sinx=p/q, find cosec2x in terms of p and q Watch

Khallil
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#21
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#21
(Original post by madnutcase434)
Thank you so much for your help. I think I've got it now. so if cos2x=1-(p2/q2), then cosx=((q2-p2)1/2)/q2 (with some rearranging) and then substitute that along with p/q (=sinx) into 1/(2sinxcosx) and voila!
Almost!

\cos^ x = 1 - \frac{p^2}{q^2} \\ \\ \implies \cos^2 x = \frac{q^2 - p^2}{q^2} \\ \\ \implies \cos x = \frac{\sqrt{q^2 - p^2}}{q}
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madnutcase434
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#22
Report Thread starter 5 years ago
#22
(Original post by Khallil)
Almost!

\cos^ x = 1 - \frac{p^2}{q^2} \\ \\ \implies \cos^2 x = \frac{q^2 - p^2}{q^2} \\ \\ \implies \cos x = \frac{\sqrt{q^2 - p^2}}{q}
oh yes! Got it now, thanks a lot!! Just a small silly mistake!
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