Cos^2 3pi/8 how do i find the exact value of the trig function with the denominator being a 4? i’m not using half angle formula.

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Answer 1 · 2018-06-09T15:14:24

#cos ^2 ({3pi}/8)= 1/2 (1 + cos ({3pi}/4)) = 1/4(2 - sqrt{2})#

We'll use the double angle formula, probably cheating because it's pretty much the same as the half angle formula.

#cos 2 theta = cos(theta + theta) = cos theta cos theta - sin theta sin theta = 2 cos ^2 theta - 1 #

#2cos^2 theta = 1 + cos 2 theta#

#cos ^2 theta = 1/2 ( 1 + cos 2 theta) #

#cos ^2 ({3pi}/8)= 1/2 (1 + cos ({3pi}/4)) = 1/2(1 + -sqrt{2}/2) = 1/4(2 - sqrt{2})#

There's your denominator.

Answer 2 · 2018-06-09T15:37:13

# cos^2((3pi)/8)=(2-sqrt2)/4#

We have to find exact value of #cos^2((3pi)/8)# ,without using half angle formula.

So, we use :

#color(blue)(sinC-sinD=2cos((C+D)/2)sin((C-D)/2)#

Subst. #C=pi/2 and D=pi/4#

#:.sin(pi/2)-sin(pi/4)=2cos((pi/2+pi/4)/2)sin((pi/2-pi/4)/2)#

#=>1-1/sqrt2=2cos(pi/4+pi/8)sin(pi/4-pi/8)#

#=>1-sqrt2/2 =2cos((3pi)/8)color(red)(sin(pi/8)#

#=>(2-sqrt2)/2=2cos((3pi)/8)color(red)(cos(pi/2-pi/8)#

#=>(2-sqrt2)/(2*2)=cos((3pi)/8)color(red)(cos((3pi)/8)#

#=>(2-sqrt2)/4=cos^2((3pi)/8)#

#i.e. cos^2((3pi)/8)=(2-sqrt2)/4#

Note:

#sintheta=cos(pi/2-theta)# ,take #theta=pi/8#

#=>color(red)(sin(pi/8)=cos(pi/2-pi/8)=cos((4pi-pi)/8)=cos((3pi)/8)#