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Question: If \[\sin A + \sin B = C\], \[\cos A + \cos B = D\], then the value of \[\sin (A + B) = \] A. \[CD...

If sin⁡A+sin⁡B=C\sin A + \sin B = C, cos⁡A+cos⁡B=D\cos A + \cos B = D, then the value of sin⁡(A+B)=\sin (A + B) =
A. CDCD
B. CD(C2+D2)\dfrac{{CD}}{{\left( {{C^2} + {D^2}} \right)}}
C. (C2+D2)2CD\dfrac{{\left( {{C^2} + {D^2}} \right)}}{{2CD}}
D. 2CD(C2+D2)\dfrac{{2CD}}{{\left( {{C^2} + {D^2}} \right)}}

Explanation

Solution

To solve this we need to know all the trigonometric identities. First we square sin⁡A+sin⁡B=C\sin A + \sin B = C and cos⁡A+cos⁡B=D\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⁡2A+cos⁡2A=1{\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)\sin A + \sin B = C - - - (1)
cos⁡A+cos⁡B=D−−−(2)\cos A + \cos B = D - - - (2)
Now squaring on both sides for equation (1) and (2).
(sin⁡A+sin⁡B)2=C2{\left( {\sin A + \sin B} \right)^2} = {C^2}
(cos⁡A+cos⁡B)2=D2{\left( {\cos A + \cos B} \right)^2} = {D^2}
Now we know the algebraic identity (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab. Applying this we have
sin⁡2A+sin⁡2B+2sin⁡A.sin⁡B=C2−−−(3){\sin ^2}A + {\sin ^2}B + 2\sin A.\sin B = {C^2} - - - (3)
cos⁡2A+cos⁡2B+2cos⁡A.cos⁡B=D2−−−−(4){\cos ^2}A + {\cos ^2}B + 2\cos A.\cos B = {D^2} - - - - (4)
Now adding equation (3) and (4) we have
sin⁡2A+sin⁡2B+2sin⁡A.sin⁡B+cos⁡2A+cos⁡2B+2cos⁡A.cos⁡B=C2+D2{\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
(sin⁡2A+cos⁡2A)+(sin⁡2B+cos⁡2B)+2(sin⁡A.sin⁡B+cos⁡A.cos⁡B)=C2+D2\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⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1 and sin⁡2B+cos⁡2B=1{\sin ^2}B + {\cos ^2}B = 1. Applying this we have
1+1+2(sin⁡A.sin⁡B+cos⁡A.cos⁡B)=C2+D21 + 1 + 2\left( {\sin A.\sin B + \cos A.\cos B} \right) = {C^2} + {D^2}
We also know cos⁡(A−B)=sin⁡A.sin⁡B+cos⁡A.cos⁡B\cos \left( {A - B} \right) = \sin A.\sin B + \cos A.\cos B. Then we have
2+2(cos⁡(A−B))=C2+D22 + 2\left( {\cos \left( {A - B} \right)} \right) = {C^2} + {D^2}
2(cos⁡(A−B))=C2+D2−22\left( {\cos \left( {A - B} \right)} \right) = {C^2} + {D^2} - 2
Or
⇒cos⁡(A−B)=C2+D2−22−−−(5)\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=(sin⁡A+sin⁡B)(cos⁡A+cos⁡B)CD = \left( {\sin A + \sin B} \right)\left( {\cos A + \cos B} \right)
Expanding the brackets we have
=sin⁡A(cos⁡A+cos⁡B)+sin⁡B(cos⁡A+cos⁡B)= \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= \sin A.\cos A + \sin A.\cos B + \sin B.\cos A + \sin B.\cos B
We know sin⁡2A=2sin⁡A.cos⁡A\sin 2A = 2\sin A.\cos A and sin⁡2B=2sin⁡B.cos⁡B\sin 2B = 2\sin B.\cos B, then using this we have
=sin⁡2A2+(sin⁡A.cos⁡B+sin⁡B.cos⁡A)+sin⁡2B2= \dfrac{{\sin 2A}}{2} + \left( {\sin A.\cos B + \sin B.\cos A} \right) + \dfrac{{\sin 2B}}{2}
Also we know sin⁡(A+B)=sin⁡A.cos⁡B+sin⁡B.cos⁡A\sin \left( {A + B} \right) = \sin A.\cos B + \sin B.\cos A, applying we have
=sin⁡2A2+sin⁡(A+B)+sin⁡2B2= \dfrac{{\sin 2A}}{2} + \sin \left( {A + B} \right) + \dfrac{{\sin 2B}}{2}
=sin⁡2A2+sin⁡2B2+sin⁡(A+B)= \dfrac{{\sin 2A}}{2} + \dfrac{{\sin 2B}}{2} + \sin \left( {A + B} \right)
Taking 12\dfrac{1}{2} common we have
=12(sin⁡2A+sin⁡2B)+sin⁡(A+B)= \dfrac{1}{2}\left( {\sin 2A + \sin 2B} \right) + \sin \left( {A + B} \right)
We know (sin⁡2A+sin⁡2B)=2sin⁡(A+B)cos⁡(A−B)\left( {\sin 2A + \sin 2B} \right) = 2\sin \left( {A + B} \right)\cos \left( {A - B} \right) then
=12(2sin⁡(A+B)cos⁡(A−B))+sin⁡(A+B)= \dfrac{1}{2}\left( {2\sin \left( {A + B} \right)\cos \left( {A - B} \right)} \right) + \sin \left( {A + B} \right)
Simplifying we have
=(sin⁡(A+B)cos⁡(A−B))+sin⁡(A+B)= \left( {\sin \left( {A + B} \right)\cos \left( {A - B} \right)} \right) + \sin \left( {A + B} \right)
Now taking sin⁡(A+B)\sin \left( {A + B} \right) common we have
=sin⁡(A+B)(cos⁡(A−B)+1)= \sin \left( {A + B} \right)\left( {\cos \left( {A - B} \right) + 1} \right)
Now from equation (5) we have cos⁡(A−B)=C2+D2−22\cos \left( {A - B} \right) = \dfrac{{{C^2} + {D^2} - 2}}{2}
=sin⁡(A+B)(C2+D2−22+1)= \sin \left( {A + B} \right)\left( {\dfrac{{{C^2} + {D^2} - 2}}{2} + 1} \right)
Taking LCM and simplifying we have
=sin⁡(A+B)(C2+D2−2+22)= \sin \left( {A + B} \right)\left( {\dfrac{{{C^2} + {D^2} - 2 + 2}}{2}} \right)
Thus we have
⇒CD=sin⁡(A+B)(C2+D22)\Rightarrow CD = \sin \left( {A + B} \right)\left( {\dfrac{{{C^2} + {D^2}}}{2}} \right)
Multiplying two on both sides we have
2CD=sin⁡(A+B)(C2+D2)2CD = \sin \left( {A + B} \right)\left( {{C^2} + {D^2}} \right)
⇒sin⁡(A+B)=C2+D22CD\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⁡2A+cos⁡2B{\sin ^2}A + {\cos ^2}B , then it is not equal to 1. Which goes the same for the sine double angle formula.