Trigonometry! #104

Geometry Level 3

If sin ⁡ α + csc ⁡ α = 2 \sin\alpha+\csc\alpha=2 then sin ⁡ n α + csc ⁡ n α = ? \sin^{n}\alpha+\csc^{n}\alpha=?

This problem is part of the set Trigonometry .

n n n ( sin ⁡ α + csc ⁡ α ) n(\sin\alpha+\csc\alpha) n sin ⁡ α + csc ⁡ α \frac{n}{\sin\alpha+\csc\alpha} 2 2

Omkar Kulkarni by Omkar Kulkarni , 22, 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

Pranjal Jain
Feb 10, 2015

Let x = sin ⁡ α x=\sin\alpha

x + 1 x = 2 ⇒ x 2 − 2 x + 1 = 0 ⇒ x = 1 x+\dfrac{1}{x}=2\\\Rightarrow x^2-2x+1=0\\\Rightarrow x=1

sin ⁡ n α + csc ⁡ n α = 1 n + 1 n = 1 + 1 = 2 \sin^n\alpha+\csc^n\alpha=1^n+1^n=1+1=\boxed{2}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...