Inverse Circular Functions: Sum & difference of two or more ICFs

Inverse Circular Functions: Sum & Difference of Two or More ICFs

Inverse Circular Functions (ICFs) are the inverse functions of the trigonometric functions sine, cosine, tangent, cosecant, secant, and cotangent. These functions allow us to find the angle whose trigonometric value is known.

In this topic, we will explore the sum and difference of two or more ICFs. We will discuss the formulas, important points, and provide examples to help understand this concept.

Table of Contents

  1. Sum of Two ICFs
  2. Difference of Two ICFs
  3. Sum and Difference of Multiple ICFs
  4. Important Points
  5. Examples

1. Sum of Two ICFs

The sum of two ICFs can be found using the following formula:

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

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

$$\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$$

$$\csc^{-1}(x) + \csc^{-1}(y) = \csc^{-1}\left(\frac{x\sqrt{1-y^2} + y\sqrt{1-x^2}}{xy - \sqrt{(1-x^2)(1-y^2)}}\right)$$

$$\sec^{-1}(x) + \sec^{-1}(y) = \sec^{-1}\left(\frac{x\sqrt{1-y^2} - y\sqrt{1-x^2}}{xy + \sqrt{(1-x^2)(1-y^2)}}\right)$$

$$\cot^{-1}(x) + \cot^{-1}(y) = \cot^{-1}\left(\frac{xy - \sqrt{(1-x^2)(1-y^2)}}{x\sqrt{1-y^2} + y\sqrt{1-x^2}}\right)$$

2. Difference of Two ICFs

The difference of two ICFs can be found using the following formula:

$$\sin^{-1}(x) - \sin^{-1}(y) = \sin^{-1}(x\sqrt{1-y^2} - y\sqrt{1-x^2})$$

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

$$\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$$

$$\csc^{-1}(x) - \csc^{-1}(y) = \csc^{-1}\left(\frac{x\sqrt{1-y^2} - y\sqrt{1-x^2}}{xy + \sqrt{(1-x^2)(1-y^2)}}\right)$$

$$\sec^{-1}(x) - \sec^{-1}(y) = \sec^{-1}\left(\frac{x\sqrt{1-y^2} + y\sqrt{1-x^2}}{xy - \sqrt{(1-x^2)(1-y^2)}}\right)$$

$$\cot^{-1}(x) - \cot^{-1}(y) = \cot^{-1}\left(\frac{xy + \sqrt{(1-x^2)(1-y^2)}}{x\sqrt{1-y^2} - y\sqrt{1-x^2}}\right)$$

3. Sum and Difference of Multiple ICFs

The sum and difference of multiple ICFs can be found by applying the sum and difference formulas iteratively. For example, to find the sum of three ICFs, we can use the formula:

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

Similarly, the difference of three ICFs can be found using the formula:

$$\sin^{-1}(x) - \sin^{-1}(y) - \sin^{-1}(z) = \sin^{-1}\left(\sin^{-1}(x) - \sin^{-1}(y)\right) - \sin^{-1}(z)$$

4. Important Points

  • The sum and difference of ICFs can be found using specific formulas for each trigonometric function.
  • It is important to remember the domain and range of each ICF to ensure the validity of the calculations.
  • The sum and difference formulas can be applied iteratively to find the sum and difference of multiple ICFs.

5. Examples

Example 1:

Find the value of $\sin^{-1}\left(\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{3}\right)$.

Using the sum formula for $\sin^{-1}$, we have:

$$\sin^{-1}\left(\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{3}\right) = \sin^{-1}\left(\frac{1}{2}\sqrt{1-\left(\frac{1}{3}\right)^2} + \frac{1}{3}\sqrt{1-\left(\frac{1}{2}\right)^2}\right)$$

Simplifying the expression inside the $\sin^{-1}$ function:

$$\sin^{-1}\left(\frac{1}{2}\sqrt{\frac{8}{9}} + \frac{1}{3}\sqrt{\frac{3}{4}}\right)$$

$$\sin^{-1}\left(\frac{\sqrt{2} + \sqrt{3}}{6}\right)$$

Therefore, $\sin^{-1}\left(\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{3}\right) = \sin^{-1}\left(\frac{\sqrt{2} + \sqrt{3}}{6}\right)$.

Example 2:

Find the value of $\tan^{-1}(1) + \tan^{-1}(2)$.

Using the sum formula for $\tan^{-1}$, we have:

$$\tan^{-1}(1) + \tan^{-1}(2) = \tan^{-1}\left(\frac{1+2}{1-1(2)}\right)$$

Simplifying the expression inside the $\tan^{-1}$ function:

$$\tan^{-1}\left(\frac{3}{-1}\right)$$

Since $\tan^{-1}$ is an odd function, we can rewrite the expression as:

$$-\tan^{-1}(3)$$

Therefore, $\tan^{-1}(1) + \tan^{-1}(2) = -\tan^{-1}(3)$.

In conclusion, the sum and difference of two or more ICFs can be found using specific formulas for each trigonometric function. It is important to understand these formulas and apply them correctly to solve problems involving ICFs.