Let - - be such that - - - - - - 3 - If sin -sin -21-65 and cos - - cos - -27-65-then the value of cos -2 is

Sin α+sin β = 21/65, cos α + cos β = 27/65 Now (sin α+sin β)^2+(cos α + cos β)^2 = ( 21/65 )^2+ ( 27/65 )^2 ⇒ 2+2 sin α sin β+2 cos α cos β=441/65^2+729/65^ ...

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Byju's Answer

Standard XIII

Mathematics

Trigonometric Ratios of Multiples of an Angle

Let α, β be s...

Question

$L e t α, β b e s u c h t h a t π < α - β < 3 π . I f s i n α + s i n β = \frac{21}{65} a n d c o s α + c o s β = - \frac{27}{65}, t h e n t h e v a l u e o f c o s \frac{α - β}{2} i s$

\frac{- 6}{65}

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\frac{3}{\sqrt{130}}

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\frac{6}{65}

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- \frac{3}{\sqrt{130}}

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Solution

The correct option is D $- \frac{3}{\sqrt{130}}$
$s i n α + s i n β = - \frac{21}{65}, c o s α + c o s β = - \frac{27}{65} N o w (s i n α + s i n β)^{2} + (c o s α + c o s β)^{2} = {(\frac{- 21}{65})}^{2} + {(\frac{- 27}{65})}^{2} \Rightarrow 2 + 2 s i n α s i n β + 2 c o s α c o s β = \frac{441}{65^{2}} + \frac{729}{65^{2}} \Rightarrow 2 + 2 [c o s (α - β)] = \frac{1170}{(65)^{2}} \Rightarrow 2.2 c o s^{2} (\frac{α + β}{2}) = \frac{1170}{(65)^{2}} \Rightarrow c o s (\frac{α - β}{2}) = \frac{3 \sqrt{130}}{130} = \frac{3}{\sqrt{130}} T h e r e f o r e c o s (\frac{α - β}{2}) = \frac{- 3}{\sqrt{130}}, {∵ \frac{π}{2} < \frac{α - β}{2} < \frac{3 π}{2}} .$

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Similar questions

Q. Let

α

and

β

be such that

π < α - β < 3 π

. If

sin α + sin β = - \frac{21}{65}

and

cos α + cos β = - \frac{27}{65}

, then the value of

cos \frac{(α - β)}{2}

Q. Let

α, β

be such that

0 < α - β - π < 2 π .

sin α + sin β = - \frac{21}{65}

and

cos α + cos β = - \frac{27}{65} .

Then the value of cos

\frac{α - β}{2}

is a

Q. Let

α, β

be such that

π < α - β < 3 π

. If

sin α + sin β = - \frac{21}{65}

and

cos α + cos β = - \frac{27}{65}

, then the value of

cos \frac{α - β}{2}

Q. Let

α, β

be such that

π < α - β < 3 π

. If

sin α + sin β = - \frac{21}{65} a n d cos α + cos β = - \frac{27}{65}

, then value of

cos \frac{α - β}{2}

equals

π < α - β < 3 π, sin α + sin β = \frac{- 21}{65},

cos α + cos β = \frac{- 27}{65}

. The value of

cos (\frac{α - β}{2})

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Let α, β be such that π<α−β<3π.Ifsin α+sin β=2165 and cos α+cos β=−2765,then the value of cosα−β2 is

$L e t α, β b e s u c h t h a t π < α - β < 3 π . I f s i n α + s i n β = \frac{21}{65} a n d c o s α + c o s β = - \frac{27}{65}, t h e n t h e v a l u e o f c o s \frac{α - β}{2} i s$