Prove that: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]

\[LHS = \left( cos \alpha + cos\beta \right)^2 + \left( sin\alpha + sin\beta \right)^2 \] \[ = \cos^2 \alpha + \cos^2 \beta + 2cos\alpha cos\beta + \sin^2 \alpha + \sin^2 \beta + 2sin\alpha sin\beta\] \[ = ( \cos^2 \alpha + \sin^2 \alpha) + ( \cos^2 \beta + \sin^2 \beta) + 2\left( cos\alpha cos\beta + sin\alpha sin\beta \right)\] \[ = 1 + 1 + 2\cos(\alpha - \beta)\] \[ = 2\left\{ 1 + \cos(\alpha - \beta) \right\}\] \[ = 2\left\{ 2 \cos^2 \left( \frac{\alpha - \beta}{2} \right) \right\}\] \[ = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right) = RHS\] \[\text{ Hence proved } .\]

Prove That: ( Cos α + Cos β 2 ) + ( Sin α + Sin β ) 2 = 4 Cos 2 ( α − β 2 ) - Mathematics

प्रश्न

Prove that: $(\cos α + \cos β^{2}) + {(\sin α + \sin β)}^{2} = 4 \cos^{2} (\frac{α - β}{2})$

संख्यात्मक

उत्तर

$L H S = {(c o s α + c o s β)}^{2} + {(s i n α + s i n β)}^{2}$

$= \cos^{2} α + \cos^{2} β + 2 c o s α c o s β + \sin^{2} α + \sin^{2} β + 2 s i n α s i n β$

$= (\cos^{2} α + \sin^{2} α) + (\cos^{2} β + \sin^{2} β) + 2 (c o s α c o s β + s i n α s i n β)$

$= 1 + 1 + 2 \cos (α - β)$

$= 2 {1 + \cos (α - β)}$

$= 2 {2 \cos^{2} (\frac{α - β}{2})}$

$= 4 \cos^{2} (\frac{α - β}{2}) = R H S$

$Hence proved .$

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 11 Q 10 Q 12

पाठ 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [पृष्ठ २८]

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आरडी शर्मा Mathematics [English] Class 11

पाठ 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 11 | पृष्ठ २८

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Prove That: ( Cos α + Cos β 2 ) + ( Sin α + Sin β ) 2 = 4 Cos 2 ( α − β 2 ) - Mathematics

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Prove That: ( Cos α + Cos β 2 ) + ( Sin α + Sin β ) 2 = 4 Cos 2 ( α − β 2 ) - Mathematics

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