Ivy and Andrey were asked to find an explicit formula for the sequence -100,-50,0,50,...−100,−50,0,50,...minus, 100, comma, minu

Response:

Detailed breakdown:

Answer:

Step-by-step explanation:

The trigonometric expression given is:

11 cos² x +3 sin² x + 6sin x cos x + 5

Alternatively, we can present it as:

(9 cos² x + 2 cos² x) + (2 sin² x + sin² x) + 6sin x cos x + 5

After rearranging, the expression can be rewritten as:

(9 cos² x + sin² x + 6sin x cos x) + (2 cos² x + 2 sin² x) + 5

If we factor the highlighted part similarly to a polynomial:

(9 cos² x + sin² x + 6sin x cos x) + (2 cos² x + 2 sin² x) + 5

This results in:

(3 cos x + sin x)² + 2 (cos² x + sin² x) + 5

Notably, the term in the second set of brackets (cos² x + sin² x) is a well-known trigonometric identity and equals one.

Consequently, the expression simplifies to:

(3 cos x + sin x)² + 7

The maximum and minimum values of the entire expression rely on the max and min values of (3 cos x + sin x), which follows the format (a cos x + b sin x).

The max and min values can be easily determined.

I've included a screenshot from related material below:

Here, a=3 and b=1, thus, R= √10

As the cosine value for any angle ranges between -1 and 1, the value of cos(x − α) will also fall within this range.

This means the max and min for (a cos x + b sin x) will be -R to R, and all resultant values will be between those limits.

In this case, we observe that it falls between (-√10) and √10.

Returning to our original expression:

(3 cos x + sin x)² + 7

The bracketed term ranges between (-√10) and √10.

Nevertheless, since squaring a negative value yields a positive result, we cannot use a negative value for determining the minimum. The minimum occurs at the lowest non-negative value, which is zero.

Thus, the minimum value is:

(0)² + 7 = 7

And the maximum is:

(√10)² + 7 = 17