Solution of Inverse Trigonometric Equations

Inverse trigonometric functions are the inverse functions of the trigonometric functions with restricted domains. They are used to obtain an angle from any of the trigonometric function's values. Solving inverse trigonometric equations involves finding the value of the angle that satisfies the given equation.

Understanding Inverse Trigonometric Functions

Before we delve into solving equations, let's understand the inverse trigonometric functions and their principal values.

Function	Notation	Domain	Range (Principal Value Branch)
Arcsine	$\sin^{-1}x$ or $\arcsin x$	$-1 \leq x \leq 1$	$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
Arccosine	$\cos^{-1}x$ or $\arccos x$	$-1 \leq x \leq 1$	$0 \leq y \leq \pi$
Arctangent	$\tan^{-1}x$ or $\arctan x$	$-\infty < x < \infty$	$-\frac{\pi}{2} < y < \frac{\pi}{2}$
Arccotangent	$\cot^{-1}x$ or $\arccot x$	$-\infty < x < \infty$	$0 < y < \pi$
Arcsecant	$\sec^{-1}x$ or $\arcsec x$	$x \leq -1$ or $x \geq 1$	$0 \leq y \leq \pi$, $y \neq \frac{\pi}{2}$
Arccosecant	$\csc^{-1}x$ or $\arccsc x$	$x \leq -1$ or $x \geq 1$	$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$, $y \neq 0$

Solving Inverse Trigonometric Equations

To solve an inverse trigonometric equation, you need to isolate the inverse trigonometric function on one side of the equation and then apply the corresponding trigonometric function to both sides to find the angle.

General Steps for Solving Inverse Trigonometric Equations

1. Isolate the Inverse Trigonometric Function: Make sure that the inverse trigonometric function is by itself on one side of the equation.
2. Apply the Trigonometric Function: Apply the corresponding trigonometric function to both sides of the equation.
3. Solve for the Angle: Use the range of the principal values to find the solution for the angle.
4. Consider Additional Solutions: Depending on the context, you may need to consider additional solutions outside the principal value range.

Examples

Example 1: Solving a Basic Arcsine Equation

Solve the equation $\sin^{-1}(x) = \frac{\pi}{6}$.

Solution:

Apply the sine function to both sides:

$$\sin(\sin^{-1}(x)) = \sin\left(\frac{\pi}{6}\right)$$

Since $\sin(\sin^{-1}(x)) = x$ and $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, we have:

$$x = \frac{1}{2}$$

Example 2: Solving an Arctangent Equation

Solve the equation $\tan^{-1}(2x) = \frac{\pi}{4}$.

Solution:

Apply the tangent function to both sides:

$$\tan(\tan^{-1}(2x)) = \tan\left(\frac{\pi}{4}\right)$$

Since $\tan(\tan^{-1}(2x)) = 2x$ and $\tan\left(\frac{\pi}{4}\right) = 1$, we have:

$$2x = 1$$

Solving for $x$ gives us:

$$x = \frac{1}{2}$$

Example 3: Solving an Equation with Multiple Angles

Solve the equation $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{2}$.

Solution:

Apply the sine function to both sides:

$$\sin(\sin^{-1}(x) + \sin^{-1}(y)) = \sin\left(\frac{\pi}{2}\right)$$

Using the sine addition formula:

$$\sin(\sin^{-1}(x))\cos(\sin^{-1}(y)) + \cos(\sin^{-1}(x))\sin(\sin^{-1}(y)) = 1$$

Since $\sin(\sin^{-1}(x)) = x$ and $\sin(\sin^{-1}(y)) = y$, and knowing that $\cos(\sin^{-1}(z)) = \sqrt{1 - z^2}$ for any $z$, we get:

$$x\sqrt{1 - y^2} + y\sqrt{1 - x^2} = 1$$

This equation can be solved for $x$ and $y$ given additional constraints or context.

Important Points to Remember

• The solutions to inverse trigonometric equations are angles.
• The principal value range must be considered when finding solutions.
• Additional solutions may exist outside the principal value range, depending on the context.
• Trigonometric identities and formulas can be useful in solving more complex equations.

By understanding the properties of inverse trigonometric functions and following the general steps outlined above, you can effectively solve inverse trigonometric equations. Practice with various examples to become proficient in this topic.