Question

If $x=\sin {70}^o.\sin {50}^o$ and $y=\cos {60}^o.\cos {80}^o$, then what is xy equal to?

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{16}$

$x=\sin \left({70}^0\right).\sin \left({50}^0\right)$

We know that $\sin \left(\theta \right)=\cos ({90}^0-\theta )$

$\therefore x=\sin \left({70}^0\right).\sin \left({50}^0\right)=\cos \left({90}^{-}{70}^0\right).\cos \left({90}^{-}{50}^0\right)=\cos \left({20}^0\right).\cos \left({40}^0\right)$

$y=\cos \left({60}^0\right).\cos \left({80}^0\right)$

We know that $\cos \left(\theta \right)=\sin ({90}^0-\theta )$

$\therefore y=\cos \left({60}^0\right).\cos \left({80}^0\right)=\sin \left({90}^{-}{60}^0\right).\sin \left({90}^{-}{80}^0\right)=\sin \left({30}^0\right).\sin \left({10}^0\right)$

$\therefore xy=\cos \left({20}^0\right).\cos \left({40}^0\right).\sin \left({30}^0\right).\sin \left({10}^0\right)$

Multiply both sides of equation with $2\cos \left({10}^0\right)$

$\therefore xy\left(2\cos \right({10}^0\left)\right)=\cos \left({20}^0\right).\cos \left({40}^0\right).\sin \left({30}^0\right).\sin \left({10}^0\right).2\cos \left({10}^0\right)$

Now, we will use the formula $2\sin \theta \cos \theta =\sin 2\theta$

$\therefore xy\left(2\cos \right({10}^0\left)\right)=\cos \left({20}^0\right).\cos \left({40}^0\right).\sin \left({30}^0\right).\left(2\sin \right({10}^0).\cos ({10}^0\left)\right)$

$\therefore 2xy.\cos \left({10}^0\right)=\cos \left({20}^0\right).\cos \left({40}^0\right).\sin \left({30}^0\right).\left(\sin {20}^0\right)$

Multiply $2$ on both sides of above equation

$\therefore 4xy.\cos \left({10}^0\right)=\left(2\right(\sin {20}^0).\cos ({20}^0\left)\right).\cos \left({40}^0\right).\sin \left({30}^0\right)$

$\therefore 4xy.\cos \left({10}^0\right)=\left(\sin \right({40}^0\left)\right).\cos \left({40}^0\right).\sin \left({30}^0\right)$

Multiply $2$ on both sides of above equation

$\therefore 8xy.\cos \left({10}^0\right)=\left(2\sin \right({40}^0).\cos ({40}^0\left)\right).\sin \left({30}^0\right)$

$\therefore 8xy.\cos \left({10}^0\right)=\sin \left({80}^0\right).\sin \left({30}^0\right)$

$\therefore 8xy.\cos \left({10}^0\right)=\cos \left({10}^0\right).\sin \left({30}^0\right)$

$\therefore 8xy=\sin \left({30}^0\right)=\frac{1}{2}$

$\therefore xy=\frac{1}{16}$

So, the answer is option (A)