How do you use the sum or difference identities to find the exact value of #csc((5pi)/12)#?

1 Answer
turksvids
Dec 7, 2017

#csc((5pi)/12)=sqrt(6)-sqrt(2)#

Explanation:

First we need to figure out what sums to #(5pi)/12#:
#(5pi)/(12) = pi/6 + pi/4#

(I acutally find it easier to think of that in degrees: #75^circ=30^\circ+45^circ#.)

Now we'll find #sin((5pi)/12)# and then take its reciprocal.

#sin(x+y) = sin(x)cos(y)+cos(x)sin(y)#

#sin(pi/6 + pi/4) = sin(pi/6)cos(pi/4)+cos(pi/6)sin(pi/4)#
#= (1/2)(sqrt(2)/2)+(sqrt(3)/2)(sqrt(2)/2)#
#=sqrt(2)/4+sqrt(6)/4#
#sin((5pi)/12)=(sqrt(2)+sqrt(6))/4#

So the final thing to do is take the reciprocal:
#csc((5pi/12))=1/sin((5pi)/12)#
#=1/(((sqrt(2)+sqrt(6))/4))#
#=4/(sqrt(2)+sqrt(6))# (which might be enough)
#=(4(sqrt(2)-sqrt(6)))/((sqrt(2)+sqrt(6))(sqrt(2)-sqrt(6)))#
#=(4(sqrt(2)-sqrt(6)))/(2-6)#
#=-(sqrt(2)-sqrt(6))#
#=sqrt(6)-sqrt(2)#

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