# How do you express sin^4theta-cos^3theta  in terms of non-exponential trigonometric functions?

##### 1 Answer
Feb 7, 2016

$\frac{1}{8} \cos \left(4 \theta\right) - \frac{1}{4} \cos \left(3 \theta\right) - \frac{1}{2} \cos \left(2 \theta\right) - \frac{3}{4} \cos \left(\theta\right) + \frac{3}{8}$

#### Explanation:

We can use the trig identities:

${\sin}^{4} \theta = \frac{3}{8} - \frac{1}{2} \cos \left(2 \theta\right) + \frac{1}{8} \cos \left(4 \theta\right)$
And

${\cos}^{3} \left(\theta\right) = \frac{3}{4} \cos \left(\theta\right) + \frac{1}{4} \cos \left(3 \theta\right)$

Now ${\sin}^{4} \left(\theta\right) - {\cos}^{3} \left(\theta\right)$
$= \frac{3}{8} - \frac{1}{2} \cos \left(2 \theta\right) + \frac{1}{8} \cos \left(4 \theta\right) - \frac{3}{4} \cos \left(\theta\right) - \frac{1}{4} \cos \left(3 \theta\right)$

$= \frac{1}{8} \cos \left(4 \theta\right) - \frac{1}{4} \cos \left(3 \theta\right) - \frac{1}{2} \cos \left(2 \theta\right) - \frac{3}{4} \cos \left(\theta\right) + \frac{3}{8}$

I understand these identities are a bit random. If you have a good understanding of complex numbers and binomial theorem then here is a trick that can be used to obtain any of these types of trig identities.

It may be a bit tedious at first but with a bit of practice you'll learn in no time and when you do you'll never forget it.

The trick is to express the trig function in terms of its complex exponential and then expand that term using the binomial theorem to the appropriate power. After which you can then re-arrange the terms back into the non power trig functions like above. If we use ${\sin}^{4} x$ as an example:

If you are versed in complex numbers, remember that:

$\cos \left(n x\right) + i \sin \left(n x\right) = {e}^{i n x}$

From this it can be shown:

$\sin \left(n x\right) = \frac{{e}^{i n x} - {e}^{- i n x}}{2 i}$

and

$\cos \left(n x\right) = \frac{{e}^{i n x} + {e}^{- i n x}}{2}$

So it follows that:

${\sin}^{4} \left(n x\right) = {\left(\frac{{e}^{i n x} - {e}^{- i n x}}{2 i}\right)}^{4}$

Now applying the binomial theorem to expand this we get:

$\frac{1}{16 {i}^{4}} \left({\left({e}^{i x}\right)}^{4} - 4 {\left({e}^{i x}\right)}^{3} \left({e}^{- i x}\right) + 6 {\left({e}^{i x}\right)}^{2} {\left({e}^{- i x}\right)}^{2} - 4 \left({e}^{i x}\right) {\left({e}^{- i x}\right)}^{3} + {\left({e}^{- i x}\right)}^{4}\right)$

Bringing the powers into the exponents then tidying this up a little:

$\frac{1}{16} \left(\left({e}^{4 i x}\right) - 4 \left({e}^{4 i x - i x}\right) + 6 \left({e}^{2 i x - 2 i x}\right) - 4 \left({e}^{i x - 3 i x}\right) + \left({e}^{- 4 i x}\right)\right)$

$= \frac{1}{16} \left({e}^{4 i x} + {e}^{- 4 i x} - 4 \left({e}^{2 i x} + {e}^{- 2 i x}\right) + 6\right)$

$= \frac{1}{8} \frac{{e}^{4 i x} + {e}^{- 4 i x}}{2} - \frac{4}{8} \frac{{e}^{2 i x} + {e}^{- 2 i x}}{2} + \frac{3}{8}$

And now we see have the exponential forms for $\cos \left(n x\right)$ so we can re write these as:

$\frac{1}{8} \cos \left(4 x\right) - \frac{1}{2} \cos \left(2 x\right) + \frac{3}{8} = {\sin}^{4} \left(x\right)$

The exact same method can be applied to ${\cos}^{3} \left(x\right)$ to obtain identity. In fact this method can be used to obtain any identity of ${\cos}^{n} \left(x\right)$ or ${\sin}^{n} \left(x\right)$and may come in useful should you choose to study fields such as vector calculus.

I hope this helped.