Trigonometry
posted by Dirk .
I need to prove that the following is true. Thanks.
csc^2(A/2)=2secA/secA1
Right Side=(2/cosA)/(1/cosA  1)
= (2/cosA)/[(1cosA)/cosA]
=2/cosA x (cosA)/(1cosA)
=2/(1cosA)
now recall cos 2X = cos^2 X  sin^2 X and we could say
cos A = cos^2 A/2  sin^2 A/2 and of course sin^2 A/2 + cos^2 A/2=1
so the above 2/(1  cosA)
=2/[sin^2 A/2 + cos^2 A/2  (cos^2 A/2  sin^2 A/2)]
= 2/(2sin^2 A/2)
= csc^2 A/2
= Left Side
Ok, so just using a different form of the double/halfangle formula allowed you to get the sin and therein getting csc.
Thanks Reiny! For a detailed how to and detailed steps. I will be using this service in the future. Thanks again.
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