Inverse Trigonometric Functions
Solved Problems
Example 1.
Find the domain of the function \[y = \arcsin \left( {x - 1} \right).\]
Solution.
We know that the inverse sine function is defined in the closed interval \(\left[ { - 1,1} \right].\) Then we have the following inequality:
\[ - 1 \le x - 1 \le 1.\]
Adding \(1\) to all parts of the double inequality yields:
\[ - 1 \le x - 1 \le 1, \Rightarrow - 1 + 1 \le x - 1 + 1 \le 1 + 1, \Rightarrow 0 \le x \le 2.\]
Hence, the domain of the function is \(x \in \left[ {0,2} \right].\)
Example 2.
Find the domain of the function \[y = \arccos \frac{{1 - x}}{2}.\]
Solution.
The inverse cosine function is defined on the interval \(\left[ { - 1,1} \right].\) So we can write:
\[ - 1 \le \frac{{1 - x}}{2} \le 1.\]
Multiply all parts of the inequality by \(2:\)
\[ - 2 \le 1 - x \le 2.\]
Substract \(1:\)
\[ - 3 \le - x \le 1.\]
Now we multiply all sides by \(-1.\) Remember that multiplying an inequality by a negative number reverses this inequality. Therefore
\[ - 3 \le - x \le 1, \Rightarrow - 1 \le x \le 3.\]
Hence, the domain of the function is \(x \in \left[ { - 1,3} \right].\)
Example 3.
Find the range of the function \[y = 2\,{\arctan ^2}\frac{x}{2}.\]
Solution.
The range of the inverse tangent function \(y = \arctan x\) is the open interval \(\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right).\)
The function \(\arctan \frac{x}{2}\) has the same range \(\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right).\)
When squared, the interval \(\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\) is transformed into \(\left[ {0,\frac{{{\pi ^2}}}{4}} \right).\)
Multiplying by \(2\) we get the following answer:
\[\text{codom}\left( y \right) = \left[ {0,\frac{{{\pi ^2}}}{2}} \right).\]
Example 4.
Find the range of the function \[y = \frac{1}{2}{\text{arccot}^3}x.\]
Solution.
The basic function \(y = \text{arccot }x\) has the range \(\left( {0,\pi } \right).\)
For the cubic function \(y = {\text{arccot}^3}x,\) we get the range \(\left( {0,\pi^3 } \right).\)
Hence, the range of the function \(y = \frac{1}{2}{\text{arccot}^3}x\) is given by
\[\text{codom}\left( y \right) = \left({0,\frac{{{\pi ^3}}}{2}} \right).\]
Example 5.
Evaluate \[{\sin ^2}\left( {\arccos \frac{4}{5}} \right).\]
Solution.
By the Pythagorean theorem,
\[{\sin ^2}\left( {\arccos \frac{4}{5}} \right) = 1 - {\cos ^2}\left( {\arccos \frac{4}{5}} \right) = 1 - {\left( {\cos \left( {\arccos \frac{4}{5}} \right)} \right)^2} = 1 - {\left( {\frac{4}{5}} \right)^2} = 1 - \frac{{16}}{{25}} = \frac{9}{{25}}.\]
Example 6.
Evaluate \[\sin \left( {\arccos \frac{3}{5}} \right).\]
Solution.
Let \(\alpha = \arccos \frac{3}{5}.\) By the Pythagorean identity,
\[{\sin ^2}\alpha + {\cos ^2}\alpha = 1.\]
Hence we can write
\[\sin \alpha = \pm\sqrt {1 - {{\cos }^2}\alpha } = \pm\sqrt {1 - {{\cos }^2}\left( {\arccos \frac{3}{5}} \right)} = \pm\sqrt {1 - {{\left( {\cos \left( {\arccos \frac{3}{5}} \right)} \right)}^2}} = \pm\sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = \pm\sqrt {1 - \frac{9}{{25}}} = \pm\sqrt {\frac{{16}}{{25}}} = \pm\frac{4}{5}.\]
Note that the angle \(\alpha = \arccos \frac{3}{5}\) is in the \(1\text{st}\) quadrant where the sine has positive sign, so the answer is
\[\sin \left( {\arccos \frac{3}{5}} \right) = \frac{4}{5}.\]
Example 7.
Evaluate \[\sin \left( {\frac{1}{2}\arccos \frac{3}{5}} \right).\]
Solution.
We know that
\[\sin \frac{\alpha }{2} = \pm \sqrt {\frac{{1 - \cos \alpha }}{2}} .\]
The angle \({\frac{1}{2}\arccos \frac{3}{5}}\) lies in the \(1\text{st}\) quadrant where the sine is positive. Therefore we take the positive sign on the square root. This gives
\[\sin \left( {\frac{1}{2}\arccos \frac{3}{5}} \right) = \sqrt {\frac{{1 - \cos \left( {\arccos \frac{3}{5}} \right)}}{2}} = \sqrt {\frac{{1 - \frac{3}{5}}}{2}} = \sqrt {\frac{{\frac{2}{5}}}{2}} = \frac{1}{{\sqrt 5 }}.\]
Example 8.
Evaluate \[\cot \left( {\text{arcsec}\left( { - 2} \right)} \right).\]
Solution.
It follows from the Pythagorean identity that
\[{\tan ^2}\alpha = {\sec ^2}\alpha - 1.\]
Let \(\alpha = \text{arcsec} \left( { - 2} \right).\) The angle \(\alpha\) lies in the \(2\text{nd}\) quadrant where \(\tan\alpha \lt 0.\) Then we have
\[\tan \alpha = - \sqrt {{{\sec }^2}\alpha - 1} .\]
The cotangent is the reciprocal function of tangent. Hence
\[\cot \alpha = \frac{1}{{\tan \alpha }} = - \frac{1}{{\sqrt {{{\sec }^2}\alpha - 1} }} = - \frac{1}{{\sqrt {{{\sec }^2}\left( {\text{arcsec} \left( { - 2} \right)} \right) - 1} }} = - \frac{1}{{\sqrt {{{\left( { - 2} \right)}^2} - 1} }} = - \frac{1}{{\sqrt {4 - 1} }} = - \frac{1}{{\sqrt 3 }}.\]
The answer is
\[\cot \left( {\text{arcsec} \left( { - 2} \right)} \right) = - \frac{1}{{\sqrt 3 }}.\]
Example 9.
Evaluate \[A = \frac{2}{\pi }\left( {\arcsin \frac{3}{4} + \arccos \frac{3}{4}} \right).\]
Solution.
We denote
\[z = \arcsin \frac{3}{4} + \arccos \frac{3}{4}.\]
Using the sine addition formula, we have
\[\sin z = \sin \left( {\arcsin \frac{3}{4} + \arccos \frac{3}{4}} \right) = \sin \left( {\arcsin \frac{3}{4}} \right)\cos \left( {\arccos \frac{3}{4}} \right) + \cos \left( {\arcsin \frac{3}{4}} \right)\sin \left( {\arccos \frac{3}{4}} \right).\]
Calculate all terms separately:
\[\sin \left( {\arcsin \frac{3}{4}} \right) = \frac{3}{4};\]
\[\cos \left( {\arccos \frac{3}{4}} \right) = \frac{3}{4};\]
\[\cos \left( {\arcsin \frac{3}{4}} \right) = \sqrt {1 - {{\sin }^2}\left( {\arcsin \frac{3}{4}} \right)} = \sqrt {1 - {{\left( {\frac{3}{4}} \right)}^2}} = \sqrt {1 - \frac{9}{{16}}} = \sqrt {\frac{7}{{16}}} = \frac{{\sqrt 7 }}{4};\]
\[\sin \left( {\arccos \frac{3}{4}} \right) = \sqrt {1 - {{\cos }^2}\left( {\arccos \frac{3}{4}} \right)} = \sqrt {1 - {{\left( {\frac{3}{4}} \right)}^2}} = \sqrt {1 - \frac{9}{{16}}} = \sqrt {\frac{7}{{16}}} = \frac{{\sqrt 7 }}{4}.\]
Then
\[\sin z = \frac{3}{4} \cdot \frac{3}{4} + \frac{{\sqrt 7 }}{4} \cdot \frac{{\sqrt 7 }}{4} = \frac{9}{{16}} + \frac{7}{{16}} = \frac{{16}}{{16}} = 1.\]
Find the original expression \(A:\)
\[\sin z = 1, \Rightarrow z = \frac{\pi }{2}, \Rightarrow A = \frac{2}{\pi }z = \frac{2}{\pi } \cdot \frac{\pi }{2} = 1.\]
Example 10.
Evaluate \[B = \frac{8}{\pi }\left( {\arctan \frac{1}{2} + \arctan \frac{1}{3}} \right).\]
Solution.
We use the tangent addition formula:
\[\tan \left( {\alpha + \beta } \right) = \frac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}.\]
Let \(z = {\arctan \frac{1}{2} + \arctan \frac{1}{3}}.\) Then
\[\tan z = \tan \left( {\arctan \frac{1}{2} + \arctan \frac{1}{3}} \right) = \frac{{\tan \left( {\arctan \frac{1}{2}} \right) + \tan \left( {\arctan \frac{1}{3}} \right)}}{{1 - \tan \left( {\arctan \frac{1}{2}} \right)\tan \left( {\arctan \frac{1}{3}} \right)}} = \frac{{\frac{1}{2} + \frac{1}{3}}}{{1 - \frac{1}{2} \cdot \frac{1}{3}}} = \frac{{\frac{5}{6}}}{{1 - \frac{1}{6}}} = \frac{{\cancel{{\frac{5}{6}}}}}{{\cancel{{\frac{5}{6}}}}} = 1.\]
Now we can easily find the value of the expression \(B:\)
\[\tan z = 1, \Rightarrow z = \frac{\pi }{4}, \Rightarrow B = \frac{8}{\pi }z = \frac{8}{\pi } \cdot \frac{\pi }{4} = 2.\]
