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in Trigonometry by (45.0k points)
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The value of cos2\(\left(\frac{\pi}{6}+x \right)\) - sin2\(\left(\frac{\pi}{6}-x \right)\) is

A. \(\frac{1}{2}\) cos 2x

B. 0

C. \(-\frac{1}{2}\) cos 2x

D. \(\frac{1}{2}\)

1 Answer

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by (47.4k points)
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Best answer

Correct answer is A.

Given expression is  cos2\(\left(\frac{\pi}{6}+x \right)\) - sin2\(\left(\frac{\pi}{6}-x \right)\)

[using the identity sin2 x + cos2 x = 1]

[using the formula a2 + b2 = (a + b)2 – 2ab]

[Using the identity cos (A + B) – cos (A - B) = -2sinA sinB ]

[multiplying and dividing the term cos2 x with 2]

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