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Expand cos(A + B + C). Hence prove that cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B, if A + B + C = π2

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प्रश्न

Expand cos(A + B + C). Hence prove that cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B, if A + B + C = `pi/2`

बेरीज
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उत्तर

Taking A + B = X and C = Y

We get cos(X + Y) = cos X cos Y – sin X sin Y

(i.e) cos(A + B + C) = cos(A + B) cos C – sin(A + B) sin C

= (cos A cos B – sin A sin B) cos C – [sin A cos B + cos A sin B] sin C

cos(A + B + C) = cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C

If (A + B + C) = `π/2` then cos(A + B + C) = 0

⇒ cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C = 0

⇒ cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin

C sin A cos B

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Trigonometric Functions and Their Properties
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometry - Exercise 3.4 [पृष्ठ १०९]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 3 Trigonometry
Exercise 3.4 | Q 8 | पृष्ठ १०९

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