Which of these is equivalent?

Geometry Level 1

4 cos ⁡ 2 θ 2 sin ⁡ 2 θ 2 cos ⁡ 2 θ \large \dfrac { { 4\cos ^{ 2 }{ \frac { \theta }{ 2 } } \sin ^{ 2 }{ \frac { \theta }{ 2 } } } }{ \cos ^{ 2 }{ \theta } }

Simplify the trigonometric expression above.

cot ⁡ 2 θ \cot^{2} {\theta} tan ⁡ 3 θ \tan^{3} {\theta} 1 − sin ⁡ 2 θ 1-\sin^{2} {\theta} tan ⁡ 2 θ \tan^{2} {\theta} csc ⁡ 2 θ \csc^{2} {\theta} sin ⁡ 2 θ \sin^{2} {\theta}

Icaro Buscarini by Icaro Buscarini , 26, Brazil

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.

2 solutions

4 cos ⁡ 2 θ 2 sin ⁡ 2 θ 2 cos ⁡ 2 θ = ( 2 cos ⁡ θ 2 sin ⁡ θ 2 ) 2 cos ⁡ 2 θ = sin ⁡ 2 θ cos ⁡ 2 θ = tan ⁡ 2 θ \large \displaystyle \begin{aligned} \frac{4 \cos^2 \frac{\theta}{2} \sin^2 \frac{\theta}{2}}{\cos^2 \theta} &= \frac{\left(2 \cos \frac{\theta}{2} \sin \frac{\theta}{2} \right)^2}{\cos^2 \theta}\\ \large \displaystyle &= \frac{\sin^2 \theta}{\cos^2 \theta}\\ \large \displaystyle &= \tan^2 \theta \end{aligned}

Formula Used: sin ⁡ 2 θ = 2 × sin ⁡ θ × cos ⁡ θ . \large \displaystyle \text{Formula Used: } \sin 2 \theta = 2 \times \sin \theta \times \cos \theta.

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution. Keep it up!

Sandeep Bhardwaj - 5 years ago

Log in to reply

Thank You! ⌣ ¨ \Large \ddot\smile

Samara Simha Reddy - 5 years ago
Icaro Buscarini
May 18, 2016

2 Helpful 0 Interesting 0 Brilliant 0 Confused

@Icaro Buscarini

Nice problem. :)

The solution can be improved by using the latex at the place of the lonely image. ;) You can find the latex guide here . Thanks!

Sandeep Bhardwaj - 5 years ago

Log in to reply

Thank you! ;)

Icaro Buscarini - 5 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...