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Trigonometry
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If sinA=5/6, where angle A is in Quadrant 1, find the exact value of sin (1/2A)
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\[Sin(a)=5/6\] opp/hyp, and \[c^2 = a^2+b^2\] from pythagorean theorem, so \[a = \sqrt(c^2-b^2)\] or in this case \[\sqrt(6^2-5^2) = \sqrt(11)\ and\ \cos(\theta)=adj/hyp\ \ thus\ \ \sqrt(11)/6\] with all that in mind, use the half-angle identify for SIN
i never learned the half-angle O.O is there another way :( ... this is from my take-home test
\[sin(A/2)\] = \[\sqrt{(1-\cos(A))/2}\]
that's what you need cos() for and thus the adjacent side value from the pythagorean theorem
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