How to use double angle identities to find $\sin x$ and $\cos x$ from $\sin 2x$?

If $\sin 2x =\frac{5}{13}$ and $0^\circ < x < 45^\circ$, find $\sin x$ and $\cos x$.

The answers should be $\frac{\sqrt{26}}{26}$ and $\frac{5\sqrt{26}}{26}$

Ideas

The idea is to use double angle identities. One such identity is $\sin 2x=2\sin x\cos x$.

It's easy to use it to find $\sin 2x$ from known $\sin x$ and $\cos x$. But here it's the other way around.

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Do you know the double angle identities? If you write them out, they give you a formula for $\sin(2x)$ in terms of $\sin(x)$ and $\cos(x)$. You can put $\cos(x)$ in terms of $\sin(x$ using the identity $\sin^2+\cos^2=1$, then solve for $\sin$. – Potato Jun 21 '13 at 21:43

Because we know $\sin(2x) = 2\sin(x)\cos(x)$, it is like solving an equation: $u^2+v^2 = 1$ and $2uv = 5/13$, $u = \sin(x)$ and $v = \cos(x)$. Hope this helps.

EDIT: oh don't forget to take only the positive roots.

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Thanks That is where I was getting stuck though.... I know that as I work out the problem, I get to where (5/26) = sin(x) cos(x). Do I continue to make it (5/26) = sin(x) (sqrt 1- sin^2(x))? I'm not sure how to solve from there. Appreciate the help! – Mike Jun 21 '13 at 23:21
@Mike: That works fine. Now if you square it, you get a quadratic in $\sin^2 x$. Please, watch the parentheses-when you write "sqrt 1- sin^2(x)" you mean "sqrt (1- sin^2(x))" or $\sqrt {1- \sin^2(x)}$ but it looks like only the $1$ would be under the square root sign. You can see here for some hints on formatting equations. – Ross Millikan Jun 22 '13 at 3:17

HINT: $\cos ^2(x) = \dfrac{1}{2}(1 + \cos (2x))$, and $\cos (2x) = \dfrac{12}{13}$

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(Some might say that this is more of a "half angle" approach, but the two are intimately linked!) – The Chaz 2.0 Jun 21 '13 at 21:57
Thanks That is where I was getting stuck though.... I know that as I work out the problem, I get to where (5/26) = sin(x) cos(x). Do I continue to make it (5/26) = sin(x) (sqrt 1- sin^2(x))? I'm not sure how to solve from there. Appreciate the help! – Mike Jun 22 '13 at 0:09
That approach is more in line with the other answer/hint. For mine, just take the square root of $\dfrac{1}{2}(1+\dfrac{12}{13}))$. Then do something similar to find the sine. – The Chaz 2.0 Jun 22 '13 at 5:31

Suppose we had a right triangle with an angle $2x$, and $\sin2x=\frac{5}{13}$. Further suppose that the hypotenuse of the triangle was 13. We can deduce that the side oppoites $2x$ must be 5. Applying the Pythagorean theorem to find the other side we have $$13^2=5^2+(\text{adjacent side})^2$$ $$169-25=(\text{adjacent side})^2$$ $$144=(\text{adjacent side})^2$$ implying that the side opposite angle $2x$ is $12$. This allows us to state that $$\cos 2x=\frac{12}{13}$$ Which is easier to work with because $$\cos2x=\cos^2x-\sin^2x=2\cos^2x-1$$ Substituting we have $$\frac{12}{13}=2\cos^2x-1$$ $$\cos^2x=\frac{25}{26}$$ $$\cos x=\frac{5}{\sqrt{26}}=\frac{5\sqrt{26}}{26}$$ Applying the Pythagorean identity we have $$\cos^2x+\sin^2x=1$$ $$\frac{25}{26}+\sin^2x=1$$ $$\sin^2x=\frac{1}{26}$$ $$\sin x=\frac{1}{\sqrt{26}}=\frac{\sqrt{26}}{26}$$

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