# Converting one inverse trig function to another

• cr7einstein
In summary, the problem asks for an expression of $$2\arctan\left(\sqrt\frac{a-b}{a+b}\tan\left(\frac\theta 2\right)\right)$$ in terms of inverse cosine. The attempted solution involves a substitution and the use of inverse trigonometric identities. However, a simpler solution is to let ##x = 2\arctan\left(\sqrt\frac{a-b}{a+b}\tan\left(\frac\theta 2\right)\right)## and express ##\tan\frac x2## in terms of ##\cos x##.
cr7einstein

## Homework Statement

Express $$2 arctan (\sqrt\frac{a-b}{a+b} tan (\theta/2))$$ in terms of inverse cosine

## Homework Equations

I realize it amounts to find a smart substitution, but I can't find one.

## The Attempt at a Solution

I tried ##b/a=tan \theta## , but I can't find any way to get rid of the other tangent term. I would be really nice if the solution is not a direct substitution of the enormous formula for the inverse trig functions, as it only gets messy. Thanks in advance!

@Math_QED I mentioned that I don't want to just substitute into the formulae as they are very messy...No problem, you can look into it when you have the time.

cr7einstein said:

## Homework Statement

Express $$2 arctan (\sqrt\frac{a-b}{a+b} tan (\theta/2))$$ in terms of inverse cosine

## Homework Equations

I realize it amounts to find a smart substitution, but I can't find one.

## The Attempt at a Solution

I tried ##b/a=tan \theta## , but I can't find any way to get rid of the other tangent term. I would be really nice if the solution is not a direct substitution of the enormous formula for the inverse trig functions, as it only gets messy. Thanks in advance!
Hint: Let ##x = 2 \arctan \left[\sqrt\frac{a-b}{a+b} \tan \left(\frac \theta 2\right)\right]##. Then ##\tan \frac x2 = \sqrt\frac{a-b}{a+b} \tan \left(\frac \theta 2\right)##. You can express ##\tan\frac x2## in terms of ##\cos x##.

cr7einstein
Done! Thanks a lot @vela !

## 1. How do I convert an inverse trig function to another?

To convert an inverse trig function to another, you can use the trigonometric identities and algebraic manipulations. For example, you can use the identity sin(x) = 1/csc(x) to convert from inverse sine to inverse cosecant.

## 2. What are the common inverse trig functions?

The most common inverse trig functions are inverse sine, inverse cosine, inverse tangent, inverse cosecant, inverse secant, and inverse cotangent. These functions are used to find the angle measure given the ratio of sides in a right triangle.

## 3. Can I express all inverse trig functions in terms of each other?

Yes, it is possible to express all inverse trig functions in terms of each other using trigonometric identities and algebraic manipulations. This allows for easy conversion from one inverse trig function to another.

## 4. How can I remember the conversions between inverse trig functions?

One way to remember the conversions between inverse trig functions is to use the acronym "SOHCAHTOA" which stands for "sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, tangent is opposite over adjacent." This can help you remember the ratios and their corresponding inverse trig functions.

## 5. Are there any limitations to converting inverse trig functions?

There are a few limitations to converting inverse trig functions. One limitation is that some ratios may not correspond to a valid angle measure, which means the conversion would not be possible. Additionally, the values of the inverse trig functions may be undefined for certain input values, such as dividing by zero.

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