How do you evaluate #\sin ( \frac { 3\pi } { 8} ) + \sin ( \frac { 3\pi } { 8} )#?

2 Answers
Jim G.
Jul 4, 2017

#sqrt(2+sqrt2)#

Explanation:

#sin((3pi)/8)+sin((3pi)/8)=2sin((3pi)/8)#

#"using the "color(blue)"half angle formula for sin"#

#•color(white)(x)sin(theta/2)=+-sqrt((1-costheta)/2)#

#"here " theta/2=(3pi)/8rArrtheta=(6pi)/8=(3pi)/4#

#•color(white)(x)cos((3pi)/4)=-cos(pi/4)#

#sin((3pi)/8)=+-sqrt((1-cos((3pi)/4))/2)#

We only require the positive root since #(3pi)/8# is in the first quadrant.

#=sqrt((1+cos(pi/4))/2)=sqrt((1+1/sqrt2)/2#

#=sqrt((sqrt2+1)/(2sqrt2))=sqrt((2+sqrt2)/4)=(sqrt(2+sqrt2))/2#

#rArr2sin((3pi)/8)=cancel(2)xxsqrt(2+sqrt2)/cancel(2)=sqrt(2+sqrt2)#

Nghi N
Jul 5, 2017

#sqrt(2 + sqrt2)#

Explanation:

Find sin ((3pi)/8) by using trig identity:
#2sin^2 a = 1 - cos 2a#
In this case --> #cos 2a = cos ((3pi)/4) = - sqrt2/2#
#2sin^2 ((3pi)/8) = 1 - cos ((3pi)/4) = 1 + sqrt2/2 = (2 + sqrt2)/2#
#sin^2 ((3pi)/8) = (2 + sqrt2)/4#
#sin ((3pi)/8) = +- sqrt(2 + sqrt2)/2#.
Since #sin ((3pi)/8)# is positive, take the positive value.
#sin ((3pi)/8) = sqrt(2 + sqrt2)/2#
Finally,
#sin ((3pi)/8) + sin ((3pi)/8) = (2(2 + sqrt2))/2 = sqrt(2 + sqrt2#

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