Sin2a*cos5a-sina*cos6a

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вопрос

To simplify the expression sin(2a) cos(5a) - sin(a) cos(6a), we can apply the product-to-sum identities and difference-to-product identities for trigonometric functions.

sin(2a) = 2sin(a)cos(a
cos(5a) = cos(4a + a) = cos(4a)cos(a) - sin(4a)sin(a) = cos(a)cos(4a) - sin(a)sin(4a
cos(6a) = cos(5a + a) = cos(5a)cos(a) - sin(5a)sin(a) = (cos(a)cos(4a) - sin(a)sin(4a))cos(a) - sin(a)sin(4a)cos(a)

Now, substitute these values into the expression:

2sin(a)cos(a) (cos(a)cos(4a) - sin(a)sin(4a)) - sin(a) ((cos(a)cos(4a) - sin(a)sin(4a))cos(a) - sin(a)sin(4a)cos(a))

Expand and simplify the expression further:

2sin(a)cos(a) cos(a)cos(4a) - 2sin(a)cos(a) sin(a)sin(4a) - sin(a) cos(a)cos(4a) + sin(a)^2sin(4a) cos(a
= 2sin(a)cos^2(a)cos(4a) - 2sin(a)sin(4a)cos(a) - sin(a)cos(a)cos(4a) + sin(a)^2sin(4a)cos(a
= 2sin(a)cos(4a)cos^2(a) - 2sin(a)sin(4a)cos(a) - sin(a)cos(4a)cos(a) + sin(a)^2sin(4a)cos(a
= cos(4a)(2sin(a)cos^2(a) - sin(a)cos(a)) - sin(4a)(2sin(a)cos(a) - sin(a)cos(a)
= cos(4a)sin(a) - sin(4a)sin(a
= sin(a)[cos(4a) - sin(4a)]

Therefore, sin(2a) cos(5a) - sin(a) cos(6a) simplifies to sin(a)[cos(4a) - sin(4a)].