How do you solve #Csc^2(a/2)=2 Sec(a)#?

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Answer 1 · 2016-07-04T20:14:09

Let us solve the trigonometric equation by using trigonometric identities.

First of all, let us change cosecant and secant for inverses of sine and cosine, respectively:

#1/{sin^2 (a/2)} = 2 1/{cos(a)} rightarrow sin^2 (a/2) = 1/2 cos (a) rightarrow#
#rightarrow 2 sin^2 (a/2) = cos (a)#

Now, we shall use the definition of the half-angle sine:

#sin (alpha/2) = pm sqrt{{1-cos(alpha)}/2}#

So:

#sin^2 (alpha/2) = {1-cos(alpha)}/2#

And thus, our equation remains:

#cancel 2 cdot {1-cos(a)}/{cancel 2} = cos (a) rightarrow 1 - cos(a) = cos(a)#

And finally:

#1 = 2 cos(a) rightarrow cos(a) = 1/2 rightarrow #
#rightarrow a = cos^{-1} (1/2) = 60º equiv pi/3 " rad"#

Tip: let us check it:

#csc^2 (60/2) = 2 cdot sec (60) rightarrow 2^2 = 2 cdot 2 rightarrow 4 = 4#