Question

If \[\sin A + \sin B = C\], \[\cos A + \cos B = D\], then the value of \[\sin (A + B) = \]
A. \[CD\]
B. \[\dfrac{{CD}}{{\left( {{C^2} + {D^2}} \right)}}\]
C. \[\dfrac{{\left( {{C^2} + {D^2}} \right)}}{{2CD}}\]
D. \[\dfrac{{2CD}}{{\left( {{C^2} + {D^2}} \right)}}\]

Answer 1

Hint: To solve this we need to know all the trigonometric identities. First we square \[\sin A + \sin B = C\] and \[\cos A + \cos B = D\], after we add both of them. Then we multiply C and D and on further simplification we will get the answer. To simplify this we need to know the Pythagoras identity \[{\sin ^2}A + {\cos ^2}A = 1\] and also the sum and difference formula of sine and cosine function.

Complete step by step answer:
Given
\[\sin A + \sin B = C - - - (1)\]
\[\cos A + \cos B = D - - - (2)\]
Now squaring on both sides for equation (1) and (2).
\[{\left( {\sin A + \sin B} \right)^2} = {C^2}\]
\[{\left( {\cos A + \cos B} \right)^2} = {D^2}\]
Now we know the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]. Applying this we have
\[{\sin ^2}A + {\sin ^2}B + 2\sin A.\sin B = {C^2} - - - (3)\]
\[{\cos ^2}A + {\cos ^2}B + 2\cos A.\cos B = {D^2} - - - - (4)\]
Now adding equation (3) and (4) we have
\[{\sin ^2}A + {\sin ^2}B + 2\sin A.\sin B + {\cos ^2}A + {\cos ^2}B + 2\cos A.\cos B = {C^2} + {D^2}\]
Now grouping we have
\[\left( {{{\sin }^2}A + {{\cos }^2}A} \right) + \left( {{{\sin }^2}B + {{\cos }^2}B} \right) + 2\left( {\sin A.\sin B + \cos A.\cos B} \right) = {C^2} + {D^2}\]
We know the Pythagoras identity \[{\sin ^2}A + {\cos ^2}A = 1\] and \[{\sin ^2}B + {\cos ^2}B = 1\]. Applying this we have
\[1 + 1 + 2\left( {\sin A.\sin B + \cos A.\cos B} \right) = {C^2} + {D^2}\]
We also know \[\cos \left( {A - B} \right) = \sin A.\sin B + \cos A.\cos B\]. Then we have
\[2 + 2\left( {\cos \left( {A - B} \right)} \right) = {C^2} + {D^2}\]
\[2\left( {\cos \left( {A - B} \right)} \right) = {C^2} + {D^2} - 2\]
Or
\[ \Rightarrow \cos \left( {A - B} \right) = \dfrac{{{C^2} + {D^2} - 2}}{2} - - - (5)\]
Now let us multiply C and D then we have
\[CD = \left( {\sin A + \sin B} \right)\left( {\cos A + \cos B} \right)\]
Expanding the brackets we have
\[ = \sin A\left( {\cos A + \cos B} \right) + \sin B\left( {\cos A + \cos B} \right)\]
\[ = \sin A.\cos A + \sin A.\cos B + \sin B.\cos A + \sin B.\cos B\]
We know \[\sin 2A = 2\sin A.\cos A\] and \[\sin 2B = 2\sin B.\cos B\], then using this we have
\[ = \dfrac{{\sin 2A}}{2} + \left( {\sin A.\cos B + \sin B.\cos A} \right) + \dfrac{{\sin 2B}}{2}\]
Also we know \[\sin \left( {A + B} \right) = \sin A.\cos B + \sin B.\cos A\], applying we have
\[ = \dfrac{{\sin 2A}}{2} + \sin \left( {A + B} \right) + \dfrac{{\sin 2B}}{2}\]
\[ = \dfrac{{\sin 2A}}{2} + \dfrac{{\sin 2B}}{2} + \sin \left( {A + B} \right)\]
Taking \[\dfrac{1}{2}\] common we have
\[ = \dfrac{1}{2}\left( {\sin 2A + \sin 2B} \right) + \sin \left( {A + B} \right)\]
We know \[\left( {\sin 2A + \sin 2B} \right) = 2\sin \left( {A + B} \right)\cos \left( {A - B} \right)\] then
\[ = \dfrac{1}{2}\left( {2\sin \left( {A + B} \right)\cos \left( {A - B} \right)} \right) + \sin \left( {A + B} \right)\]
Simplifying we have
\[ = \left( {\sin \left( {A + B} \right)\cos \left( {A - B} \right)} \right) + \sin \left( {A + B} \right)\]
Now taking \[\sin \left( {A + B} \right)\] common we have
\[ = \sin \left( {A + B} \right)\left( {\cos \left( {A - B} \right) + 1} \right)\]
Now from equation (5) we have \[\cos \left( {A - B} \right) = \dfrac{{{C^2} + {D^2} - 2}}{2}\]
\[ = \sin \left( {A + B} \right)\left( {\dfrac{{{C^2} + {D^2} - 2}}{2} + 1} \right)\]
Taking LCM and simplifying we have
\[ = \sin \left( {A + B} \right)\left( {\dfrac{{{C^2} + {D^2} - 2 + 2}}{2}} \right)\]
Thus we have
\[ \Rightarrow CD = \sin \left( {A + B} \right)\left( {\dfrac{{{C^2} + {D^2}}}{2}} \right)\]
Multiplying two on both sides we have
\[2CD = \sin \left( {A + B} \right)\left( {{C^2} + {D^2}} \right)\]
\[ \Rightarrow \sin \left( {A + B} \right) = \dfrac{{{C^2} + {D^2}}}{{2CD}}\].

So, the correct answer is “Option C”.

Note: We need to be careful in the simplification part. While applying the Pythagoras identity both the angle of sine and cosine should be the same. If we have \[{\sin ^2}A + {\cos ^2}B\] , then it is not equal to 1. Which goes the same for the sine double angle formula.