Video Transcript
Use de Moivre’s theorem to express
sin of four 𝜃 in terms of powers of sin 𝜃 and cos 𝜃.
We begin by recalling what de
Moivre’s theorem tells us. It says that 𝑒 to the power of
𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃. Now, of course, we’re looking to
express sin of four 𝜃 in terms of powers of sin 𝜃 and cos 𝜃. So, how do we achieve this? Well, by simply multiplying 𝜃 by
four, we see that we can rewrite de Moivre’s theorem as 𝑒 to the power of four 𝑖𝜃
equals cos of four 𝜃 plus 𝑖 sin of four 𝜃. It’s a good start, but we still
have 𝑒 to the power of four 𝑖𝜃.
So, we’re going to go back to the
original expression, 𝑒 to the 𝑖𝜃 equals cos 𝜃 plus 𝑖 sin 𝜃, and we’re going to
raise both sides as a power of four. And we can see that 𝑒 to the 𝑖𝜃
all to the power of four is actually the same as 𝑒 to the power of four 𝑖𝜃. And this in turn means that cos of
four 𝜃 plus 𝑖 sin of four 𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃 to the fourth
power. And that’s great because we can now
use the binomial theorem to distribute the parentheses on the right-hand side. This says that 𝑎 plus 𝑏 to the
𝑛th power is equal to the sum from 𝑘 equals zero to 𝑛 of 𝑛 choose 𝑘 times 𝑎 to
the power of 𝑛 minus 𝑘 times 𝑏 to the power of 𝑘.
When 𝑛 is equal to four, we have
𝑎 plus 𝑏 to the fourth power equals 𝑎 to the fourth power plus four choose one 𝑎
cubed 𝑏 plus four choose two 𝑎 squared 𝑏 squared plus four choose three 𝑎𝑏 plus
𝑏 to the fourth power. Now, in fact, four choose one and
four choose three are four, and four choose two is equal to six. And so, this is the formula we’re
going to use to distribute our parentheses. The first term is cos 𝜃 to the
fourth power. The second is four cos cubed 𝜃
times 𝑖 sin 𝜃, although convention dictates that we write this as four 𝑖 cos
cubed 𝜃 sin 𝜃.
We then have six cos squared 𝜃 𝑖
sin 𝜃 squared. And since we can write 𝑖 sin 𝜃
squared as 𝑖 squared sin squared 𝜃 and 𝑖 squared equals to negative one, this
whole term becomes negative six cos squared 𝜃 sin squared 𝜃. Our fourth term is four cos 𝜃 𝑖
sin 𝜃 cubed. And then, we write 𝑖 sin 𝜃 cubed
as 𝑖 cubed sin cubed 𝜃. We then write 𝑖 cubed as 𝑖 times
𝑖 squared. We said that 𝑖 squared was equal
to negative one, so this is simply negative 𝑖. And then, this time is simply
negative four 𝑖 cos 𝜃 sin cubed 𝜃.
Our final term is 𝑖 sin 𝜃 to the
fourth power. We distribute the parentheses and
write this as 𝑖 to the fourth power times sin 𝜃 to the fourth power. 𝑖 to the fourth power, though, is
𝑖 squared times 𝑖 which is negative one times negative one, which is simply
one. This means the final term in the
distribution of the parentheses cos 𝜃 plus 𝑖 sin 𝜃 to the fourth power is sin 𝜃
to the fourth power. Now, going back to the question, we
want to express sin four 𝜃 in terms of powers of sin 𝜃 and cos 𝜃. And if we notice on the left-hand
side that sin 𝜃 is the coefficient of 𝑖, that gives us a pretty good hint of what
we’re going to do next. We’re going to equate the imaginary
parts on both sides of our equation.
We just saw that the imaginary part
on the left-hand side, the coefficient of 𝑖, is sin four 𝜃. On the right-hand side, we have
four cos cubed 𝜃 sin 𝜃 minus four cos 𝜃 sin cubed 𝜃. And so, we can say that the
coefficients of 𝑖 on the left-hand side must be equal to the coefficient of 𝑖 on
the right. That is, sin four 𝜃 equals four
cos cubed 𝜃 sin 𝜃 minus four cos 𝜃 sin cubed 𝜃. And so, we’re done. We’ve used de Moivre’s theorem to
express sin four 𝜃 in terms of powers of sin and cos. It’s sin four 𝜃 equals four cos
cubed 𝜃 sin 𝜃 minus four cos 𝜃 sin cubed 𝜃.
