# If eight theta is equal to pi show that cos 7 theta+ cot theta is equal to zero

Vikas TU
14149 Points
4 years ago
Dear Student,
The question should be cos 7 theta+cos theta=0.
8*theta=pi, so theta=pi/8.
Now, from half angle formula cos(theta/2) =± sqrt((1+ cos theta)/2).
Here pi/8 = ½ (pi/4) so we can apply half angle formula and
get cos(pi/8) = +sqrt((1+cos pi/4)/2)
= sqrt((1+(1/sqrt 2))/2
)=sqrt((2+sqrt2)/2) –(1)
Similarly for 7pi/8, applying half angle formula.
Here 7pi/8 = ½ (7pi/4) and
cos(7pi/4)=cos(2pi-pi/4)
=cos(pi/4
)=1/sqrt(2)
So cos(7pi/8) = -sqrt((1+cos 7pi/4)/2)[negative as 7pi/8 lies in second quadrant where cos is negative] =
-sqrt((1+(1/sqrt 2))/2)=
-sqrt((2+sqrt2)/2)—(2)

therefore , from (1) and (2) we get cos 7pi/4 + cos pi/4 =0
or cos 7theta + cos theta=0 (proved)
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
21 Points
4 years ago
Q. Should be Cos 7 theta+ cos theta=Cos (8 theta- theta) + cos theta=(Cos 8 theta × cos theta - sin 8 theta × sin theta) + cos theta=(Cos π × cos theta- sin π ×sin theta) +cos theta=(-1.cos theta- 0.sin theta) + cos theta=-cos theta +cos theta=0
Soumendu Majumdar
159 Points
4 years ago

theta=pi/8
so cos 7theta = cos(pi-pi/8) = -cos(pi/8) =-root{(1 + cos(pi/4))/2} =-root{(2root2 + 1)/2root2}
now cos theta = cos pi/8 =root{(2root2 + 1)/2root2}
therefore cos7theta + cos theta =0
P.S--The question is wrong...it would be cos theta instead of cot theta!
Hope it helps!