By De Moivre's theorem,
[tex]\left(\cos\dfrac{3\pi}5+i\sin\dfrac{3\pi}5\right)^3=\boxed{\cos\dfrac{9\pi}5+i\sin\dfrac{9\pi}5}[/tex]
We can stop here ...
# # #
... but we can also express these trig ratios in terms of square roots. Let [tex]x=\dfrac\pi5[/tex] and let [tex]c=\cos x[/tex]. Then recall that
[tex]\cos5x=c^5-10c^3\sin^2x+5c\sin^4x[/tex]
[tex]\cos5x=c^5-10c^3(1-c^2)+5c(1-2c^2+c^4)[/tex]
[tex]\cos5x=16c^5-20c^3+5c[/tex]
On the left, [tex]5x=\pi[/tex] so that
[tex]16c^5-20c^3+5c+1=(c+1)(4c^2-2c-1)^2=0[/tex]
Since [tex]c=\cos\dfrac\pi5\neq1[/tex], we're left with
[tex]4c^2-2c-1=0\implies c=\dfrac{1+\sqrt5}4[/tex]
because we know to expect [tex]\cos x>0[/tex]. Then from the Pythagorean identity, and knowing to expect [tex]\sin x>0[/tex], we get
[tex]\sin x=\sqrt{1-c^2}=\sqrt{\dfrac58-\dfrac{\sqrt5}8}[/tex]
Both [tex]\cos[/tex] and [tex]\sin[/tex] are [tex]2\pi[/tex]-periodic, so that
[tex]\cos\dfrac{9\pi}5=\cos\left(\dfrac{9\pi}5-2\pi\right)=\cos\left(-\dfrac\pi5\right)=\cos\dfrac\pi5[/tex]
and
[tex]\sin\dfrac{9\pi}5=\sin\left(-\dfrac\pi5\right)=-\sin\dfrac\pi5[/tex]
so that the answer we left in trigonometric form above is equal to
[tex]\cos\dfrac\pi5-i\sin\dfrac\pi5=\boxed{\dfrac{1+\sqrt5}4+i\sqrt{\dfrac58+\dfrac{\sqrt5}8}}[/tex]
Answer:
Answer is (cos(9pi/5)+isin(9pi/5))
Step-by-step explanation:
took the test
