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If Sin α + Sin β = a and Cos α + Cos β = B , Prove that (I) Sin ( α + β ) = 2 a B a 2 + B 2 - Mathematics

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प्रश्न

If \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha + \cos \beta = b\] , prove that 
(i)\[\sin \left( \alpha + \beta \right) = \frac{2ab}{a^2 + b^2}\]

संख्यात्मक
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उत्तर

The given equations are \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha + \cos \beta = b\]

(i) \[\because \text{ sin } C + \text{ sin } D = 2\sin\frac{C + D}{2}\cos\frac{C - D}{2}\]
\[ \therefore 2\sin\frac{\alpha + \beta}{2}\cos\frac{\alpha - \beta}{2} = a . . . (1)\] 

Now, using the identity

\[\text{ sin } C + \text{ sin } D = 2\sin\frac{C + D}{2}\cos\frac{C - D}{2}\]  for the LHS of \[\cos \alpha + \cos \beta = b\] , we get
\[2\cos\frac{\alpha + \beta}{2}\cos\frac{\alpha - \beta}{2} = b . . . (2)\]
On dividing (1) by (2), we get
\[\tan\frac{\alpha + \beta}{2} = \frac{a}{b}\]
We know,
\[sin\theta = \frac{2\tan\frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}\]
\[\therefore \sin\left( \alpha + \beta \right) = \frac{2\tan\left( \frac{\alpha + \beta}{2} \right)}{1 + \tan^2 \left( \frac{\alpha + \beta}{2} \right)}\]
\[ \Rightarrow \sin\left( \alpha + \beta \right) = \frac{2 \times \frac{a}{b}}{1 + \frac{a^2}{b^2}} = \frac{2ab}{a^2 + b^2}\]
 
 

 

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [पृष्ठ २९]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 38.1 | पृष्ठ २९

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