But I Don't Know Sin(20)!

Geometry Level 1

What is the value of $\sin 20^{\circ}(\tan 10^{\circ}+\cot 10^{\circ})$ ?

\dfrac{3}{2}

\dfrac{1}{3}

\dfrac{1}{2}

2 solutions

Danish Ahmed
May 12, 2015

$( \sin 20)(\tan 10+\cot 10)$

$=(2\sin 10\cos 10)(\dfrac{\sin 10}{\cos 10}+\dfrac{\cos 10}{\sin 10})$

$=2\sin 10\cos 10(\dfrac{\sin^210+\cos^210}{\sin 10\cos 10})$

$2(\sin^210+\cos^210)=2$

Simple and easy to understand, well done!

Hence, you can generalize it:

$\sin(2x)\bigg(\tan(x)+\cot(x)\bigg)=2~\forall~x\in\Bbb{R}\setminus\left\{\frac{n\pi}{2}\right\}~\forall~n\in\Bbb{Z}$

Here, $x$ is considered in radians.

Prasun Biswas - 6 years, 1 month ago

yeah!Nice... :P

Nihar Mahajan - 6 years, 1 month ago

I sincerely hope that isn't sarcasm.

Prasun Biswas - 6 years, 1 month ago

@Prasun Biswas – Not at all! It was a compliment!

Nihar Mahajan - 6 years, 1 month ago

@Prasun Biswas – u r a genius>!

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – Now, that was definitely sarcasm.

Prasun Biswas - 6 years, 1 month ago

@Prasun Biswas – You can simplify your constraint: $\ldots \mathbb R \setminus\left\{ \frac{n \pi}2 \right\}$ for all integers $n$ .

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – Ah yes! Thanks!

Prasun Biswas - 6 years ago

@Prasun Biswas – Actually you can drop the $\mathbb R$ completely as well, you didn't account for complex numbers.

Pi Han Goh - 6 years ago

Michael Fuller
Jul 22, 2015

$\sin { 20 } (\tan { 10 } +\cot { 10 })$

$=\sin { 20 } \left(\dfrac { \tan ^{ 2 }{ 10 }+1 }{ \tan { 10 } } \right)$

$=\sin { 20 } \left(\dfrac { \sec ^{ 2 }{ 10 } }{ \tan { 10 } } \right)$

$=2\sin { 10 } \cos { 10 } \left (\dfrac { \dfrac { 1 }{ \cos ^{ 2 }{ 10 } } }{ \dfrac { \sin { 10 } }{ \cos { 10 } } } \right)$

$=2\sin { 10 } \cos { 10 } \left(\dfrac { \frac { 1 }{ \cos { 10 } } }{ \sin { 10 } } \right)$

$=\dfrac { 2\cos { 10 } }{ \cos { 10 } }$

$=~\large\color{#20A900}{\boxed { 2 }}$