Question

Sin5a*cos4a-cos5a*sin4a-sina

Answer 1

Ответ:0

Объяснение:

По формуле синуса разности sin(α-β)=sinαcosβ-cosβsinα;

(sin5α·cos4α-cos5α·sin4α)  - sinα ⇔ sin(5α-4α)-sinα=sinα-sinα=0

Answer 2

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Answer 3

To simplify the expression sin(5a)*cos(4a) - cos(5a)*sin(4a) - sin(a), we can use trigonometric identities.

Let's start by expanding sin(5a) and cos(5a) using the angle sum and difference identities:

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

Now we can substitute these values back into the original expression:

sin(5a)*cos(4a) - cos(5a)*sin(4a) - sin(a) = (sin(4a)cos(a) + cos(4a)sin(a))*cos(4a) - (cos(4a)cos(a) - sin(4a)sin(a))*sin(4a) - sin(a) = sin(4a)cos(a)cos(4a) + cos(4a)sin(a)cos(4a) - cos(4a)cos(a)sin(4a) + sin(4a)sin(a)sin(4a) - sin(a) = sin(4a)cos(a)cos(4a) + cos(4a)sin(a)cos(4a) - cos(4a)cos(a)sin(4a) + sin(4a)sin(a)sin(4a) - sin(a)

Next, we can simplify further by using trigonometric identities such as cos(a)cos(b) = (1/2)(cos(a-b) + cos(a+b)) and sin(a)sin(b) = (1/2)(cos(a-b) - cos(a+b)):

= sin(4a)(1/2)(cos(0) + cos(24a)) + cos(4a)(1/2)(cos(0) + cos(24a)) - cos(4a)(1/2)(cos(0) - cos(24a)) + sin(4a)(1/2)(cos(0) - cos(24a)) - sin(a) = sin(4a)(1/2)(1 + cos(8a)) + cos(4a)(1/2)(1 + cos(8a)) - cos(4a)(1/2)(1 - cos(8a)) + sin(4a)(1/2)(1 - cos(8a)) - sin(a) = (1/2)(sin(4a)(1 + cos(8a)) + cos(4a)(1 + cos(8a)) - cos(4a)(1 - cos(8a)) + sin(4a)(1 - cos(8a))) - sin(a) = (1/2)(sin(4a) + sin(4a)cos(8a) + cos(4a) + cos(4a)cos(8a) - cos(4a) + cos(4a)cos(8a) + sin(4a) - sin(4a)cos(8a)) - sin(a) = (1/2)(2sin(4a) + 2cos(4a)cos(8a)) - sin(a) = sin(4a) + cos(4a)cos(8a) - sin(a)

Therefore, the simplified form of the expression sin(5a)*cos(4a) - cos(5a