# Deriving Trigonometric Identities (Part 2)

Hi all! This is the second installation of my "Deriving Trigonometric Identities" series. If you haven't read the first one, I suggest you go read it here. Without further ado, let's dive into the math!

Proving the sine product to sum formula

We want to find a formula for $\sin\alpha\sin\beta$. But how can we find the product of two sines with different angles? The key step is noting that we have seen this term before in our formulas; namely, $\cos(\alpha+\beta)=\cos\alpha\cos\beta - \sin\alpha\sin\beta$ and $\cos(\alpha-\beta)=\cos\alpha\cos\beta + \sin\alpha\sin\beta$. We can therefore derive the formula as follows: $\begin{array}{lrcl}& \cos\alpha\cos\beta + \sin\alpha\sin\beta&=&\cos(\alpha-\beta)\\ - & \cos\alpha\cos\beta - \sin\alpha\sin\beta&=&\cos(\alpha+\beta)\\ \hline &2\sin\alpha\sin\beta&=&\cos(\alpha-\beta)-\cos(\alpha+\beta)\\ & \sin\alpha\sin\beta&=&\boxed{\dfrac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}}\end{array}$ and we are done.

Proving the cosine product to sum formula

We want to find a formula for $\cos\alpha\cos\beta$. We can proceed much like our previous solution, but instead cancelling out the sines. $\begin{array}{lrcl}& \cos\alpha\cos\beta - \sin\alpha\sin\beta&=&\cos(\alpha+\beta)\\ +& \cos\alpha\cos\beta + \sin\alpha\sin\beta&=&\cos(\alpha-\beta)\\ \hline &2\cos\alpha\cos\beta&=&\cos(\alpha+\beta)+\cos(\alpha-\beta)\\ & \cos\alpha\cos\beta&=&\boxed{\dfrac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}}\end{array}$ and we are done.

Proving the sine-cosine product to sum formulas

We want to find a formula for $\sin\alpha\cos\beta$. But we can't add or subtract the cosine angle sum and difference formulas anymore, because they don't have a $\sin\alpha\cos\beta$ term. However, using the same line of reasoning, we try to find another trig identity with this term, and lo and behold: $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$ and $\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha$ have what we want. We do a similar calculation as before: $\begin{array}{lrcl}& \sin\alpha\cos\beta + \sin\beta\cos\alpha&=&\sin(\alpha+\beta)\\ + & \sin\alpha\cos\beta - \sin\beta\cos\alpha&=&\sin(\alpha-\beta)\\ \hline &2\sin\alpha\cos\beta&=&\sin(\alpha+\beta)+\sin(\alpha-\beta)\\ & \sin\alpha\cos\beta&=&\boxed{\dfrac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}}\end{array}$ and we are done.

Proving sine sum to product formula

We want to find a formula for finding $\sin\theta+\sin\varphi$. Seeing that in our product to sum formulas, we have a sum of sines in the formula, we can maybe reverse the process and create a sum to product formula. In particular, we would like to use the formula $\sin\alpha\cos\beta=\dfrac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}$. Making the substitution $\theta=\alpha+\beta$ and $\varphi=\alpha-\beta$: \begin{aligned}\sin\alpha\cos\beta&=\dfrac{\sin\theta+\sin\varphi}{2}\\ \text{We solve that }&\alpha=\dfrac{\theta+\varphi}{2}\text{ and }\beta=\dfrac{\theta-\varphi}{2}\\ \sin\left(\dfrac{\theta+\varphi}{2}\right)\cos\left(\dfrac{\theta-\varphi}{2}\right)&=\dfrac{\sin\theta+\sin\varphi}{2}\\\sin\theta+\sin\varphi &=\boxed{2\sin\left(\dfrac{\theta+\varphi}{2}\right)\cos\left(\dfrac{\theta-\varphi}{2}\right)}\end{aligned}

Proving sine difference to product formula

We want to find a formula for $\sin\theta-\sin\varphi$. We can simply plug in $-\varphi$ in our previous formula to obtain this formula. \begin{aligned} \sin\theta-\sin\varphi &= \sin\theta + \sin(-\varphi)\\ &= 2\sin\left(\dfrac{\theta+(-\varphi)}{2}\right)\cos\left(\dfrac{\theta-(-\varphi)}{2}\right)\\&=\boxed{2\sin\left(\dfrac{\theta-\varphi}{2}\right)\cos\left(\dfrac{\theta+\varphi}{2}\right)} \end{aligned}

In general, $\boxed{\sin\theta\pm\sin\varphi =2\sin\left(\dfrac{\theta\pm\varphi}{2}\right)\cos\left(\dfrac{\theta\mp\varphi}{2}\right)}$

Proving the cosine sum to product formula

We want to find a formula for $\cos\theta+\cos\varphi$. We want to use our previous strategy of working backwards from identities we already know. We see that $\cos\alpha\cos\beta=\dfrac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}$ serves our purpose. Substituting $\theta=\alpha+\beta$ and $\varphi=\alpha-\beta$: \begin{aligned}\cos\alpha\cos\beta &= \dfrac{\cos\theta+\cos\varphi}{2}\\ \cos\left(\dfrac{\theta+\varphi}{2}\right)\cos\left(\dfrac{\theta-\varphi}{2}\right)&= \dfrac{\cos\theta+\cos\varphi}{2}\\ \cos\theta+\cos\varphi &= \boxed{2\cos\left(\dfrac{\theta+\varphi}{2}\right)\cos\left(\dfrac{\theta-\varphi}{2}\right)}\end{aligned} and we are done.

Proving the cosine difference to sum formula

We want to find a formula for $\cos\theta-\cos\varphi$. Much like the previous few proofs, we will use the product to sum formulas to derive this formula. In particular, $\sin\alpha\sin\beta=\dfrac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}$ seems like what we should use. Substituting $\varphi=\alpha+\beta$ and $\theta=\alpha-\beta$ (note that we are not substituting $\theta=\alpha+\beta$ and $\varphi=\alpha-\beta$): \begin{aligned}\sin\alpha\sin\beta &= \dfrac{\cos\theta-\cos\varphi}{2}\\ \sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\varphi-\theta}{2}\right)&= \dfrac{\cos\theta-\cos\varphi}{2}\\ \cos\theta-\cos\varphi &= \boxed{2\sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\varphi-\theta}{2}\right)} \\ \text{or }&= \boxed{-2\sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\theta-\varphi}{2}\right)}\end{aligned} and we are done. Note by Daniel Liu
5 years, 7 months ago

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How do you manage to get time to write such long notes and that too using LaTeX ?? Don't you have school studies and homework to do !!

P.S. -- I am not scolding you, just enquiring...

- 5 years, 7 months ago

Hi Prasun,

I am in middle school, so I do not have as much homework as high schoolers. Also, you'll be amazed how fast I can type LaTeX... :)

Daniel

- 5 years, 7 months ago

You are in middle school and you already know all this.. Amazing!!

- 5 years, 7 months ago

I hated this stuff during schooling years!!

- 5 years, 7 months ago

I suppose they just gave you the formulas and never explained how or why?

- 5 years, 7 months ago

No they never gave us the proofs. But I myself proved all these so I enjoyed this.

- 5 years, 7 months ago

I APPRECIATE YOUR INTELLIGENCE IN MATHEMATICS. YOU ARE TRULY A GENIUS WITH EXTRAORDINARY SKILLS IN SOLVING AND PROVING MATHEMATICAL PROBLEMS.

- 4 years, 11 months ago