हिंदी

If cos α + cos β = 0 = sin α + sin β , then prove that cos 2 α + cos 2 β = − 2 cos ( α + β ) .

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प्रश्न

If  \[\cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta\] , then prove that \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .

 
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उत्तर

Given:  \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\]

\[\therefore \cos\alpha + \cos\beta = 0\]
Squaring on both sides, we get 

\[\cos^2 \alpha + \cos^2 \beta + 2\cos\alpha \cos\beta = 0 . . . . . \left( 1 \right)\]

Also,

\[\sin\alpha + \sin\beta = 0\] 

Squaring on both sides, we get
\[\sin^2 \alpha + \sin^2 \beta + 2\sin\alpha \sin\beta = 0 . . . . . \left( 2 \right)\]
Subtracting (2) from (1), we get
\[\left( \cos^2 \alpha + \cos^2 \beta + 2\cos\alpha \cos\beta \right) - \left( \sin^2 \alpha + \sin^2 \beta + 2\sin\alpha \sin\beta \right) = 0\]
\[ \Rightarrow \cos^2 \alpha - \sin^2 \alpha + \cos^2 \beta - \sin^2 \beta + 2\left( \cos\alpha \cos\beta - \sin\alpha \sin\beta \right) = 0\]
\[ \Rightarrow \cos2\alpha + \cos2\beta + 2\cos\left( \alpha + \beta \right) = 0\]
\[ \Rightarrow \cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\]

 

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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]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 45 | पृष्ठ ३०

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