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If sinA=5/6, where angle A is in Quadrant 1, find the exact value of sin (1/2A)

Trigonometry
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1,172cm(2)
I assume that's sin (.5A). There is a half-angle formula for sin.
so can you please help me solve the problem?

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Other answers:

\[\cos^2A+\sin^2A=1\]\[\cos A=\sqrt{1-\sin^2A}\] From this \(\cos A\) can be found u can find \(\cos A\). Then use \[\cos A=1-2\sin^2(\frac{ 1 }{ 2 }A)\]\[\sin \frac{ 1 }{ 2 }A=\sqrt{\frac{ 1-\cos A }{ 2 }}\] from this u can find \[\sin \frac{ 1 }{2}A\] Does that help?
but do u have an answer?
Mine is at the very top
did u use the half-angle formula because i'm getting a radical?
what r u getting?
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getting it?
is that the complete answer?
yup
ok thank you so much
Welcome:) do u get it?
yes :)

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