Question

if sinx=5/13 and in quadrant 1 then sin x/2=
please explain

Answer 1

\[\sin(x) = \frac{5}{13}\] graphing this, it would look like \

Answer 2

cool i got that so far

Answer 3

Therefore, to find \(\sin\left(\frac{x}{2}\right)\) we need to know the half angle identity of sine. That is, \[\sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1-\cos(x)}{2}}\]

Since we have our triangle to work with, we can use that to find \(\cos(x)\).

Remember that with right triangles we can apply the pythagorean theorem and SOHCAHTOA. Therefore \(\cos(x) = \dfrac{ \sf adj}{\sf hyp}\)

Answer 4

supp

Answer 5

Yes i knew we needed a half angle identity. I couldnt find it in my notes thanks for that.

Answer 6

Going back to our triangle, we can apply the pythagorean theorem to solve for the missing side. We're given the hypotenuse and a side, therefore, \[c^2 = a^2+b^2 \implies b^2 = c^2 -a^2 \implies b = \sqrt{13^2-5^2} = \sqrt{144} = 12\]

Answer 7

So now we can use this to find \(\cos(x)\). \[\sf \cos(x) = \frac{adj}{hyp} \implies \cos(x) = \frac{12}{13}\]

Answer 8

Soh Cah Toa c;

Answer 9

Now we just plug this into our equation and solve :) because we're given that it's in quadrant 1, we're going to take the positive answer for this. \[\sin\left(\frac{x}{2}\right) =+\sqrt{\frac{1-\cos(x)}{2}} = +\sqrt{\frac{1-\dfrac{12}{13}}{2}} = +\sqrt{\dfrac{\frac{1}{13}}{2}} =+\sqrt{\frac{1}{26}} \]

Answer 10

Awesome!

Answer 11

@Jhannybean can you help me still