Sin a - sin b - - 2165-cos a - cos b - - 2765-cos a - b 2 -

The correct option is B 3 √(130) (cosα #160; +cosβ #160; ) ^ 2 + (sinα #160; +sinβ #160; ) ^ 2 = ( 27 65 ) ^ 2 + ( 21 65 ) ^ 2 ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

sin a + sin b...

Question

$sin a + sin b = - \frac{21}{65}, cos a + cos b = - \frac{27}{65} \Rightarrow cos (\frac{a - b}{2}) =$

- \frac{6}{65}

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

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

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\frac{6}{65}

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Solution

The correct option is B $- \frac{3}{\sqrt{130}}$
${(cos α + cos β)}^{2} + {(sin α + sin β)}^{2} = {(\frac{- 27}{65})}^{2} + {(\frac{- 21}{65})}^{2} \Rightarrow 2 + 2 cos (α - β) = \frac{1170}{65 \times 65} \Rightarrow 4 cos^{2} \frac{(α - β)}{2} = \frac{1170}{65} \Rightarrow cos (\frac{α - β}{2}) = \frac{- \sqrt{1170}}{130} = \frac{- 3}{\sqrt{130}}$

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sina+sinb=−2165,cosa+cosb=−2765⇒cos(a−b2)=

The correct option is B −3√130(cosα+cosβ)2+(sinα+sinβ)2=(−2765)2+(−2165)2⇒2+2cos(α−β)=117065×65⇒4cos2(α−β)2=117065⇒cos(α−β2)=−√1170130=−3√130

$sin a + sin b = - \frac{21}{65}, cos a + cos b = - \frac{27}{65} \Rightarrow cos (\frac{a - b}{2}) =$