Question

If x and y are acute angles such that cos x + cos y = $ \dfrac{3}{2} $ ​and sin x + sin y = $ \dfrac{3}{4} $ ,​ then find the value of sin (x + y):
(a) $ \dfrac{2}{5} $
(b) $ \dfrac{3}{4} $
(c) $ \dfrac{3}{5} $
(d) $ \dfrac{4}{5} $

Answer 1

Hint: Use the formula cos x + cos y = $ 2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right) $ and sin x + sin y = $ 2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right) $ to get two equations. Divide these equations and then use the formula $ \sin \left( x+y \right)=\dfrac{2tan\left( \dfrac{x+y}{2} \right)}{1+ta{{n}^{2}}\left( \dfrac{x+y}{2} \right)} $ to get the final answer.

Complete step-by-step answer:
In this question, we are given that x and y are acute angles such that cos x + cos y = $ \dfrac{3}{2} $ and sin x + sin y = $ \dfrac{3}{4} $ .
We need to find the value of sin (x + y).
We already know that for two angles x and y, the sum of their cosines is equal to the double of the product of the cosine of half of their sum and the cosine of half of their difference.
i.e. cos x + cos y = $ 2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right) $
Using this property, we will get the following:
 cos x + cos y = $ \dfrac{3}{2} $
 $ 2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)=\dfrac{3}{2} $
 $ \cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)=\dfrac{3}{4} $ …(1)
We also know that for two angles x and y, the sum of their sines is equal to the double of the product of the sine of half of their sum and the cosine of half of their difference.
i.e. sin x + sin y = $ 2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right) $
Using this property, we will get the following:
sin x + sin y = $ \dfrac{3}{4} $
 $ 2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)=\dfrac{3}{4} $
 $ \sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)=\dfrac{3}{8} $ …(2)
Dividing equation (2) by equation (1), we will get the following:
 $ \dfrac{\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)}{\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)}=\dfrac{\dfrac{3}{8}}{\dfrac{3}{4}} $
 $ tan\left( \dfrac{x+y}{2} \right)=\dfrac{1}{2} $
Now, we know the formula: $ \sin \left( x+y \right)=\dfrac{2tan\left( \dfrac{x+y}{2} \right)}{1+ta{{n}^{2}}\left( \dfrac{x+y}{2} \right)} $
Substituting $ tan\left( \dfrac{x+y}{2} \right)=\dfrac{1}{2} $ in the above formula, we will get the following:
 $ \sin \left( x+y \right)=\dfrac{2\left( \dfrac{1}{2} \right)}{1+\dfrac{1}{4}}=\dfrac{4}{5} $
Hence, $ \sin \left( x+y \right)=\dfrac{4}{5} $
So, option (d) is correct.

Note: For solving such problems, students must first figure out the formulae that are to be used. The signs in the formulae must be taken care of and the formulae must be remembered properly, as any mistake in sign can lead to wrong answers.Here we try to use the transformation formulae of trigonometric functions.