Question:
The range of \(8\sin(\theta) + 6\cos(\theta) + 2\) is:
The range of \(8\sin(\theta) + 6\cos(\theta) + 2\) is:
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Adjust range calculations by considering the transformations applied to the trigonometric functions (scaling and translation).
Updated On: Mar 26, 2025
- (0,2)
- [2,102]
- (\(-\infty,\infty\))
- (2,1)
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The Correct Option is B
Solution and Explanation
For the trigonometric expression \(8\sin(\theta) + 6\cos(\theta)\), using the range:
\[
\sqrt{8^2 + 6^2} = 10
\]
\[
-10 \leq 8\sin(\theta) + 6\cos(\theta) \leq 10
\]
Adding 2:
\[
-8 \leq 8\sin(\theta) + 6\cos(\theta) + 2 \leq 12
\]
Thus, the range is \([2, 12]\).
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