Question
Question: What is \[{\sin ^6}\theta \] in terms of non-exponential trigonometric function?...
What is sin6θ in terms of non-exponential trigonometric function?
Solution
We use some concepts of complex numbers and their properties to solve this problem. We use some algebraic identities like
(a+b)6=a6+6a5b+15a4b2+20a3b3+15a2b4+6ab5+b6
(a+b)4=a4+4a3b+6a2b2+4ab3+b4
Complete step by step answer:
A complex number is an imaginary number which has no definite value. All the rational and irrational numbers come under these complex numbers. It is written as a+ib where a is the real part and b is the imaginary part. Some examples of complex numbers are 1,−75,(3+4i) and so on.
We all know the De-Moivre’s theorem which tells us that, (cos(θ)+isin(θ))n=cos(nθ)+isin(nθ)
Here, let the value of n be 6.
So, (cos(θ)+isin(θ))6=cos(6θ)+isin(6θ)
Let cosθ=c and sinθ=s for our convenience. So, that implies c2+s2=1 -----(1)
So, now let’s evaluate the left-hand part, which is (cos(θ)+isin(θ))6=(c+is)6
So, (c+is)6=c6+6ic5s−15c4s2−20ic3s3+15c2s4+6ics5−s6
⇒(c+is)6=(c6−15c4s2+15c2s4−s6)+i(6c5s−20c3s3+6cs5)
So, we can conclude that,
{\left( {\cos (\theta ) + i\sin (\theta )} \right)^6} = \cos (6\theta ) + i\sin (6\theta )$$$$ = {(c + is)^6} = ({c^6} - 15{c^4}{s^2} + 15{c^2}{s^4} - {s^6}) + i(6{c^5}s - 20{c^3}{s^3} + 6c{s^5})
Now, equating the real parts, we get cos(6θ)=c6−15c4s2+15c2s4−s6
Now, from (1), we get, c2=1−s2 and substitute this in above.
⇒cos(6θ)=(1−s2)3−15(1−s2)2s2+15(1−s2)s4−s6
⇒cos(6θ)=(1−3s2+3s4−s6)−15(1−2s2+s4)s2+15(1−s2)s4−s6
Now, let us simplify further.
⇒cos(6θ)=1−3s2+3s4−s6−(15s2−30s4+15s6)+(15s4−15s6)−s6
⇒cos(6θ)=1−18s2+48s4−32s6 ------(2)
Now in the same way, we will evaluate cos(4θ)+isin(4θ)=(c+is)4
⇒(c+is)4=c4+4ic3s−6c2s2−4ics3+s4
⇒cos(4θ)+isin(4θ)=(c4−6c2s2+s4)+i(4c3s−4cs3)
Equating the real parts, we get,
cos(4θ)=c4−6c2s2+s4
⇒cos(4θ)=(1−s2)2−6(1−s2)s2+s4
So, we get,
⇒cos(4θ)=(1−2s2+s4)−6(s2−s4)+s4
⇒cos(4θ)=1−8s2+8s4 ------(3)
Now let us evaluate cos(2θ)+isin(2θ)=(c+is)2
⇒(c+is)2=(c2−s2)+i(2cs)
Equating real parts,
⇒cos(2θ)=c2−s2=(1−s2)−s2
⇒cos(2θ)=1−2s2 -----(4)
Now, derive the value of s2 from this equation.
s2=21−cos(2θ)
Now, substitute this value in equation (3), and find the value of s4
⇒cos(4θ)=1−8(21−cos(2θ))+8s4
⇒s4=8cos(4θ)+3−4cos(2θ)
Now, substitute this value in equation (2) and find the value of s6
⇒cos(6θ)=1−18(21−cos(2θ))+48(8cos(4θ)+3−4cos(2θ))−32s6
⇒cos(6θ)=1−9(1−cos(2θ))+6(cos(4θ)+3−4cos(2θ))−32s6
So, we get it as,
⇒cos(6θ)=1−9+9cos(2θ)+6cos(4θ)+18−24cos(2θ)−32s6
⇒cos(6θ)=6cos(4θ)+10−15cos(2θ)−32s6
So, we finally get,
⇒s6=321(6cos(4θ)+10−15cos(2θ)−cos(6θ))
And we know that, s6=sin6θ
So, we can conclude that,
sin6θ=321(6cos(4θ)−15cos(2θ)−cos(6θ)+10)
Note:
Here, i is a complex value, and it is i=−1. So, it can be written as i2=−1.
Consider the value in. If n is a multiple of 4, then in=1
If n is an even number other than multiples of 4, then in=−1.
An imaginary number is written as a+ib where a is the real part and b is the imaginary part.