# Solve the problem that involves ##\cos^{-1} x + \cos^{-1}y##

• chwala
chwala
Gold Member
Homework Statement
See attached.
Relevant Equations
Trigonometry
In my approach (using a right angled triangle) i let,

##\cos^{-1} x = C## ⇒##\cos C = \sqrt{1-y^2}##

and

##\cos^{-1} y= A## ⇒ ##\cos A= \sqrt{1-x^2}##

Also, ##A+C = \dfrac{π}{2}##

and ##\cos \dfrac{π}{2}= 0##

##xy - \sqrt{(y^2) ⋅(x^2)}=xy-xy=0##

It follows that,

##\cos^{-1} [xy - \sqrt{(1-x^2)(1-y^2)}]= \cos^{-1}[ xy - \sqrt{(y^2) ⋅(x^2)}]##

##=\cos^{-1} (xy - \sqrt{y^2}⋅ \sqrt{x^2})=\cos^{-1} (xy-xy)=\cos^{-1} (0)= \dfrac{π}{2}##

Cos of the LHS =
$$\cos(\cos^{-1}x+\cos^{-1}y)=\cos(\cos^{-1}x)\cos(\cos^{-1}y)-\sin(\cos^{-1}x)\sin(\cos^{-1}y)$$
$$=xy-\sqrt{1-x^2}\sqrt{1-y^2}$$

chwala and SammyS
chwala said:
Homework Statement: See attached.
Relevant Equations: Trigonometry

View attachment 331694In my approach (using a right angled triangle) i let,

##\cos^{-1} x = C## ⇒##\cos C = \sqrt{1-y^2}##

and

##\cos^{-1} y= A## ⇒ ##\cos A= \sqrt{1-x^2}##

Also, ##A+C = \dfrac{π}{2}##
You seem to be assuming that ##\displaystyle x^2+y^2=1 \, , \ ## as if ##(x,\, y) \ ## is an ordered pair on the unit circle. That assumption is not necessary.

If ## \displaystyle \cos^{-1} x = C \, , \ ## then ##\displaystyle \cos C = x \ ## and ##\displaystyle \sin C = \sqrt{ 1-x^2} \ . \ ##

etc.

See Post #2 by @anuttarasammyak .

Last edited:
chwala
anuttarasammyak said:
Cos of the LHS =
$$\cos(\cos^{-1}x+\cos^{-1}y)=\cos(\cos^{-1}x)\cos(\cos^{-1}y)-\sin(\cos^{-1}x)\sin(\cos^{-1}y)$$
$$=xy-\sqrt{1-x^2}\sqrt{1-y^2}$$
ok i see the logic now

##\cos(A+B)=\cos A \cos B - \sin A \sin B##.

Cheers Man!

## What is the range of values for which the expression ##\cos^{-1} x + \cos^{-1} y## is defined?

The expression ##\cos^{-1} x + \cos^{-1} y## is defined when both ##x## and ##y## are in the range $$[-1, 1]$$. This is because the inverse cosine function, ##\cos^{-1}##, is only defined for inputs within this interval.

## Can ##\cos^{-1} x + \cos^{-1} y## be simplified to a single inverse trigonometric function?

No, generally ##\cos^{-1} x + \cos^{-1} y## cannot be simplified to a single inverse trigonometric function. However, there are trigonometric identities and specific values for which the sum can be evaluated or simplified further.

## What is the sum of ##\cos^{-1} x + \cos^{-1} y## if ##x = y = 0##?

If ##x = y = 0##, then ##\cos^{-1} 0 = \frac{\pi}{2}##. Therefore, ##\cos^{-1} 0 + \cos^{-1} 0 = \frac{\pi}{2} + \frac{\pi}{2} = \pi##.

## Is there a trigonometric identity involving ##\cos^{-1} x + \cos^{-1} y##?

Yes, one useful identity is that if ##x## and ##y## are such that ##x^2 + y^2 \leq 1##, then:$\cos^{-1} x + \cos^{-1} y = \cos^{-1} (xy - \sqrt{(1 - x^2)(1 - y^2)}).$This identity can sometimes be used to simplify expressions involving the sum of inverse cosines.

## How can we interpret the sum ##\cos^{-1} x + \cos^{-1} y## geometrically?

Geometrically, ##\cos^{-1} x## and ##\cos^{-1} y## represent the angles whose cosines are ##x## and ##y##, respectively. The sum ##\cos^{-1} x + \cos^{-1} y## can be interpreted as the sum of these two angles. If these angles are part of a triangle or other geometric figure, their sum can provide insights into the properties of that figure.

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