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 sinαsinβ\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)=cosαcosβsinαsinβ\cos(\alpha+\beta)=\cos\alpha\cos\beta - \sin\alpha\sin\beta and cos(αβ)=cosαcosβ+sinαsinβ\cos(\alpha-\beta)=\cos\alpha\cos\beta + \sin\alpha\sin\beta. We can therefore derive the formula as follows: cosαcosβ+sinαsinβ=cos(αβ)cosαcosβsinαsinβ=cos(α+β)2sinαsinβ=cos(αβ)cos(α+β)sinαsinβ=cos(αβ)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 cosαcosβ\cos\alpha\cos\beta. We can proceed much like our previous solution, but instead cancelling out the sines. cosαcosβsinαsinβ=cos(α+β)+cosαcosβ+sinαsinβ=cos(αβ)2cosαcosβ=cos(α+β)+cos(αβ)cosαcosβ=cos(α+β)+cos(αβ)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 sinαcosβ\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 sinαcosβ\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)=sinαcosβ+sinβcosα\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha and sin(αβ)=sinαcosβsinβcosα\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha have what we want. We do a similar calculation as before: sinαcosβ+sinβcosα=sin(α+β)+sinαcosβsinβcosα=sin(αβ)2sinαcosβ=sin(α+β)+sin(αβ)sinαcosβ=sin(α+β)+sin(αβ)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 sinθ+sinφ\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 sinαcosβ=sin(α+β)+sin(αβ)2\sin\alpha\cos\beta=\dfrac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}. Making the substitution θ=α+β\theta=\alpha+\beta and φ=αβ\varphi=\alpha-\beta: sinαcosβ=sinθ+sinφ2We solve that α=θ+φ2 and β=θφ2sin(θ+φ2)cos(θφ2)=sinθ+sinφ2sinθ+sinφ=2sin(θ+φ2)cos(θφ2)\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 sinθsinφ\sin\theta-\sin\varphi. We can simply plug in φ-\varphi in our previous formula to obtain this formula. sinθsinφ=sinθ+sin(φ)=2sin(θ+(φ)2)cos(θ(φ)2)=2sin(θφ2)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, sinθ±sinφ=2sin(θ±φ2)cos(θφ2)\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 cosθ+cosφ\cos\theta+\cos\varphi. We want to use our previous strategy of working backwards from identities we already know. We see that cosαcosβ=cos(α+β)+cos(αβ)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=αβ\varphi=\alpha-\beta: cosαcosβ=cosθ+cosφ2cos(θ+φ2)cos(θφ2)=cosθ+cosφ2cosθ+cosφ=2cos(θ+φ2)cos(θφ2)\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 cosθcosφ\cos\theta-\cos\varphi. Much like the previous few proofs, we will use the product to sum formulas to derive this formula. In particular, sinαsinβ=cos(αβ)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=αβ\theta=\alpha-\beta (note that we are not substituting θ=α+β\theta=\alpha+\beta and φ=αβ\varphi=\alpha-\beta): sinαsinβ=cosθcosφ2sin(θ+φ2)sin(φθ2)=cosθcosφ2cosθcosφ=2sin(θ+φ2)sin(φθ2)or =2sin(θ+φ2)sin(θφ2)\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...

Prasun Biswas - 5 years, 7 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 - 5 years, 7 months ago

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

Shabarish Ch - 5 years, 7 months ago

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I hated this stuff during schooling years!!

Venture HI - 5 years, 7 months ago

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

Daniel Liu - 5 years, 7 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 - 5 years, 7 months ago

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

Jimbo Magno - 4 years, 11 months ago

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