(a) By DeMoivre's theorem, we have
![(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta](https://tex.z-dn.net/?f=%28%5Ccos%5Ctheta%2Bi%5Csin%5Ctheta%29%5E5%3D%5Ccos5%5Ctheta%2Bi%5Csin5%5Ctheta)
On the LHS, expanding yields
![\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^4\theta](https://tex.z-dn.net/?f=%5Ccos%5E5%5Ctheta%2B5i%5Ccos%5E4%5Ctheta%5Csin%5Ctheta-10%5Ccos%5E3%5Ctheta%5Csin%5E2%5Ctheta-10i%5Ccos%5E2%5Ctheta%5Csin%5E3%5Ctheta%2B5%5Ccos%5Ctheta%5Csin%5E4%5Ctheta%2Bi%5Csin%5E4%5Ctheta)
Matching up real and imaginary parts, we have for (i) and (ii),
![\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta](https://tex.z-dn.net/?f=%5Ccos5%5Ctheta%3D%5Ccos%5E5%5Ctheta-10%5Ccos%5E3%5Ctheta%5Csin%5E2%5Ctheta%2B5%5Ccos%5Ctheta%5Csin%5E4%5Ctheta)
![\sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta](https://tex.z-dn.net/?f=%5Csin5%5Ctheta%3D5%5Ccos%5E4%5Ctheta%5Csin%5Ctheta-10%5Ccos%5E2%5Ctheta%5Csin%5E3%5Ctheta%2B%5Csin%5E5%5Ctheta)
(b) By the definition of the tangent function,
![\tan5\theta=\dfrac{\sin5\theta}{\cos5\theta}](https://tex.z-dn.net/?f=%5Ctan5%5Ctheta%3D%5Cdfrac%7B%5Csin5%5Ctheta%7D%7B%5Ccos5%5Ctheta%7D)
![=\dfrac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B5%5Ccos%5E4%5Ctheta%5Csin%5Ctheta-10%5Ccos%5E2%5Ctheta%5Csin%5E3%5Ctheta%2B%5Csin%5E5%5Ctheta%7D%7B%5Ccos%5E5%5Ctheta-10%5Ccos%5E3%5Ctheta%5Csin%5E2%5Ctheta%2B5%5Ccos%5Ctheta%5Csin%5E4%5Ctheta%7D)
![=\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B5%5Ctan%5Ctheta-10%5Ctan%5E3%5Ctheta%2B%5Ctan%5E5%5Ctheta%7D%7B1-10%5Ctan%5E2%5Ctheta%2B5%5Ctan%5E4%5Ctheta%7D)
![=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7Bt%5E5-10t%5E3%2B5t%7D%7B5t%5E4-10t%5E2%2B1%7D)
(c) Setting
![\theta=\dfrac\pi5](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cdfrac%5Cpi5)
, we have
![t=\tan\dfrac\pi5](https://tex.z-dn.net/?f=t%3D%5Ctan%5Cdfrac%5Cpi5)
and
![\tan5\left(\dfrac\pi5\right)=\tan\pi=0](https://tex.z-dn.net/?f=%5Ctan5%5Cleft%28%5Cdfrac%5Cpi5%5Cright%29%3D%5Ctan%5Cpi%3D0)
. So
![0=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}](https://tex.z-dn.net/?f=0%3D%5Cdfrac%7Bt%5E5-10t%5E3%2B5t%7D%7B5t%5E4-10t%5E2%2B1%7D)
At the given value of
![t](https://tex.z-dn.net/?f=t)
, the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.
![0=t^5-10t^3+5t\implies0=t^4-10t^2+5](https://tex.z-dn.net/?f=0%3Dt%5E5-10t%5E3%2B5t%5Cimplies0%3Dt%5E4-10t%5E2%2B5)
Remember, this is saying that
![0=\tan^4\dfrac\pi5-10\tan^2\dfrac\pi5+5](https://tex.z-dn.net/?f=0%3D%5Ctan%5E4%5Cdfrac%5Cpi5-10%5Ctan%5E2%5Cdfrac%5Cpi5%2B5)
If we replace
![\tan^2\dfrac\pi5](https://tex.z-dn.net/?f=%5Ctan%5E2%5Cdfrac%5Cpi5)
with a variable
![x](https://tex.z-dn.net/?f=x)
, then the above means
![\tan^2\dfrac\pi5](https://tex.z-dn.net/?f=%5Ctan%5E2%5Cdfrac%5Cpi5)
is a root to the quadratic equation,
![x^2-10x+5=0](https://tex.z-dn.net/?f=x%5E2-10x%2B5%3D0)
Also, if
![\theta=\dfrac{2\pi}5](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cdfrac%7B2%5Cpi%7D5)
, then
![t=\tan\dfrac{2\pi}5](https://tex.z-dn.net/?f=t%3D%5Ctan%5Cdfrac%7B2%5Cpi%7D5)
and
![\tan5\left(\dfrac{2\pi}5\right)=\tan2\pi=0](https://tex.z-dn.net/?f=%5Ctan5%5Cleft%28%5Cdfrac%7B2%5Cpi%7D5%5Cright%29%3D%5Ctan2%5Cpi%3D0)
. So by a similar argument as above, we deduce that
![\tan^2\dfrac{2\pi}5](https://tex.z-dn.net/?f=%5Ctan%5E2%5Cdfrac%7B2%5Cpi%7D5)
is also a root to the quadratic equation above.
(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write
![x^2-10x+5=\left(x-\tan^2\dfrac\pi5\right)\left(x-\tan^2\dfrac{2\pi}5\right)](https://tex.z-dn.net/?f=x%5E2-10x%2B5%3D%5Cleft%28x-%5Ctan%5E2%5Cdfrac%5Cpi5%5Cright%29%5Cleft%28x-%5Ctan%5E2%5Cdfrac%7B2%5Cpi%7D5%5Cright%29)
Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy
![5=\tan^2\dfrac\pi5\tan^2\dfrac{2\pi}5\implies\tan\dfrac\pi5\tan\dfrac{2\pi}5=\pm\sqrt5](https://tex.z-dn.net/?f=5%3D%5Ctan%5E2%5Cdfrac%5Cpi5%5Ctan%5E2%5Cdfrac%7B2%5Cpi%7D5%5Cimplies%5Ctan%5Cdfrac%5Cpi5%5Ctan%5Cdfrac%7B2%5Cpi%7D5%3D%5Cpm%5Csqrt5)
But
![\tanx>0](https://tex.z-dn.net/?f=%5Ctanx%3E0)
for all
![0](https://tex.z-dn.net/?f=0%3Cx%3C%5Cdfrac%5Cpi2)
, as is the case for
![x=\dfrac\pi5](https://tex.z-dn.net/?f=x%3D%5Cdfrac%5Cpi5)
and
![x=\dfrac{2\pi}5](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B2%5Cpi%7D5)
, so we choose the positive root.