# First time with trigonometric identities

• Mar 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 $\displaystyle [0,2\pi]$ such that $\displaystyle 2sin^2x=1$

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

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

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

Another way is to use the trig identity $\displaystyle \cos(2x) = 1- 2 \sin^2x$
The equation is equivalent to $\displaystyle \cos(2x)=0$ that you surely know how to solve
• Mar 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 $\displaystyle [0,2\pi]$ such that $\displaystyle 2sin^2x=1$

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

Any help would be appreciated.

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

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

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

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

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

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

$\displaystyle x = \frac{7\pi}{4}$
• Mar 7th 2009, 04:34 AM
mattty
Quote:

Originally Posted by running-gag
Another way is to use the trig identity $\displaystyle \cos(2x) = 1- 2 \sin^2x$
The equation is equivalent to $\displaystyle \cos(2x)=0$ 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.
• Mar 7th 2009, 04:41 AM
mattty
Quote:

Originally Posted by skeeter
$\displaystyle \sin^2{x} = \frac{1}{2}$

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

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?
• Mar 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 $\displaystyle x$, $\displaystyle \sqrt{x^2} = \left|x\right| = \begin{cases} x, & \mbox{if }x \ge 0 \\ -x, & \mbox{if }x < 0. \end{cases}$.
• Mar 7th 2009, 04:55 AM
skeeter
another way to look at it ...

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

$\displaystyle 2\sin^2{x} - 1 = 0$

factor ...

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

set each factor equal to 0 ...

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

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

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

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.
• Mar 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
$\displaystyle \cos (x) = \cos (a)$ are
$\displaystyle x = a +2k \pi$ and $\displaystyle x = -a +2k \pi$

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

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

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

Finally the solutions between 0 and $\displaystyle 2\pi$ are
$\displaystyle \frac{\pi}{4} ; \frac{5\pi}{4}$ and $\displaystyle \frac{3\pi}{4} ; \frac{7\pi}{4}$
• Mar 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 $\displaystyle sin^2x$ and what it means. I guess i thought if i took the square root of both sides then the answer would be $\displaystyle sin\sqrt{x} =\frac{ 1}{\sqrt{2}}$ rather than $\displaystyle sinx =\frac{ 1}{\sqrt{2}}$

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