# Inverse trigonometry prove this

• utkarshakash
In summary, the given equation can be simplified using trigonometric identities to show that the left hand side is equal to the right hand side. This involves using the sum of angles formula and simplifying using the cosine and sine functions.
utkarshakash
Gold Member

## Homework Statement

If $cos^{-1}\frac{x}{a}+cos^{-1}\frac{y}{b} = \alpha$ then show that $\frac{x^2}{a^2}-\frac{2xy}{ab}cos \alpha + \frac{y^2}{b^2} = sin^2 \alpha$

## The Attempt at a Solution

I assume inverse functions to be θ and β respectively. So in the second equation for LHS I can convert it like this

$cos^2\theta + cos^2\beta - 2cos \alpha cos\theta cos\beta$

Similarly for RHS
$sin^2 \alpha = sin^2(\theta + \beta)$

Taking the inverse cosine of a cosine gives the argument to the inverse cosine. Then you would use the sum of angles trigonometric formula.

utkarshakash said:

## Homework Statement

If $cos^{-1}\frac{x}{a}+cos^{-1}\frac{y}{b} = \alpha$ then show that $\frac{x^2}{a^2}-\frac{2xy}{ab}cos \alpha + \frac{y^2}{b^2} = sin^2 \alpha$

## The Attempt at a Solution

I assume inverse functions to be θ and β respectively. So in the second equation for LHS I can convert it like this

$cos^2\theta + cos^2\beta - 2cos \alpha cos\theta cos\beta$

Similarly for RHS
$sin^2 \alpha = sin^2(\theta + \beta)$
So you need to show that $\ \ \cos^2(\theta) + \cos^2(\beta) - 2\cos(\theta + \beta) \cos(\theta)\cos(\beta)= \sin^2(\theta + \beta)\ .$

$\cos(\theta+\beta)=\cos(\theta)\cos( \beta)-\sin(\theta)\sin(\beta)$

$\displaystyle \sin^2(\theta+\beta)=\left(\sin(\theta)\cos(\beta)+\cos(\theta)\sin(\beta) \right)^2\\ \quad\quad\quad=\sin^2(\theta)\cos^2(\beta)+2\sin(\theta)\cos(\beta)\cos(\theta)\sin(\beta)+\cos^2( \theta)\sin^2(\beta)$

So you can get some cancellation.

BTW: Didn't you post this very same question previously?

SammyS said:
So you need to show that $\ \ \cos^2(\theta) + \cos^2(\beta) - 2\cos(\theta + \beta) \cos(\theta)\cos(\beta)= \sin^2(\theta + \beta)\ .$

$\cos(\theta+\beta)=\cos(\theta)\cos( \beta)-\sin(\theta)\sin(\beta)$

$\displaystyle \sin^2(\theta+\beta)=\left(\sin(\theta)\cos(\beta)+\cos(\theta)\sin(\beta) \right)^2\\ \quad\quad\quad=\sin^2(\theta)\cos^2(\beta)+2\sin(\theta)\cos(\beta)\cos(\theta)\sin(\beta)+\cos^2( \theta)\sin^2(\beta)$

So you can get some cancellation.

BTW: Didn't you post this very same question previously?

No. Not exactly like this.

## 1. What is inverse trigonometry?

Inverse trigonometry is a branch of mathematics that deals with finding the angle measures of a right triangle given the lengths of its sides. It is the opposite of regular trigonometry, which finds the side lengths given the angle measures.

## 2. Why is it important to prove inverse trigonometric identities?

Proving inverse trigonometric identities is important because it allows us to establish the relationships between different trigonometric functions and understand how they are related to one another. This can help us in solving more complex trigonometric equations and problems.

## 3. How do you prove an inverse trigonometric identity?

To prove an inverse trigonometric identity, we start with one side of the equation and use algebraic manipulation and trigonometric identities to transform it into the other side of the equation. This process involves using basic trigonometric identities, such as the Pythagorean identities and the sum and difference identities.

## 4. What are some common inverse trigonometric identities?

Some common inverse trigonometric identities include the arcsine identity, arccosine identity, and arctangent identity. These identities relate the inverse trigonometric functions (arcsine, arccosine, arctangent) to their corresponding trigonometric functions (sine, cosine, tangent).

## 5. How can I use inverse trigonometry to solve real-world problems?

Inverse trigonometry can be used to solve real-world problems involving right triangles, such as finding the height of a building or the distance between two objects. By using the inverse trigonometric functions, we can determine the angle measures needed to solve these problems and find the missing side lengths of a triangle.

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