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

[chwala]

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,

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

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

 

[anuttarasammyak]

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}

 

[SammyS]

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 .

 

[chwala]

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!