Author Topic: Maximum, minimum of trig functions (Read 1675 times) Tweet Share

TrueTears · « **on:** December 25, 2008, 07:27:48 pm »

Well my book says to do this:

eg, find the maximum and minimum values of $cos^2 x + 2cos x +6$ .

first, let a = $cos x$ so the equation becomes: $a^2+2a+6$ . Then by completing the square we get $(a+1)^2 + 5$ . Sub $cos x$ back in, and we get $( cos x +1 ) ^2 +5$

Now -1<= cos x <= 1 so we sub the max value of $cos x$ , which is 1 into the equation
$( cos x +1 ) ^2 +5$ and we get 9. hence that's the max value of $cos^2 x + 2cos x +6$

Same process for minimum value, sub in -1 into $( cos x +1 ) ^2 +5$ and we get 5, which is the minimum value of $cos^2 x + 2cos x +6$ .

However there is a question which this method doesn't seem to work.
Find the max and min value of $sin^2 x +4$ . Let a = $sin x$ so equation becomes $a^2 + 4$

Now -1<= $sin x$ <=1 so sub 1 into $a^2 + 4$ and we get 5 as the max value. Then we sub -1 in and we also get 5 o.O. However the minimum value is 4. How do i arrive at the answer? Is there a way to do it analytically? Because graphically, i have not learnt how to sketch $sin^2 x$ without calc.

Many thanks.

phagist_ · « **Reply #1 on:** December 25, 2008, 07:30:31 pm »

The min value of $sin^2(x)$ is zero.

TrueTears · « **Reply #2 on:** December 25, 2008, 07:35:24 pm »

Quote from: phagist_ on December 25, 2008, 07:30:31 pm

The min value of $sin^2(x)$ is zero.

yeah but how do u work that out without sketching it on a calc. My book says to work out the max/min value of what is under the ^2. ie for $(sin x)^2 + 4$ , u would work out the max/min value for $sin x$ only and then plug those back in the equation. However here u get the same number for both max and min lol

humph · « **Reply #3 on:** December 25, 2008, 07:38:43 pm »

The range of $\sin x$ is $[-1,1]$ , so when you square any of these values, you'll get some number in the interval $[0,1]$ . That is, if $y \in [0,1]$ , then $y^2 \in [0,1]$ , and if $y \in [-1,0]$ , then $y^2 \in [0,1]$ .

phagist_ · « **Reply #4 on:** December 25, 2008, 07:40:49 pm »

You just have to know the max and min of the $sin^2(x)$ graph right?

and you know $sin(x)$ range is from $-1\le x\le1$ so if that function is squared you know (hopefully) that the new range is from $[0, 1]$ and thus you have your min/max.

humph · « **Reply #5 on:** December 26, 2008, 09:15:18 pm »

Quote from: Mao on December 26, 2008, 08:38:46 pm

Quote from: humph on December 25, 2008, 07:38:43 pm
The range of $\sin x$ is $[-1,1]$ , so when you square any of these values, you'll get some number in the interval $[0,1]$ . That is, if $y \in [0,1]$ , then $y^2 \in [0,2]$ , and if $y \in [-1,0]$ , then $y^2 \in [0,1]$ .

did you mean, 'if $y \in [0,1]$ , then $y^2 \in [0,\textbf{2}]$ '

Thanks, corrected

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Author Topic: Maximum, minimum of trig functions (Read 1675 times) Tweet Share

TrueTears

Maximum, minimum of trig functions

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TrueTears

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