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
