2cos^2(2х+пи/3)-sin^2(х+пи/6)=2

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вопрос

To simplify this expression, we will use the trigonometric identities for cosine and sine.

Start with the given expression: 2cos^2(2x+π/3) - sin^2(x+π/6) = 2

Use the double angle identity for cosine: cos(2θ) = 2cos^2(θ) - 1
2cos^2(2x+π/3) = cos(2(2x+π/3)) = cos(4x + 2π/3) = 2cos^2(2x) - 1

Use the angle sum identity for sine: sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
sin(x+π/6) = sin(x)cos(π/6) + cos(x)sin(π/6) = sin(x)√3/2 + cos(x)1/2

Substituting these simplifications back into the original expression:
2cos^2(2x) - 1 - [sin(x)(√3/2) + cos(x)(1/2)]^2 = 2

Simplify the expression further:
2cos^2(2x) - 1 - [3sin^2(x)/4 + cos^2(x)/4 + √3sin(x)cos(x)/2] = 2

Putting it all together and simplifying further:
2cos^2(2x) - 1 - 3sin^2(x)/4 - cos^2(x)/4 - √3sin(x)cos(x)/2 = 2

Combining like terms and rearranging:
2cos^2(2x) - cos^2(x)/4 - 3sin^2(x)/4 - √3sin(x)cos(x)/2 = 3

Therefore, the simplified form of the given expression is:
2cos^2(2x) - cos^2(x)/4 - 3sin^2(x)/4 - √3sin(x)cos(x)/2 = 3