Question

Prove that $\sin 20^\circ \sin 70^\circ - \cos 20^\circ \cos 70^\circ = 0$ .

Answer 1

Hint : Here, we will use the formula that $\cos$ over $a + b$ is equal to sum of product of $\cos a$ and $\cos b$ and product of $- \sin a$ and $\sin b$ . Then substitute the value of $\cos$ to get the required answer.

Complete step-by-step answer :
The given equation is $\sin 20^\circ \sin 70^\circ - \cos 20^\circ \cos 70^\circ = 0$ .
To prove the value of given equation, we will take left hand side term, that is,
$\sin 20^\circ \sin 70^\circ - \cos 20^\circ \cos 70^\circ$
Let us now use a known formula that is $\sin a\sin b - \cos a\cos b = \cos \left( {a + b} \right)$ .
On substituting the values of $a = 20^\circ$ and $b = 70^\circ$ , we get,
$\sin \left( {20^\circ } \right)\sin \left( {70^\circ } \right) - \cos \left( {20^\circ } \right)\cos \left( {70^\circ } \right) = \cos \left( {20^\circ + 70^\circ } \right)$
Since, we can write $\cos \left( {20^\circ + 70^\circ } \right)$ in the form of,
$\cos \left( {20^\circ + 70^\circ } \right) = \cos 90^\circ$
Now, we know the value for $\cos 90^\circ$ that is $\cos \left( {90} \right) = 0$ .
So, we have taken the left hand side and proved that it is equal to the right hand side.
Hence, the value for the equation is $\sin 20^\circ \sin 70^\circ - \cos 20^\circ \cos 70^\circ = 0$ .

Note : Make sure the formula for the equation will not be same for all the equations. This can also be done by another method. If we take $\cos \left( {20} \right)$ as $\cos \left( {90 - 70} \right)$ and take $\cos \left( {70} \right)$ as $\cos \left( {90 - 20} \right)$ then by substituting and using the property that $\cos \left( {90 - \theta } \right)$ is equal to $\sin \left( \theta \right)$ we can prove the equation.