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Question

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


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