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## Use the half angle formulas to determine.

A. sin22.5 degrees
and
B. tanpi/12 degrees

A) sin(22.5)degrees=sin(45/2)degrees
sin(45/2)=+sqrt[(1-cos(45))/2]

=+sqrt[(1-1/sqrt(2))/2]

=+sqrt[{(sqrt(2)-1)/sqrt(2)}/2]
=+sqrt[{2-sqrt(2)/2}/2]
=+sqrt[{2-sqrt(2)}/4]
=+sqrt[{2-1.414213}/2
=+sqrt{0.585786}/2
=+0.765366/2
=0.382683

B) tan pi/12 degrees
1 degree=pi/180
15pi/(15)(12)=15*(pi/180)=15 degrees
tan15 degrees=0.26794919
getting the answer using half-angle formulas gives us the following:
tan(30/2)=sqrt[(1-cos30)/(1+cos30)]
=sqrt[(1-sqrt(3)/2)/(1+sqrt(3)/2)]
simplify 1-sqrt(3)/2 and 1+sqrt(3)/2
1-sqrt(3)/2=[2-sqrt(3)]/2
1+sqrt(3)/2=[2+sqrt(3)]/2
divide[2-sqrt(3)]/2 by [2+sqrt(3)]/2
we get [2-sqrt(3)]/[2+sqrt(3)]
rationalize the denominator
we get [2-sqrt(3)][2-sqrt(3)]/[2-sqrt(3)][2+sqrt(3)]
this gives us [4-4sqrt(3)+3]/[4-3] which simplifies to
[7-4sqrt(3)]
[7-4sqrt(3)]=7-4(1.7320508)=7-6.9282032=0.0717968
now we take the square root of 0.0717968
sqrt(0.0717968)=0.26794919 which we got before
tan(pi/12)=0.26794919
cos45=1/√2
1-2sin222.5=cos45=1/√2

sin222.5=(2-√2)/2

sin22.5=√(2-√2)/√2

2tan(pi/12)/(1-tan2(pi/12))=tan(pi/6)=1/√3

1-tan2(pi/12)=2√3tan(pi/12)

tan2(pi/12)+2√3tan(pi/12)-1=0

tan(pi/12)=2-√3