan^{1}(1/x)] \) without using a calculator, giving your answer in terms of \(x\).
Solution
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\( \sin[\arctan(u/3)] \)
Problem Statement
Evaluate \( \sin[\arctan(u/3)] \) expressing your answer in terms of \(u\).
Solution
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\(\displaystyle{ \csc \left[ \cos^{1} \left( \frac{x}{\sqrt{x^2+9}} \right) \right] }\)
Problem Statement
Evaluate \(\displaystyle{ \csc \left[ \cos^{1} \left( \frac{x}{\sqrt{x^2+9}} \right) \right] }\) without using a calculator, giving your answer in terms of \(x\).
Solution
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\(\tan(\arcsin(x))\)
Problem Statement
Evaluate (simplify) \(\tan(\arcsin(x))\) without using a calculator and give your answer in exact, simplified form.
Solution
video by PatrickJMT 

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\(\csc(\arctan(2x))\)
Problem Statement
Evaluate (simplify) \(\csc(\arctan(2x))\) without using a calculator and give your answer in exact, simplified form.
Solution
video by PatrickJMT 

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