Inverse of Cosine

Geometry Level 2

Let g ( x ) g(x) be the inverse function of f ( x ) = cos ⁡ x ( 0 < x < π ) f(x)=\cos x\ (0 < x < \pi) such that g ( 5 13 ) = α , g ( 12 13 ) = β . g\left(\frac{5}{13}\right)=\alpha, g\left(\frac{12}{13}\right)=\beta. What is the value of f ( α + β ) ? f(\alpha+\beta)?

-1 0 1 2 \frac{1}{2} 1

Ajay Dutta by Ajay Dutta , 24, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tom Engelsman
Jan 31, 2017

Let g ( x ) = a r c c o s ( x ) ⇒ α = a r c c o s ( 5 13 ) , β = a r c c o s ( 12 13 ) . g(x) = arccos(x) \Rightarrow \alpha = arccos(\frac{5}{13}), \beta = arccos(\frac{12}{13}). Upon observation, α \alpha and β \beta are complementary angles from the 5-12-13 right triangle. Hence f ( α + β ) = c o s ( π 2 ) = 0 . f(\alpha + \beta) = cos(\frac{\pi}{2}) = \boxed{0}.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...