3ctg60(sin310cos70-sin70cos310)

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вопрос

To simplify the given expression, we can first use the formula "sin(A-B) = sinAcosB - cosAsinB":

3ctg60(sin(310)cos(70) - sin(70)cos(310))

Now, we can use the fact that cotangent is the reciprocal of tangent, i.e., ctg(x) = 1/tan(x):

3(1/tan(60))(sin(310)cos(70) - sin(70)cos(310))

Since tan(60) = sqrt(3), we can simplify further:

3(sqrt(3))(sin(310)cos(70) - sin(70)cos(310))

Now, we use the fact that sin(310) = sin(360-50) = -sin(50) and cos(310) = cos(360-50) = cos(50):

3(sqrt(3))(-sin(50)cos(70) - sin(70)cos(50))

Now, we can rewrite sin(50)cos(70) as cos(70)sin(50) using the commutative property of multiplication:

3(sqrt(3))(-cos(70)sin(50) - sin(70)cos(50))

Therefore, the simplified expression is:

-3sqrt(3)(cos(70)sin(50) + sin(70)cos(50))