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If Sin α = 4 5 and Cos β = 5 13 , Prove that Cos α − β 2 = 8 √ 65 - Mathematics

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प्रश्न

If  \[\sin \alpha = \frac{4}{5} \text{ and }  \cos \beta = \frac{5}{13}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \frac{8}{\sqrt{65}}\]

 
संख्यात्मक
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उत्तर

Given:

\[\sin \alpha = \frac{4}{5} \]

\[\cos \beta = \frac{5}{13}\] .

Now,

\[\cos\alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left( \frac{4}{5} \right)^2} = \frac{3}{5}\]
\[\text{ And } , \]
\[sin\beta = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left( \frac{5}{13} \right)^2} = \frac{12}{13}\]

Now,

\[\cos\left( \alpha - \beta \right) = cos\alpha \times cos\beta + sin\alpha \times sin\beta\]
\[ \Rightarrow \cos\left( \alpha - \beta \right) = \frac{3}{5} \times \frac{5}{13} \times \frac{4}{5} \times \frac{12}{13} = \frac{63}{65}\]
\[\text{ Thus } , \]
\[\cos\frac{\alpha - \beta}{2} = \sqrt{\frac{1 + \cos\left( \alpha - \beta \right)}{2}}\]
\[ = \sqrt{\frac{1 + \frac{63}{65}}{2}}\]
\[ = \frac{8}{\sqrt{65}}\]

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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 ३०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 43 | पृष्ठ ३०

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