What is the maximum and minimum value of 3cosx +4sinx+5
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-√a² + b² ≤ a sin A + b cos A ≤ √a² + b²
= - √4² + 3² ≤ 4 sin x + 3 cos x ≤√4² + 3²
= - 5 ≤ 4 sin x + 3 cos x ≤ 5
= -5 + 5 ≤ 4 sin x + 3 cos x ≤ 5 + 5
= 0 ≤ 4 sin x + 3 cos x ≤ 10
so the maximum value is 10 and the minimum value is 0.
= - √4² + 3² ≤ 4 sin x + 3 cos x ≤√4² + 3²
= - 5 ≤ 4 sin x + 3 cos x ≤ 5
= -5 + 5 ≤ 4 sin x + 3 cos x ≤ 5 + 5
= 0 ≤ 4 sin x + 3 cos x ≤ 10
so the maximum value is 10 and the minimum value is 0.
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Answer:
Step-by-step explanation:
How did the 5+5 came I m not understanding
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