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 \(\varphi=\alpha-\beta\): \[\begin{align*}\sin\alpha\cos\beta&=\dfrac{\sin\theta+\sin\varphi}{2}\\ \text{We solve that }&\alpha=\dfrac{\theta+\varphi}{2}\text{ and }\beta=\dfrac{\theta-\varphi}{2}\\ \sin\left(\dfrac{\theta+\varphi}{2}\right)\cos\left(\dfrac{\theta-\varphi}{2}\right)&=\dfrac{\sin\theta+\sin\varphi}{2}\\\sin\theta+\sin\varphi &=\boxed{2\sin\left(\dfrac{\theta+\varphi}{2}\right)\cos\left(\dfrac{\theta-\varphi}{2}\right)}\end{align*}\]

**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. \[\begin{align*} \sin\theta-\sin\varphi &= \sin\theta + \sin(-\varphi)\\ &= 2\sin\left(\dfrac{\theta+(-\varphi)}{2}\right)\cos\left(\dfrac{\theta-(-\varphi)}{2}\right)\\&=\boxed{2\sin\left(\dfrac{\theta-\varphi}{2}\right)\cos\left(\dfrac{\theta+\varphi}{2}\right)} \end{align*}\]

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 \(\varphi=\alpha-\beta\): \[\begin{align*}\cos\alpha\cos\beta &= \dfrac{\cos\theta+\cos\varphi}{2}\\ \cos\left(\dfrac{\theta+\varphi}{2}\right)\cos\left(\dfrac{\theta-\varphi}{2}\right)&= \dfrac{\cos\theta+\cos\varphi}{2}\\ \cos\theta+\cos\varphi &= \boxed{2\cos\left(\dfrac{\theta+\varphi}{2}\right)\cos\left(\dfrac{\theta-\varphi}{2}\right)}\end{align*}\] 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 \(\varphi=\alpha-\beta\)): \[\begin{align*}\sin\alpha\sin\beta &= \dfrac{\cos\theta-\cos\varphi}{2}\\ \sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\varphi-\theta}{2}\right)&= \dfrac{\cos\theta-\cos\varphi}{2}\\ \cos\theta-\cos\varphi &= \boxed{2\sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\varphi-\theta}{2}\right)} \\ \text{or }&= \boxed{-2\sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\theta-\varphi}{2}\right)}\end{align*}\] and we are done.

## Comments

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TopNewestHow 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...

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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

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

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I hated this stuff during schooling years!!

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

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

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

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