# How to simplify the trigonometric functions sine and cosine in a derivative?

I am learning calculus and came across the following problem:

Find $$\frac{dx}{d\theta}$$ when $$x=\sin^2\theta \cos^3\theta$$

I solved this using the product rule.

$$u= \sin^2\theta \qquad v=\cos^3\theta$$

$$u'= \sin2\theta \qquad v'=-3\cos^2\theta \sin\theta$$

\begin{align} \frac{dx}{d\theta}&= \sin^2\theta(-3\cos^2\theta \sin\theta) +\cos^3\theta(\sin2\theta) \\ &= -3\cos^2\theta \sin^3\theta + \cos^3\theta \sin2\theta \end{align}

The problem is that this does not match any of the options the exercise provides, so how can I simplify it further? Are there common methods for simplification?

• Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Feb 22 at 20:46
• There are often many ways to write trig expressions, and many interpretations of "simplified". It would help if you gave the options from the exercise, to focus on showing how to recognize which matches your solution. ... A suggestion here is to avoid mixing multiple-angle arguments ($\theta$ vs $2\theta$). Expanding $\sin 2\theta$, you can factor: $$-3\cos^2\theta\sin^3\theta+\cos^3\theta\cdot 2\sin\theta\cos\theta = \sin\theta\cos^2\theta\left(-3\sin^2\theta+2\cos^2\theta\right)$$ From there, a sum of even powers of sine and cosine could/should be written in terms of one or the other. – Blue Feb 22 at 21:02
• @Marcus: $\sin2\theta=2\sin\theta\cos\theta$ as a special case of the angle-sum rule: $$\sin 2\theta = \sin(\theta+\theta)=\sin\theta\cos\theta+\cos\theta\sin\theta=2\sin\theta\cos\theta$$ Note that the hardest part about learning Calculus is often applying stuff you're supposed to remember from before Calculus. While you don't have to have every trig identity memorized (just have a list handy), you'll find that simpler things like $\sin2\theta$ arise fairly often. – Blue Feb 22 at 21:21
• Just out of curiosity, I am wondering about this: if you were unaware that $\sin(2\theta) = 2\sin\theta\cos\theta$, how did you know that the derivative of $\sin^2 \theta$ is $\sin(2\theta)$? – David K Feb 22 at 22:52
• OK, that explains $\sin(2\theta)$. It seems that getting help from the calculator actually worked against you in this case, because if you had used the chain rule you would have gotten a more useful form of the derivative. – David K Feb 23 at 21:39

With

$$u = \sin^2\theta, \; v = \cos^3 \theta, \tag 1$$

the chain rule yields

$$u'(\theta) = \dfrac{d\sin^2 \theta}{d\theta} = \dfrac{d\sin^2 \theta}{d\sin \theta}\dfrac{d\sin \theta}{d\theta} = 2(\sin \theta)(\cos\theta), \tag 2$$

and

$$v'(\theta) = \dfrac{d\cos^3\theta}{d\cos \theta} \dfrac{d\cos \theta}{d\theta} = 3(\cos^2 \theta)(-\sin \theta) = -3(\cos^2 \theta)(\sin \theta); \tag 3$$

then

$$\dfrac{d((\sin^2 \theta )(\cos^3 \theta)) }{d\theta} = \dfrac{d(uv)}{d\theta} = u'(\theta)v(\theta) + u(\theta)v'(\theta)$$ $$= (2(\sin \theta)(\cos \theta))(\cos^3 \theta) + (\sin^2 \theta)((-3\cos^2\theta)(\sin \theta))$$ $$= 2(\sin \theta)(\cos^4 \theta) -3(\sin^3\theta)(\cos^2 \theta)$$ $$= (\sin \theta)(\cos^2 \theta)(2\cos^2 \theta - 3\sin^2 \theta), \tag 4$$

which agrees with the expression given by Blue in his/her comment to the question itself.