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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$\begingroup$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.
$\endgroup$ 2 $\begingroup$HINT: $\cos ^2(x) = \dfrac{1}{2}(1 + \cos (2x))$, and $\cos (2x) = \dfrac{12}{13}$
$\endgroup$ 3 $\begingroup$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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