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Find the maximum and minimum value of $$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$.

i simplified and reach to expression as follows :

$5 + 2\sin(x)[\sin(x)-3\cos(x)]$. How do i go from here?

Thanks

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    Can we use differentiation? – 2017-02-08
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    take $\frac{d}{dx}(\text{Your last expression})=0$ – 2017-02-08
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    If not, there is a clear way to convert it into a linear combination of sin(2x) and cos(2x). – 2017-02-08

4 Answers 4

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The expression can be simplified to $$\cos^2 x + \sin^2 x +2\sin^2 x +2 -3\sin 2x $$ $$=3 -3\sin 2x + 2\frac {1-\cos 2x}{2} $$ $$=4 - 3\sin 2x - \cos 2x $$

Now we know that $$-\sqrt{a^2 +b^2} \leq a\sin \alpha + b \cos \alpha \leq \sqrt {a^2 +b^2} $$

Hope you can take it from here.

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    But how do i proceed from my solution or without changing to sin2x or cos2x – 2017-02-08
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$$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$ $$-3\sin(2x)+2\dfrac{1-\cos2x}{2}+3$$ $$-3\sin(2x)-\cos(2x)+4$$ now use $$|a\sin\alpha+b\cos\alpha|\leq\sqrt{a^2+b^2}$$

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HINT:

Let $y=\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$

Divide both sides by $\cos^2x$ and use $\sec^2x=1+\tan^2x$ to form a Quadratic Equation in $\tan x$

Now as $\tan x$ is real, the discriminant must be $\not<0$

Or divide both sides by $\sin^2x$ to form a Quadratic Equation in $\cot x$

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    Aha, nice idea :) – 2017-02-08
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    I get y $\in [4+\sqrt10,4-\sqrt10]$. but how do i find max and min value in this interval? – 2017-02-09
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    @TaylorTed, You mean $$4-\sqrt{10}\le y\le4+\sqrt{10}$$ right? So, these the extreme two values of $y$ – 2017-02-09
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    okay i was thinking if there was some point in middle of domian where y attains max. but i found range itself – 2017-02-09
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$$ \begin{align} \frac{d}{dx} \cos^2x-6\sin(x)\cos(x)+3\sin^2x+2 &= -6 \cos(2 x) + 2 \sin(2 x) \end{align} $$

as a result of a simple differential calculation; now you'd like the $RHS$, namely $ -6 \cos(2 x) + 2 \sin(2 x)$, to be $0$ as to find the local minima and maxima of the function.

$$ \begin{align} -6 \cos(2 x) + 2 \sin(2 x) &= 0 \iff \\ x &= \pi n + \arctan \big( \frac{1}{3} (-1 - \sqrt{10} \big) \bigvee \\ x &= \pi n + \arctan \big( \frac{1}{3} (-1 + \sqrt{10} \big) \\ \forall &n \in \mathbb{Z} \end{align} $$

From here you can find the absolute minima and maxima by several methods.