Question

# If $$\sin \theta + \sin \phi = a$$ and $$\cos \theta + \cos \phi = b$$, then

A
cosθϕ2=±12a2+b2
B
cosθϕ2=±12a2b2
C
tanθϕ2=±4a2b2a2+b2
D
cos(θϕ)=a2+b222

Solution

## The correct options are A $$\cos \dfrac {\theta - \phi}{2} = \pm \dfrac {1}{2} \sqrt {a^{2} + b^{2}}$$ B $$\cos (\theta - \phi) = \dfrac {a^{2} + b^{2} - 2}{2}$$ D $$\tan \dfrac {\theta - \phi}{2} = \pm \sqrt {\dfrac {4 - a^{2} - b^{2}}{a^{2} + b^{2}}}$$$$\sin { \theta } +\sin { \phi } =a$$ and $$\cos { \theta } +\cos { \phi } =b$$Squaring both equations we get$$\sin ^{ 2 }{ \theta } +\sin ^{ 2 }{ \phi } +2\sin { \theta } \sin { \phi } ={ a }^{ 2 }$$ ....... $$(i)$$$$\cos ^{ 2 }{ \theta } +\cos ^{ 2 }{ \phi } +2\cos { \theta } \cos { \phi } ={ b }^{ 2 }$$ ....... $$(ii)$$Adding equations $$(i)$$ and $$(ii)$$, we get$$\left( { \sin ^{ 2 }{ \theta } }+{ \cos ^{ 2 }{ \theta } } \right) +\left( { \sin ^{ 2 }{ \phi } }+{ \cos ^{ 2 }{ \phi } } \right) +2\sin { \theta } \sin { \phi } +2\cos { \theta } \cos { \phi } ={ a }^{ 2 }+{ b }^{ 2 }$$$$2+2\left[ \cos { \theta } \cos { \phi } +\sin { \theta } \sin { \phi } \right] ={ a }^{ 2 }+{ b }^{ 2 }$$$$2+2\left[ \cos { \left( \theta -\phi \right) } \right] ={ a }^{ 2 }+{ b }^{ 2 }$$ ....... $$(iii)$$$$2+2\left[ 2\cos ^{ 2 }{ \left( \cfrac { \theta -\phi }{ 2 } \right) -1 } \right] = { a }^{ 2 }+{ b }^{ 2 }\left[ \because \cos { 2\theta } =2\cos ^{ 2 }{ \theta } -1 \right]$$$$4\cos^{ 2 }\left( \cfrac { \theta -\phi }{ 2 } \right) = { a }^{ 2 }+{ b }^{ 2 }$$$$\cos { \left( \cfrac { \theta -\phi }{ 2 } \right) =\pm \cfrac { 1 }{ 2 } } \sqrt { { a }^{ 2 }+{ b }^{ 2 } }$$ which is option (A)Using equation $$(iii)$$$$2+2\cos { \left( \theta -\phi \right) } ={ a }^{ 2 }+{ b }^{ 2 }$$$$\cos { \left( \theta -\phi \right) } =\cfrac { { a }^{ 2 }+{ b }^{ 2 }-2 }{ 2 }$$ which is option (D)$$\sin^{ 2 } \left( \cfrac { \theta -\phi }{ 2 } \right) = 1-\cos^{ 2 }\left( \cfrac { \theta -\phi }{ 2 } \right) =1-\cfrac { 1 }{ 4 } \left( { a }^{ 2 }+{ b }^{ 2 } \right)$$$$\sin { \left( \cfrac { \theta -\phi }{ 2 } \right) = } \pm \cfrac { 1 }{ 2 } \sqrt { 4-{ a }^{ 2 }-{ b }^{ 2 } }$$$$\tan { \left( \cfrac { \theta -\phi }{ 2 } \right) =\cfrac { \sin { \left( \cfrac { \theta -\phi }{ 2 } \right) } }{ \cos { \left( \cfrac { \theta -\phi }{ 2 } \right) } } = } \cfrac { \pm \cfrac { 1 }{ 2 } \sqrt { 4-{ a }^{ 2 }-{ b }^{ 2 } } }{ \pm \cfrac { 1 }{ 2 } \sqrt { { a }^{ 2 }+b^{ 2 } } }$$$$\tan { \left( \cfrac { \theta -\phi }{ 2 } \right) = } \pm \sqrt { \cfrac { 4-{ a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 }+b^{ 2 } } }$$ which is option (C)Mathematics

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