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Prove that cosθ cos ⁢𝜃2 −cos⁡3⁢𝜃⁢cos ⁢9⁢𝜃2 = sin 7θ sin 8θ. [Hint: Express L.H.S. = 12⁢[2⁢cos⁡𝜃⁢cos ⁢𝜃2−2⁢cos⁡3⁢𝜃⁢cos ⁢9⁢𝜃2] - Mathematics

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Question

Prove that cosθ `cos  theta/2 - cos 3theta cos  (9theta)/2` = sin 7θ sin 8θ.

[Hint: Express L.H.S. = `1/2[2costheta cos  theta/2 - 2 cos 3theta cos  (9theta)/2]`

Sum
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Solution

`LHS = costheta cos theta/2 - cos3theta cos 9theta/2 = 1/2 [2costhetacos  theta/2 - 2 cos 3theta cos (9theta)/2]`

Use 2 cos ⁡A cos ⁡B = cos⁡(A + B) + cos⁡(A − B):

`2 cos theta cos theta/2 = cos  (3theta)/2 + cos  theta/2`

`2 cos 3theta cos  (9theta)/2 = cos  (15theta)/2 + cos  (3theta)/2`

`LHS = 1/2 [cos  theta/2 - cos  (15theta)/2]`

Now use `cos C - cos D = -2sin  (C+D)/2 sin  (C-D)/2`

`LHS = 1/2 [-2sin ((theta/2+(15theta)/2)/2) sin  ((theta/2-(15theta)/2)/2)] = sin(4theta) sin ((7theta)/2)`

`cos theta cos  theta/2 - cos 3theta cos  (9theta)/2 = sin (4theta) sin ((7theta)/2)`

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Trigonometric Functions
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Chapter 3: Trigonometric Functions - Exercise [Page 53]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 3 Trigonometric Functions
Exercise | Q 6 | Page 53
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