Trigonometric Functions

Geometry Level 1

If x = sin ( cos 1 y ) x=\sin \left(\cos^{-1} y\right) then what is the solution of x 2 + y 2 x^2+y^2 ?

5 3 6 2 4 1

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1 solution

Chew-Seong Cheong
Sep 15, 2017

x = sin ( cos 1 y ) = 1 cos 2 ( cos 1 y ) = 1 y 2 x 2 = 1 y 2 \begin{aligned} x & = \sin \left(\cos^{-1} y \right) \\ & = \sqrt{1-\cos^2 \left(\cos^{-1} y \right) } \\ & = \sqrt{1- y^2} \\ x^2 & = 1-y^2 \end{aligned}

x 2 + y 2 = 1 \begin{aligned} \implies x^2+y^2 & = \boxed{1} \end{aligned}

@Sakib Nazmus , you should use a backslash " \ " for functions such as \sin sin \sin , \cos cos \cos , \tan tan \tan , \ln ln \ln , \gcd gcd \gcd , \int \int , \sqrt x x \sqrt x , \sum \sum . Then the space after the function is automatic (always remember LaTex creators are not stupid). Note that function should not appear italic which are for variables and constants. For example sin x s i n x sin x -- note that all letters are italic and stick together despite separated by a space. Now \sin x sin x \sin x -- note only x x is italic and the space.

Chew-Seong Cheong - 3 years, 8 months ago

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Thanks for the information.

Nazmus sakib - 3 years, 8 months ago

I think you mean x 2 = 1 y 2 x^2=1-y^2 instead of x 2 = 1 y 2 x^2=\sqrt{1-y^2}

Nazmus sakib - 3 years, 8 months ago

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Thanks. I have changed it.

Chew-Seong Cheong - 3 years, 8 months ago

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