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Find the value of cos 2A, A lies in the first quadrant, when cos A = 1517

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प्रश्न

Find the value of cos 2A, A lies in the first quadrant, when cos A = `15/17`

योग
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उत्तर

we know sin2A + cos2A = 1

sin2A = 1 – cos2A

= `1 - (15/17)^2`

= `1 - 225/289`

= `(289 - 225)/289`

sin2A = `64/289`

sin A = `+- sqrt(64/289)`

= `+- 8/17`

Since A lies in the first quadrant, sin A is positive

∴ sin A = `8/17`

cos 2A = cos2A – sin2A

= `(15/17)^2 - 64/289`

=`225/289 - 64/289`

= `(225- 64)/289`

= `161/289`

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Trigonometric Functions and Their Properties
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.5 [पृष्ठ ११७]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.5 | Q 1. (i) | पृष्ठ ११७

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