# How to Solve Inverse Trigonometric Functions?

Inverse trigonometric functions are defined as the inverse functions of the basic trigonometric functions. Learn how to solve inverse trigonometric functions by the following step-by-step guide.

Inverse trigonometric functions are also known as anti-trigonometric functions, arcus functions, or cyclometric functions. Inverse trigonometric functions are the inverse functions of the basic trigonometric functions sine, cosine, tangent, cotangent, secant, and cosecant functions.

## A step-by-step guide to inverse trigonometric functions

The basic inverse trigonometric formulas are as follows:

• $$\color{blue}{sin^{-1}(-x)=-sin^{-1}x, x∈[-1,1]}$$
• $$\color{blue}{tan^{-1}(-x)=-tan^{-1}x, x∈ R}$$
• $$\color{blue}{cosec^{-1}(-x)=-cosec^{-1}x, x∈ R-(-1,1)}$$
• $$\color{blue}{cos^{-1}(-x)=π-cos^{-1}x, x∈[-1,1]}$$
• $$\color{blue}{sec^{-1}(-x)=π-sec^{-1}x, x∈ R -(-1,1)}$$
• $$\color{blue}{cot^{-1}(-x)=π-cot^{-1}x, x∈ R}$$

### Inverse trigonometric function formulas for reciprocal functions

The inverse trigonometric function for reciprocal values x converts the given inverse trigonometric function to its reciprocal function. This follows from the trigonometric functions where sin and cosecant are reciprocal to each other, tangent and cotangent are reciprocal to each other, and cos and secant are reciprocal to each other.

The inverse triangular formula of inverse sine, inverse cosine, and inverse tangent can also be expressed as follows.

• $$\color{blue}{sin^{-1}(x)=cosec^{-1}\frac{1}{x}, x∈ R -(-1,1)}$$
• $$\color{blue}{cos^{-1}(x)=sec^{-1}\frac{1}{x}, x∈ R -(-1,1)}$$
• $$\color{blue}{tan^{-1}(x)=cot^{-1}\frac{1}{x}, x >0}$$
• $$\color{blue}{tan^{-1}(x)=-π+cot^{-1}x, x<0}$$

### Inverse trigonometric function formulas for complementary functions

The complementary functions, sine-cosine, tangent-cotangent, and secant-cosecant can be interpreted as:

• $$\color{blue}{sin^{-1}(x)+cos^{-1}x=\frac{\pi}{2}, x ∈ [-1,1]}$$
• $$\color{blue}{tan^{-1}(x)+cot^{-1}x=\frac{\pi}{2}, x ∈ R}$$
• $$\color{blue}{sec^{-1}(x)+cosec^{-1}x=\frac{\pi}{2}, x ∈ R -[-1,1]}$$

### Sum and difference of inverse trigonometric function formulas

The sum and difference of two inverse trigonometric functions can be combined to form an inverse function, according to the following set of formulas:

• $$\color{blue}{sin^{-1}x+sin^{-1}y=sin^{-1}(x.\sqrt{\left(1-y^2\right)}+y\sqrt{\left(1-x^2\right)})}$$
• $$\color{blue}{sin^{-1}x-sin^{-1}y=sin^{-1}(x.\sqrt{\left(1-y^2\right)}-y\sqrt{\left(1-x^2\right)})}$$
• $$\color{blue}{cos^{-1}x+cos^{-1}y=cos^{-1}\left(xy-\sqrt{\left(1-x^2\right)}.\sqrt{\left(1-y^2\right)}\right)}$$
• $$\color{blue}{cos^{-1}x-cos^{-1}y=cos^{-1}\left(xy+\sqrt{\left(1-x^2\right)}.\sqrt{\left(1-y^2\right)}\right)}$$
• $$\color{blue}{tan^{-1}x+tan^{-1}y=tan^{-1}\frac{(x+y)}{(1-xy)}, \:if\: xy<1}$$
• $$\color{blue}{tan^{-1}x+tan^{-1}y=tan^{-1}\frac{(x-y)}{(1+xy)}, \:if\: xy> -1}$$

### Double of inverse trigonometric function formulas

Twice an inverse trigonometric function can be solved to form a single trigonometric function according to the following set of formulas:

• $$\color{blue}{2sin^{-1}x=sin^{-1}\left(2x.\sqrt{\left(1-x^2\right)}\right)}$$
• $$\color{blue}{2cos^{-1}x=cos^{-1}(2x^2-1)}$$
• $$\color{blue}{2tan^-1x=tan^{-1}(\frac{2x}{1}-x^2)}$$

### Inverse Trigonometric Functions – Example 1:

Find the value of $$sin^{-1}(-1)$$.

Solution:

Use this formula: $$\color{blue}{sin^{-1}(-x)=-sin^{-1}x, x∈[-1,1]}$$

$$sin^{-1}(-1)=-sin^{-1}(1)$$

$$-sin^{-1} (1)=\frac{\pi }{2}$$

## Exercises for Inverse Trigonometric Functions

### Evaluate each of the following.

1. $$\color{blue}{tan^{-1}(\sqrt{3})-cot^{-1}(-\sqrt{3})}$$
2. $$\color{blue}{sin(cot^{-1}x)}$$
3. $$\color{blue}{tan^{-1}(1)+cos^{-1}(-\frac{1}{2})+sin^{-1}(-\frac{1}{2})}$$
4. $$\color{blue}{cos^{-1}(cos(-\frac{\pi}{3}))}$$
5. $$\color{blue}{sin\left(cos^{-1}\frac{3}{5}\right)}$$
1. $$\color{blue}{-\frac{\pi}{2}}$$
2. $$\color{blue}{\frac{\sqrt{1+x^2}}{1+x^2}}$$
3. $$\color{blue}{\frac{3\pi}{4}}$$
4. $$\color{blue}{\frac{\pi}{3}}$$
5. $$\color{blue}{\frac{4}{5}}$$

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