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Show that (cosθ-cos3θ)(sin8θ+sin2θ)(sin5θ-sinθ)(cos4θ-cos6θ) = 1 - Mathematics

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प्रश्न

Show that `((cos theta -cos 3theta)(sin 8theta + sin 2theta))/((sin 5theta - sin theta) (cos 4theta - cos 6theta))` = 1

योग
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उत्तर

`((cos theta -cos 3theta)(sin 8theta + sin 2theta))/((sin 5theta - sin theta) (cos 4theta - cos 6theta)) = ((cos 3theta -cos theta)(sin 8theta + sin 2theta))/((sin 5theta - sin theta) (cos 6theta - cos 4theta))`

= `(2sin((3theta + theta)/2) sin ((theta - 3theta)/2) * 2sin((8theta + 20)/2) cos((8theta - 2theta)/2))/((2cos((5theta + theta)/2) * sin((5theta - theta)/2) * 2sin((6theta - 4theta)/2) sin((4theta  - 6theta)/2)`

= `(sin 2theta* sin(- theta) *sin 5theta * cos 3theta)/(cos3theta * sin 2theta * sin 5theta * sin (- theta))`

= 1

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Trigonometric Functions and Their Properties
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.6 [पृष्ठ १२१]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.6 | Q 6 | पृष्ठ १२१

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