When Trigonometry gets complicated!

Geometry Level 2

This expression: 2 n sin ⁡ θ 2 n cos ⁡ θ 2 cos ⁡ θ 4 cos ⁡ θ 8 … cos ⁡ θ 2 n = ? 2^n\sin\frac{\theta}{2^n}\cos\frac{\theta}{2}\cos\frac{\theta}{4}\cos\frac{\theta}{8}\dots\cos\frac{\theta}{2^n}=? for n = 1 , 2 , 3 , … n=1,2,3,\dots

sec ⁡ θ \sec\theta csc ⁡ θ \csc\theta sin ⁡ θ \sin\theta cos ⁡ θ \cos\theta tan ⁡ θ \tan\theta cot ⁡ θ \cot\theta

Hana Wehbi by Hana Wehbi , 51, USA

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

X X
Jul 11, 2018

First of all, sin ⁡ ( 2 θ ) = 2 sin ⁡ θ cos ⁡ θ \sin(2\theta)=2\sin\theta\cos\theta

Let f ( n ) = 2 n sin ⁡ θ 2 n cos ⁡ θ 2 n cos ⁡ θ 2 n − 1 … cos ⁡ θ 4 cos ⁡ θ 2 f(n)=2^n\sin\dfrac{\theta}{2^n}\cos\dfrac{\theta}{2^n}\cos\dfrac{\theta}{2^{n-1}}\dots\cos\dfrac{\theta}{4}\cos\dfrac{\theta}{2}

= 2 n − 1 ( 2 sin ⁡ θ 2 n cos ⁡ θ 2 n ) cos ⁡ θ 2 n − 1 … cos ⁡ θ 4 cos ⁡ θ 2 =2^{n-1}(2\sin\dfrac{\theta}{2^n}\cos\dfrac{\theta}{2^n})\cos\dfrac{\theta}{2^{n-1}}\dots\cos\dfrac{\theta}{4}\cos\dfrac{\theta}{2}

= 2 n − 1 sin ⁡ θ 2 n − 1 cos ⁡ θ 2 n − 1 … cos ⁡ θ 4 cos ⁡ θ 2 = f ( n − 1 ) =2^{n-1}\sin\dfrac{\theta}{2^{n-1}}\cos\dfrac{\theta}{2^{n-1}}\dots\cos\dfrac{\theta}{4}\cos\dfrac{\theta}{2}=f(n-1)

So, f ( n ) = f ( n − 1 ) = f ( n − 2 ) = . . . = f ( 2 ) = f ( 1 ) = 2 sin ⁡ θ 2 cos ⁡ θ 2 = sin ⁡ θ f(n)=f(n-1)=f(n-2)=...=f(2)=f(1)=2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}=\sin\theta

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...