Cos^2 π/8 + cos^2 3π/8 + Cos^2 5π/8 + cos^2 7π/8 Solve And Answer The Value ?

Question

68815 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-28T14:24:00

#rarrcos^2(pi/8)+cos^2((3pi)/8)+cos^2((5pi)/8)cos^2((7pi)/8)=2#

#rarrcos^2(pi/8)+cos^2((3pi)/8)+cos^2((5pi)/8)+cos^2((7pi)/8)#

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

#=cos^2(pi/8)+cos^2((3pi)/8)+cos^2((3pi)/8)+cos^2(pi/8)#

#=2*[cos^2(pi/8)+cos^2((3pi)/8)]#

#=2*[cos^2(pi/8)+sin^2(pi/2-(3pi)/8)]#

#=2*[cos^2(pi/8)+sin^2(pi/8)]=2*1=2#

Answer 2 · 2018-04-28T14:42:22

# 2#.

Here is another solution, using the Identity :

#1+cos2theta=2cos^2theta.............(ast)#

We know that, #cos(pi-theta)=-costheta#.

#:. cos(5/8pi)=cos(pi-3/8pi)=-cos(3/8pi),"&, likewise, "#

# cos(7/8pi)=-cos(1/8pi)#.

#"Hence, the reqd. value"=2cos^2(1/8pi)+2cos^2(3/8pi)#,

#={1+cos(2*1/8pi)}+{1+cos(2*3/8pi)}......[because, (ast)]#,

#=2+cos(1/4pi)+cos(3/4pi)#,

#=2+cos(1/4pi)+cos(pi-1/4pi)#,

#=2+cos(1/4pi)-cos(1/4pi)#,

#=2#, as Respected Abhishek K. has already derived!