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.