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Write the maximum and minimum values of 3 cos x + 4 sin x + 5. |
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Answer» the maximum value of (a cosx + bsinx) = \(\sqrt{a^2+b^2}\) So maximum value = \(\sqrt{9+16}\) = 5 the minimum value of (a cosx + b sinx) = - \(\sqrt{a^2+b^2}\) minimum value = - 5 |
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