# Thread: First time with trigonometric identities

1. ## 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.

2. 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

3. 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
$

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

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

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

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

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

$x = \frac{7\pi}{4}$

4. 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.

5. 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?

6. 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 $x$, $\sqrt{x^2} = \left|x\right| = \begin{cases} x, & \mbox{if }x \ge 0 \\ -x, & \mbox{if }x < 0. \end{cases}$.

7. another way to look at it ...

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

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

factor ...

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

set each factor equal to 0 ...

$\sin{x} = \frac{1}{\sqrt{2}}$ ... $\sin{x} = -\frac{1}{\sqrt{2}}$

8. 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.

Originally Posted by skeeter
another way to look at it ...
thanks for spelling it out, i understand now, muchly appreciated.

9. 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 $\frac{3\pi}{4} ; \frac{7\pi}{4}$

10. 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 ).