Question

The range of \(8\sin(\theta) + 6\cos(\theta) + 2\) is:

• A. (0,2)
• B. [2,102]
• C. (\(-\infty,\infty\))
• D. (2,1)

Hint:

Adjust range calculations by considering the transformations applied to the trigonometric functions (scaling and translation).

Answer 1

The Correct Option is B

The range of the trigonometric expression \(8\sin(\theta) + 6\cos(\theta)\) is determined as follows: \[ \sqrt{8^2 + 6^2} = 10 \] Therefore, the bounds for the expression are: \[ -10 \leq 8\sin(\theta) + 6\cos(\theta) \leq 10 \] When 2 is added to the expression: \[ -10 + 2 \leq 8\sin(\theta) + 6\cos(\theta) + 2 \leq 10 + 2 \] \[ -8 \leq 8\sin(\theta) + 6\cos(\theta) + 2 \leq 12 \] The resulting range for \(8\sin(\theta) + 6\cos(\theta) + 2\) is \([2, 12]\).