Question

Упростить:sin³(2a) cos(6a) + cos³(2a) sin(6a)​

Answer 1

sin³2α=sin2α·sin²2α=sin2α·(1-cos²2α)=sin2α-sin2αcos²2α

cos³2α=cos2α·cos²2α=cos2α·(1-sin²2α)=cos2α-cos2α·sin²2α

sin³2α cos6α + cos³2α sin6α=

=(sin2α-sin2αcos²2α)·cos6α+(cos2α-cos2α·sin²2α)·sin6α=

=sin2α·cos6α-sin2α·cos²2α·cos6α+cos2α·sin6α-cos2α·sin²2α·sin6α=

группируем

=(sin2α·cos6α+cos2α·sin6α)-(sin2α·cos²2α·cos6α+cos2α·sin²2α·sin6α)=

=sin8α-sin2α·cos2α·(cos2α·cos6α+sin2α·sin6α)=

=sin8α-(1/2)sin4α·cos4α=

=sin8α-(1/4)sin8α=(3/4)sin8α

​

Answer 2

Ответ:Sin³ 2a·Cos³ 2a·Cos 6a·Sin 6a = 1/8 Sin ³4a·1/2·Sin 12a= 1/16Sin³ 4a·Sin 12 a=

=1/16 Sin³4a·(3Sin 4a - 4Sin³4a) = 3/16Sin^4 4a - 1/4 Sin^6 4a

Answer 3

To simplify the given expression, we can use the trigonometric identity for the sum of cubes:

sin³(x) + cos³(x) = (sin(x) + cos(x)) * (1 - sin(x) * cos(x))

Let's apply this identity to the given expression:

sin³(2a) * cos(6a) + cos³(2a) * sin(6a)

= [sin(2a) + cos(2a)] * [1 - sin(2a) * cos(2a)] * sin(6a)

Now, we need to use another trigonometric identity:

sin(2x) = 2 * sin(x) * cos(x)

Substitute 2a for x:

sin(2a) = 2 * sin(a) * cos(a)

cos(2a) = cos²(a) - sin²(a)

Now the expression becomes:

[2 * sin(a) * cos(a) + (cos²(a) - sin²(a))] * [1 - (2 * sin(a) * cos(a)) * (cos²(a) - sin²(a))] * sin(6a)

Now let's further simplify:

[2 * sin(a) * cos(a) + cos²(a) - sin²(a)] * [1 - 2 * sin(a) * cos(a) * cos²(a) + 2 * sin(a) * cos(a) * sin²(a)] * sin(6a)

Use the trigonometric identity:

sin²(a) + cos²(a) = 1

Substitute in the expression:

[2 * sin(a) * cos(a) + cos²(a) - (1 - cos²(a))] * [1 - 2 * sin(a) * cos(a) * (1 - cos²(a)) + 2 * sin(a) * cos(a) * sin²(a)] * sin(6a)

Simplify further:

[2 * sin(a) * cos(a) + cos²(a) - 1 + cos²(a)] * [1 - 2 * sin(a) * cos(a) + 2 * sin(a) * cos(a) * sin²(a)] * sin(6a)

Combine like terms:

[4 * sin(a) * cos(a) + 2 * cos²(a) - 1] * [1 - 2 * sin(a) * cos(a) + 2 * sin(a) * cos(a) * sin²(a)] * sin(6a)

Now, the expression appears to be simplified as much as possible. If you have any further specific instructions or if you suspect there might be additional identities to apply, please let me know, and I can check it again!