Answer:
Explanation:
The product of 2 - sinx and 3 + cosx is equal to (2 - sinx)(3 + cosx). We can find the maximum value of this expression by applying the product-to-sum identity:
(2 - sinx)(3 + cosx) = 6 + 2cosx - 3sinx - sinxcosx
We can then use the double-angle formula for sine and cosine to rewrite the expression as follows:
6 + 2cosx - 3sinx - sinxcosx = 6 + 2cosx - 3sinx - (2sinxcosx)/2
= 6 + 2cosx - 3sinx/2 - sinxcosx/2
= 6 + (cosx - sinx/2) - (sinxcosx/2)
We can now apply the sum-to-product identity to the first pair of terms, and the difference-to-product identity to the second pair of terms:
= (6 + cosx - sinx/2) + (cosx + sinx/2)
= (7 + cosx) + (cosx + sinx/2)
= 14 + 2cosx + sinx/2
This expression is equal to 14 + 2cosx + sinx/2. Since the sine function has a range of [-1, 1] and the cosine function has a range of [-1, 1], the maximum possible value of this expression is 14 + 2*1 + 1/2 = 15.5. Therefore, the greatest possible value for the product of 2 - sinx and 3 + cosx is 15.5.