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Prove that 32(3)sin π48 cos π48 cos π24 cos π12 cos π6 = 3

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प्रश्न

Prove that `32(sqrt(3)) sin  pi/48  cos  pi/48  cos  pi/24  cos  pi/12  cos  pi/6` = 3

योग
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उत्तर

`32sqrt(3)[sin  pi/48 xx cos  pi/48] = 16sqrt(3)[2sin  pi/48  cos  pi/48]`

= `16sqrt(3) sin  pi/24((2pi)/48 = pi/24)`

Now `16sqrt(3)[sin  pi/24 xx cos  pi/24]`

= `8sqrt(3)[2 sin  pi/24  cos  pi/24]`

= `8sqrt(3)[sin  (2pi)/24]`

= `8sqrt(3) sin  pi/12`

Now `8sqrt(3)[sin  pi/12  cos  pi/12]`

= `4sqrt(3)[2 sin  pi/12 cos  pi/12]`

= `4sqrt(3)[sin  (2pi)/12]`

= `4sqrt(3)(sin  pi/6)`

Now `4sqrt(3) sin  pi/6 cos  pi/6 = 2sqrt(3)[2sin  pi/6 cos  pi/6]`

`2sqrt(3)[sin  (2pi)/6] = 2sqrt(3) sin  pi/3`

= `2sqrt(3) xx sqrt(3)/2`

= 3

= R.H.S

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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 ११८]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.5 | Q 11 | पृष्ठ ११८

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