3sin2a-4cos2a/5cos2a-sin2a

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вопрос

To simplify the expression (3sin(2a) - 4cos(2a)) / (5cos(2a) - sin(2a)), we can use the double angle identities for sine and cosine.

Recall that sin(2a) = 2sin(a)cos(a) and cos(2a) = cos^2(a) - sin^2(a).

Substitute these into the expression:

= 3(2sin(a)cos(a)) - 4(cos^2(a) - sin^2(a)) / 5(cos^2(a) - sin^2(a)) - sin(2a)

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

Now we can simplify further:

= 6sin(a)cos(a) - 4(cos^2(a) - sin^2(a)) / 5(cos^2(a) - sin^2(a)) - 2sin(a)cos(a)

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

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

There is no further simplification possible unless the specific values of a are provided.