First time with trigonometric identities

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March 7th 2009, 03:54 AM
mattty

First time with trigonometric identities

Hi, I've been going around in circles with this question, coming back to the original equation over and over because I can't honestly figure out what I'm supposed to be doing.
The question is:
Find all the values of x in the interval $[0,2\pi]$ such that $2sin^2x=1$

I realize that I am meant to be looking for values between 0 and $2\pi$ but I've no idea how to manipulate the trig identities in order to do that.

Any help would be appreciated.
March 7th 2009, 04:12 AM
running-gag

Hi

There are several ways to solve this equation
You can see that it is equivalent to $\sin x = \pm \frac{\sqrt{2}}{2}$
Draw a circle and you will find the solution between 0 and $2\pi$ : $\frac{\pi}{4} ; \frac{3\pi}{4} ; \frac{5\pi}{4} ; \frac{7\pi}{4}$

Another way is to use the trig identity $\cos(2x) = 1- 2 \sin^2x$
The equation is equivalent to $\cos(2x)=0$ that you surely know how to solve
March 7th 2009, 04:17 AM
skeeter

Quote:

Originally Posted by mattty

Hi, I've been going around in circles with this question, coming back to the original equation over and over because I can't honestly figure out what I'm supposed to be doing.
The question is:
Find all the values of x in the interval $[0,2\pi]$ such that $2sin^2x=1$

I realize that I am meant to be looking for values between 0 and $2\pi$ but I've no idea how to manipulate the trig identities in order to do that.

Any help would be appreciated.

$2\sin^2{x} = 1<br />$

$\sin^2{x} = \frac{1}{2}$

$\sin{x} = \pm \frac{1}{\sqrt{2}}$

$x = \frac{\pi}{4}$

$x = \frac{3\pi}{4}$

$x = \frac{5\pi}{4}$

$x = \frac{7\pi}{4}$
March 7th 2009, 04:34 AM
mattty

Quote:

Originally Posted by running-gag

Another way is to use the trig identity $\cos(2x) = 1- 2 \sin^2x$
The equation is equivalent to $\cos(2x)=0$ that you surely know how to solve

I would love to say that your absolutely right and i know exactly how to solve it from that point, but the sad fact is I have no clue at all. I know i have to use the identity method, but this was just covered in passing and pretty much assumed that we knew how to go about it. I never learnt about it though. So any basic tips or links would probably be most helpful at this point.
March 7th 2009, 04:41 AM
mattty

Quote:

Originally Posted by skeeter

$\sin^2{x} = \frac{1}{2}$

$\sin{x} = \pm \frac{1}{\sqrt{2}}$

Thanks skeeter, that's put me on the verge, if I could understand how the step I quoted is done i think it would help a lot, is that simply a square rooting of both sides? If not can you please explain it?
March 7th 2009, 04:47 AM
james_bond

Quote:

Originally Posted by mattty

Thanks skeeter, that's put me on the verge, if I could understand how the step I quoted is done i think it would help a lot, is that simply a square rooting of both sides? If not can you please explain it?

Do you know that: for all real numbers $x$ , $\sqrt{x^2} = \left|x\right| = \begin{cases} x, & \mbox{if }x \ge 0 \\ -x, & \mbox{if }x < 0. \end{cases}$ .
March 7th 2009, 04:55 AM
skeeter

another way to look at it ...

$2\sin^2{x} = 1$

$2\sin^2{x} - 1 = 0<br />$

factor ...

$(\sqrt{2} \sin{x} - 1)(\sqrt{2} \sin{x} + 1) = 0$

set each factor equal to 0 ...

$\sin{x} = \frac{1}{\sqrt{2}}$ ... $\sin{x} = -\frac{1}{\sqrt{2}}$
March 7th 2009, 05:02 AM
mattty

Quote:

Originally Posted by james_bond

Do you know that: for all real numbers $x$ , $\sqrt{x^2} = \left|x\right| = \begin{cases} x, & \mbox{if }x \ge 0 \\ -x, & \mbox{if }x < 0. \end{cases}$ .

Indeed I do, I think what I'm having trouble with is getting used to the notation of $sin^2x$ and what it means. I guess i thought if i took the square root of both sides then the answer would be $sin\sqrt{x} =\frac{ 1}{\sqrt{2}}$ rather than $sinx =\frac{ 1}{\sqrt{2}}$

Thanks for the help with the clarification.

Quote:

Originally Posted by skeeter

another way to look at it ...

thanks for spelling it out, i understand now, muchly appreciated.
March 7th 2009, 05:02 AM
running-gag

Quote:

Originally Posted by mattty

I would love to say that your absolutely right and i know exactly how to solve it from that point, but the sad fact is I have no clue at all. I know i have to use the identity method, but this was just covered in passing and pretty much assumed that we knew how to go about it. I never learnt about it though. So any basic tips or links would probably be most helpful at this point.

The solutions of
$\cos (x) = \cos (a)$ are
$x = a +2k \pi$ and $x = -a +2k \pi$

Here you have
$\cos (2x) = 0 = \cos \left(\frac{\pi}{2}\right)$

Then
$2x = \frac{\pi}{2} + 2k \pi$ or $2x = -\frac{\pi}{2} + 2k \pi$

$x = \frac{\pi}{4} + k \pi$ or $x = -\frac{\pi}{4} + k \pi$

Finally the solutions between 0 and $2\pi$ are
$\frac{\pi}{4} ; \frac{5\pi}{4}$ and $\frac{3\pi}{4} ; \frac{7\pi}{4}$
March 7th 2009, 11:34 AM
james_bond

Quote:

Originally Posted by mattty

Indeed I do, I think what I'm having trouble with is getting used to the notation of $sin^2x$ and what it means. I guess i thought if i took the square root of both sides then the answer would be $sin\sqrt{x} =\frac{ 1}{\sqrt{2}}$ rather than $sinx =\frac{ 1}{\sqrt{2}}$

$\sin^2x=\left(\sin x\right)^2$ just like $\ln^2 x=\left(\ln x\right)^2$ or $f^2(x)=\left(f(x)\right)^2$ (and of course the other trigonometric functions :)).