The minimum value of 3 cos x + 4 sin x + 5 is | vedclass

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(D) The expression is of the form $a \cos x + b \sin x + c$.We know that the range of $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$

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(D) The expression is of the form $a \cos x + b \sin x + c$.We know that the range of $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.Here,$a = 3$ and $b = 4$.Therefore,the range of $3 \cos x + 4 \sin x$ is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-\sqrt{9 + 16}, \sqrt{9 + 16}] = [-5, 5]$.The minimum value of $3 \cos x + 4 \sin x$ is $-5$.Thus,the minimum value of $3 \cos x + 4 \sin x + 5$ is $-5 + 5 = 0$.

The minimum value of $3 \cos x + 4 \sin x + 5$ is

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