If f(p) = sin^{p + 2}x + cos^{p + 2}x, then the value of 6f(2) - 4ƒ(4) + 10f(0) is:

Given:

f(p) = sin^{p + 2}x + cos^{p + 2}x

Calculation:

To find the value of 6f(2) - 4f(4) + 10f(0), first calculate f(p) for p = 2, p = 4, and p = 0:

For p = 2:

f(2) = sin⁴x + cos⁴x

For p = 4:

f(4) = sin⁶x + cos⁶x

For p = 0:

f(0) = sin²x + cos²x

We know that sin²x + cos²x = 1.

Using the identity for sin⁴x + cos⁴x:

sin⁴x + cos⁴x = (sin²x + cos²x)² - 2sin²x cos²x

sin⁴x + cos⁴x = 1 - 2sin²x cos²x

Using the identity sin²x cos²x = (1/4)sin²(2x):

sin⁴x + cos⁴x = 1 - (1/2)sin²(2x)

To find sin⁶x + cos⁶x, use the identity:

sin⁶x + cos⁶x = (sin²x + cos²x)³ - 3sin²x cos²x(sin²x + cos²x)

sin⁶x + cos⁶x = 1 - 3(sin²x cos²x)

sin⁶x + cos⁶x = 1 - (3/4)sin²(2x)

Now calculate 6f(2) - 4f(4) + 10f(0):

6f(2) = 6(sin⁴x + cos⁴x) = 6(1 - (1/2)sin²(2x))

4f(4) = 4(sin⁶x + cos⁶x) = 4(1 - (3/4)sin²(2x))

10f(0) = 10(sin²x + cos²x) = 10

Combine:

6f(2) - 4f(4) + 10f(0)

= 6[1 - (1/2)sin²(2x)] - 4[1 - (3/4)sin²(2x)] + 10

= 6 - 3sin²(2x) - 4 + 3sin²(2x) + 10

= 6 - 4 + 10

= 12

The value of 6f(2) - 4f(4) + 10f(0) is 12.