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# Show that max and min values of $8\cos \theta - 15\sin \theta$ are 17 and -17 respectively.  Answer Verified
Hint: Here, we will use the extreme values of the form $a\cos \theta + b\sin \theta$ to find the max and min values.

Given,
$8\cos \theta - 15\sin \theta \to (1)$
Let us compare the equation (1) with $a\cos \theta + b\sin \theta$, we get
$a = 8,b = - 15$
As, we know the maximum and minimum values of $a\cos \theta + b\sin \theta$ are $\sqrt {{a^2} + {b^2}}$ and -$\sqrt {{a^2} + {b^2}}$respectively.
Therefore, substituting the values of a and b, we get
$\Rightarrow \max = \sqrt {{a^2} + {b^2}} = \sqrt {{8^2} + {{( - 15)}^2}} = \sqrt {64 + 225} = \sqrt {289} = 17 \\ \Rightarrow \min = - \sqrt {{a^2} + {b^2}} = - \sqrt {{8^2} + {{( - 15)}^2}} = - \sqrt {64 + 225} = - \sqrt {289} = - 17 \\$
Hence, the maximum value of $8\cos \theta - 15\sin \theta$ is 17 and minimum value of
$8\cos \theta - 15\sin \theta$ is -17.
Note: The maximum and minimum of the $a\cos \theta + b\sin \theta$ will differ only by
the sign of the value i.e.., the maximum value will have the positive sign whereas the minimum value will have the negative sign of the same value.

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