Question

Sind cos3d+cos dsin3d

Answer 1

2sin((3d-d)/2)cos((3d+d)/2)/(-2sin((3d-d)/2)sin((3d-d)/2)) 
в результате отнимания и деления получается: 
2sind cos2d/(-2sind sin2d)= -cos2d/sin2d=-ctg2d

Answer 2

The expression you provided appears to be a combination of trigonometric functions. To simplify it, let's break it down step by step using some trigonometric identities.

1. Recall the trigonometric identity: cos(A + B) = cosA * cosB - sinA * sinB
2. In your expression, let A = 3d and B = d. We have:

cos(3d + d) = cos(3d) * cos(d) - sin(3d) * sin(d)

3. Next, we need to use the following trigonometric identities:

   • cos(2A) = cos^2(A) - sin^2(A)
   • sin(2A) = 2 * cos(A) * sin(A)

4. We'll apply the double-angle identities to sin(3d):

sin(3d) = sin(2 * 1.5d) = 2 * cos(1.5d) * sin(1.5d)

5. Now, we can substitute the double-angle identities into our original expression:

cos(3d + d) = cos(3d) * cos(d) - 2 * cos(1.5d) * sin(1.5d) * sin(d)

6. We also need to use the identity:

   • sin(A + B) = sinA * cosB + cosA * sinB

7. Applying the sum-to-product identity to cos(3d + d):

cos(3d + d) = cos(3d) * cos(d) - 2 * sin(1.5d) * sin(d) * cos(1.5d) + 2 * sin(1.5d) * sin(d) * cos(1.5d)

8. Finally, we notice that the second term cancels out with itself, and we are left with:

cos(3d + d) = cos(3d) * cos(d)

So, the simplified expression for cos(3d + d) is cos(3d) * cos(d).