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⁡αsin⁡β\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⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha+\beta)=\cos\alpha\cos\beta - \sin\alpha\sin\beta and cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha-\beta)=\cos\alpha\cos\beta + \sin\alpha\sin\beta. We can therefore derive the formula as follows: cos⁡αcos⁡β+sin⁡αsin⁡β=cos⁡(α−β)−cos⁡αcos⁡β−sin⁡αsin⁡β=cos⁡(α+β)2sin⁡αsin⁡β=cos⁡(α−β)−cos⁡(α+β)sin⁡αsin⁡β=cos⁡(α−β)−cos⁡(α+β)2\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⁡αcos⁡β\cos\alpha\cos\beta. We can proceed much like our previous solution, but instead cancelling out the sines. cos⁡αcos⁡β−sin⁡αsin⁡β=cos⁡(α+β)+cos⁡αcos⁡β+sin⁡αsin⁡β=cos⁡(α−β)2cos⁡αcos⁡β=cos⁡(α+β)+cos⁡(α−β)cos⁡αcos⁡β=cos⁡(α+β)+cos⁡(α−β)2\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⁡αcos⁡β\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⁡αcos⁡β\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⁡(α+β)=sin⁡αcos⁡β+sin⁡βcos⁡α\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha and sin⁡(α−β)=sin⁡αcos⁡β−sin⁡βcos⁡α\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha have what we want. We do a similar calculation as before: sin⁡αcos⁡β+sin⁡βcos⁡α=sin⁡(α+β)+sin⁡αcos⁡β−sin⁡βcos⁡α=sin⁡(α−β)2sin⁡αcos⁡β=sin⁡(α+β)+sin⁡(α−β)sin⁡αcos⁡β=sin⁡(α+β)+sin⁡(α−β)2\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⁡θ+sin⁡φ\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⁡αcos⁡β=sin⁡(α+β)+sin⁡(α−β)2\sin\alpha\cos\beta=\dfrac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}. Making the substitution θ=α+β\theta=\alpha+\beta and φ=α−β\varphi=\alpha-\beta: sin⁡αcos⁡β=sin⁡θ+sin⁡φ2We solve that α=θ+φ2 and β=θ−φ2sin⁡(θ+φ2)cos⁡(θ−φ2)=sin⁡θ+sin⁡φ2sin⁡θ+sin⁡φ=2sin⁡(θ+φ2)cos⁡(θ−φ2)\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⁡θ−sin⁡φ\sin\theta-\sin\varphi. We can simply plug in −φ-\varphi in our previous formula to obtain this formula. sin⁡θ−sin⁡φ=sin⁡θ+sin⁡(−φ)=2sin⁡(θ+(−φ)2)cos⁡(θ−(−φ)2)=2sin⁡(θ−φ2)cos⁡(θ+φ2)\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, sin⁡θ±sin⁡φ=2sin⁡(θ±φ2)cos⁡(θ∓φ2)\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⁡θ+cos⁡φ\cos\theta+\cos\varphi. We want to use our previous strategy of working backwards from identities we already know. We see that cos⁡αcos⁡β=cos⁡(α+β)+cos⁡(α−β)2\cos\alpha\cos\beta=\dfrac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2} serves our purpose. Substituting θ=α+β\theta=\alpha+\beta and φ=α−β\varphi=\alpha-\beta: cos⁡αcos⁡β=cos⁡θ+cos⁡φ2cos⁡(θ+φ2)cos⁡(θ−φ2)=cos⁡θ+cos⁡φ2cos⁡θ+cos⁡φ=2cos⁡(θ+φ2)cos⁡(θ−φ2)\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⁡θ−cos⁡φ\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⁡αsin⁡β=cos⁡(α−β)−cos⁡(α+β)2\sin\alpha\sin\beta=\dfrac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2} seems like what we should use. Substituting φ=α+β\varphi=\alpha+\beta and θ=α−β\theta=\alpha-\beta (note that we are not substituting θ=α+β\theta=\alpha+\beta and φ=α−β\varphi=\alpha-\beta): sin⁡αsin⁡β=cos⁡θ−cos⁡φ2sin⁡(θ+φ2)sin⁡(φ−θ2)=cos⁡θ−cos⁡φ2cos⁡θ−cos⁡φ=2sin⁡(θ+φ2)sin⁡(φ−θ2)or =−2sin⁡(θ+φ2)sin⁡(θ−φ2)\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.

#Trigonometry #TrigonometricIdentities #TrigonometricSumAndProductFormulas #CosinesGroup #Guide

Note by Daniel Liu
7 years, 2 months ago

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Comments

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...

Prasun Biswas - 7 years, 2 months ago

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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

Daniel Liu - 7 years, 2 months ago

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

Shabarish Ch - 7 years, 2 months ago

I hated this stuff during schooling years!!

Venture HI - 7 years, 2 months ago

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

Daniel Liu - 7 years, 2 months ago

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

Ashtik Mahapatra - 7 years, 1 month ago

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

Jimbo Magno - 6 years, 6 months ago
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