Deriving Trigonometric Identities (Part 1)

EDIT: check out part 2 of this series here!

Hi everyone,

Have you ever had a hard time remembering all those Trigonometric Identities, like the cosine angle sum, or sine angle difference, or half angle formulas? In this post, I aim to show you guys how to prove all of the formulas, so that if you ever forget one formula, you can just prove it again!

I am assuming you already know the very very basic formulas:

cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta

sin⁡(−θ)=−sin⁡θ\sin(-\theta)=-\sin\theta

sin⁡θ=cos⁡(90−θ)\sin\theta = \cos(90-\theta)

If you don't then I recommend learning those before reading on.

In addition, I will only prove the formulas for sine and cosine. Formulas, with their proofs, for tangent can be found in the appendix.

You will need to memorize only one formula to derive the rest: sin⁡(α+β)=sin⁡αcos⁡β+sin⁡βcos⁡α\sin(\alpha+\beta)=\sin\alpha \cos\beta+\sin\beta\cos\alpha. This one is pretty impractical to reprove in a timely manner, so you will have to lock it into your brain. In addition, try to find the proof before you read my proof if you can; this way, it is easier for you to memorize it later on. Without further ado, let's start proving!

Proving the Sine Angle Difference formula

We wish to find a formula for sin⁡(α−β)\sin(\alpha-\beta). To do this, let's refer back to the one formula we do know: sin⁡(α+β)=sin⁡αcos⁡β+sin⁡βcos⁡α\sin(\alpha+\beta)=\sin\alpha \cos\beta+\sin\beta\cos\alpha. Note that if we replace β\beta with −β-\beta, then we can derive a formula for sin⁡(α−β)\sin(\alpha-\beta). Let's do that: sin⁡(α−β)=sin⁡(α+(−β))=sin⁡αcos⁡(−β)+sin⁡(−β)cos⁡α=sin⁡αcos⁡β+(−sin⁡β)cos⁡α=sin⁡αcos⁡β−sin⁡βcos⁡α\begin{aligned} \sin(\alpha-\beta)&=\sin(\alpha+(-\beta))\\ &=\sin\alpha \cos(-\beta)+\sin(-\beta)\cos\alpha\\ &= \sin\alpha \cos\beta + (-\sin\beta)\cos\alpha\\ &= \boxed{\sin\alpha\cos\beta-\sin\beta\cos\alpha} \end{aligned} and we are done. □\Box

Proving the Cosine Angle Sum formula

We wish to find a formula for cos⁡(α+β)\cos(\alpha+\beta). Since we only know the formulas for sum and difference of angles with sine, we want to turn this into sine of something. We can use cos⁡(θ)=sin⁡(90−θ)\cos(\theta)=\sin(90-\theta) for that.

cos⁡(α+β)=sin⁡(90−(α+β))=sin⁡((90−α)−β)Now we can use Sine Angle Difference formula:=sin⁡(90−α)cos⁡(β)−sin⁡βcos⁡(90−α)=cos⁡αcos⁡β−sin⁡βsin⁡α=cos⁡αcos⁡β−sin⁡αsin⁡β\begin{aligned} \cos(\alpha+\beta)&= \sin(90-(\alpha+\beta))\\ &= \sin((90-\alpha)-\beta)\\ &\text{Now we can use Sine Angle Difference formula:}\\ &= \sin(90-\alpha)\cos(\beta)-\sin\beta\cos(90-\alpha)\\ &= \cos\alpha\cos\beta-\sin\beta\sin\alpha\\ &= \boxed{\cos\alpha\cos\beta-\sin\alpha\sin\beta} \end{aligned}

Proving the Cosine Angle Difference formula

We wish to find a formula for cos⁡(α−β)\cos(\alpha-\beta). Note that we can plug in −β-\beta instead of β\beta in the Cosine Angle Sum formula to get the formula we want.

cos⁡(α−β)=cos⁡(α+(−β))=cos⁡αcos⁡(−β)−sin⁡αsin⁡(−β)=cos⁡αcos⁡β−sin⁡α(−sin⁡β)=cos⁡αcos⁡β+sin⁡αsin⁡β\begin{aligned} \cos(\alpha-\beta)&= \cos(\alpha+(-\beta))\\ &= \cos\alpha\cos(-\beta)-\sin\alpha\sin(-\beta)\\ &=\cos\alpha\cos\beta-\sin\alpha(-\sin\beta)\\ &= \boxed{\cos\alpha\cos\beta+\sin\alpha\sin\beta} \end{aligned} and we are done proving all 4 different angle sum and difference formulas (sum and difference formula proofs for tangent can be found in the Appendix). □\Box

Proving the Sine Double angle formula

We wish to find a formula for sin⁡(2θ)\sin(2\theta). This is simple using the Sine Angle Sum formula; simply plug in α=θ\alpha=\theta and β=θ\beta=\theta.

sin⁡(2θ)=sin⁡(θ+θ)=sin⁡θcos⁡θ+sin⁡θcos⁡θ=2sin⁡θcos⁡θ\begin{aligned} \sin(2\theta)&= \sin(\theta + \theta)\\ &= \sin\theta\cos\theta+\sin\theta\cos\theta\\ &=\boxed{2\sin\theta\cos\theta} \end{aligned} and we are done. □\Box

Proving the Cosine Double-angle formula

We want to find a formula for cos⁡(2θ)\cos(2\theta). Similarly for proving the Sine Double-angle formula, we plug in α=θ\alpha=\theta and β=θ\beta=\theta in the Cosine Angle Sum formula to get our desired formula.

cos⁡(2θ)=cos⁡(θ+θ)=cos⁡θcos⁡θ−sin⁡θsin⁡θ=cos⁡2θ−sin⁡2θWe now can use sin⁡2θ+cos⁡2θ=1 to simplify this:=1−2sin⁡2θ=2cos⁡2θ−1\begin{aligned} \cos(2\theta)&=\cos(\theta+\theta)\\ &= \cos\theta\cos\theta-\sin\theta\sin\theta\\ &= \cos^2\theta-\sin^2\theta\\ &\text{We now can use }\sin^2\theta+\cos^2\theta=1\text{ to simplify this:}\\ &=\boxed{1-2\sin^2\theta}\\ &= \boxed{2\cos^2\theta-1} \end{aligned} Note that in the Sine Double-angle formula, the result has both Sine and Cosine functions, but in the Cosine Double-angle formula, it can be composed entirely of Sine or Cosine. In this way, using this formula to problem solve instead of the Sine Double-angle formula often makes things much simpler. □\Box

Proving Sine Half-angle Formula

We wish to find a formula for sin⁡(theta2)\sin\left(\frac{theta}{2}\right). Since we have the double-angle formulas already, we can express the sine or cosine of an angle in terms of sine or cosine of half that angle. In other words, substituting θ2\frac{\theta}{2} into a double-angle formula, then solving for sin⁡(θ2)\sin\left(\frac{\theta}{2}\right) will give us a formula.

sin⁡(2(θ2))=sin⁡(θ)=2sin⁡(θ2)cos⁡(θ2)\begin{aligned} \sin\left(2\left(\frac{\theta}{2}\right)\right)&= \sin(\theta)\\ &= 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \end{aligned} but wait; we can't solve for sin⁡(θ2)\sin\left(\frac{\theta}{2}\right) or cos⁡(θ2)\cos\left(\frac{\theta}{2}\right) because both of them are in the equation! Fortunately, using cos⁡(2θ)\cos(2\theta) instead of sin⁡(2θ)\sin(2\theta) will solve that problem, because as we said before, we can represent cos⁡(2θ)\cos(2\theta) entirely out of sines or cosines. In this case, we will use cos⁡(2θ)=1−2sin⁡2θ\cos(2\theta)=1-2\sin^2\theta.

cos⁡(2(θ2))=cos⁡(θ)=1−2sin⁡2(θ2)  ⟹  2sin⁡2(θ2)=1−cos⁡θ  ⟹  sin⁡2(θ2)=1−cos⁡θ2  ⟹  sin⁡(θ2)=±1−cos⁡θ2\begin{aligned} \cos\left(2\left(\frac{\theta}{2}\right)\right)&= \cos(\theta)\\ &= 1-2\sin^2\left(\frac{\theta}{2}\right)\\ &\implies 2\sin^2\left(\frac{\theta}{2}\right)=1-\cos\theta\\ &\implies \sin^2\left(\frac{\theta}{2}\right)=\dfrac{1-\cos\theta}{2}\\ &\implies \sin\left(\frac{\theta}{2}\right)=\boxed{\pm\sqrt{\dfrac{1-\cos\theta}{2}}} \end{aligned}

Note that in this case, the ±\pm doesn't mean sin⁡(θ2)\sin\left(\frac{\theta}{2}\right) has two values; it simply means that if θ2\frac{\theta}{2} is in the first or second quadrants, the sign will be positive, and if it is in the third or fourth quadrants, it will be negative. □\Box

Proving the Cosine Half-angle formula

We want to find a general formula for cos⁡θ2\cos\frac{\theta}{2}. We can use a similar strategy for this as the previous proof, but instead we use cos⁡(2θ)=2cos⁡2θ−1\cos(2\theta)=2\cos^2\theta-1. cos⁡(2(θ2))=cos⁡(θ)=2cos⁡2(θ2)−1  ⟹  2cos⁡2(θ2)=cos⁡θ+1  ⟹  cos⁡2(θ2)=cos⁡θ+12  ⟹  cos⁡(θ2)=±cos⁡θ+12\begin{aligned} \cos\left(2\left(\frac{\theta}{2}\right)\right)&= \cos(\theta)\\ &= 2\cos^2\left(\frac{\theta}{2}\right)-1\\ &\implies 2\cos^2\left(\frac{\theta}{2}\right)=\cos\theta+1\\ &\implies \cos^2\left(\frac{\theta}{2}\right)=\dfrac{\cos\theta+1}{2}\\ &\implies \cos\left(\frac{\theta}{2}\right)=\boxed{\pm\sqrt{\dfrac{\cos\theta+1}{2}}} \end{aligned} Again, the ±\pm sign doesn't mean cos⁡cos⁡θ+12\cos\dfrac{\cos\theta+1}{2} has two values, it simply means take the appropriate sign based on what quadrant θ2\frac{\theta}{2} is in. □\Box

In Part 2 of "Deriving Trigonometric Identities", I will go over how to prove the sum of two trigonometric functions and product of two trigonometric functions. Stay tuned!

APPENDIX

In this section you will also need to know the very basic identity tan⁡θ=sin⁡θcos⁡θ\tan\theta=\dfrac{\sin\theta}{\cos\theta} and tan⁡(−θ)=−tan⁡(θ)\tan(-\theta)=-\tan(\theta). In addition, the proofs of these are a bit harder than the above, and might be a little less practical to reprove. Therefore, it is to your best interest to memorize these.

Proving the Tangent Angle Sum formula

We already know the Sine and Cosine Angle Sum formulas, so using tan⁡θ=sin⁡θcos⁡θ\tan\theta=\dfrac{\sin\theta}{\cos\theta} will give us the formula for tangent.

tan⁡(α+β)=sin⁡(α+β)cos⁡(α+β)=sin⁡αcos⁡β+sin⁡βcos⁡αcos⁡αcos⁡β−sin⁡αsin⁡βThis is where the trick comes: multiply the top and bottom by 1cos⁡αcos⁡β1cos⁡αcos⁡β=sin⁡αcos⁡α+sin⁡βcos⁡β1−sin⁡αcos⁡α⋅sin⁡βcos⁡β=tan⁡α+tan⁡β1−tan⁡αtan⁡β\begin{aligned} \tan(\alpha+\beta)&= \dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\\ &= \dfrac{\sin\alpha\cos\beta+\sin\beta\cos\alpha}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}\\ &\text{This is where the trick comes: }\\ &\text{multiply the top and bottom by }\dfrac{\frac{1}{\cos\alpha\cos\beta}}{\frac{1}{\cos\alpha\cos\beta}}\\ &=\dfrac{\frac{\sin\alpha}{\cos\alpha}+\frac{\sin\beta}{\cos\beta}}{1-\frac{\sin\alpha}{\cos\alpha}\cdot\frac{\sin\beta}{\cos\beta}}\\ &=\boxed{\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}} \end{aligned} and we are done. □\Box

Proving Tangent Angle Difference formula

We can replace −β-\beta instead of β\beta in the above formula to get a Tangent Angle Difference formula:

tan⁡(α−β)=tan⁡(α+(−β))=tan⁡α+tan⁡(−β)1−tan⁡αtan⁡(−β)=tan⁡α−tan⁡β1+tan⁡αtan⁡β\begin{aligned} \tan(\alpha-\beta)&= \tan(\alpha+(-\beta))\\ &= \dfrac{\tan\alpha+\tan(-\beta)}{1-\tan\alpha\tan(-\beta)}\\ &=\boxed{\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}} \end{aligned} and we are done. □\Box.

Proving Tangent Double-angle formula

We can use the Tangent Angle Sum formula to prove this one by substituting α=θ\alpha=\theta and β=θ\beta=\theta, much like our previous two Double-angle proofs.

tan⁡(2θ)=tan⁡(θ+θ)=tan⁡θ+tan⁡θ1−tan⁡θtan⁡θ=2tan⁡θ1−tan⁡2θ\begin{aligned} \tan(2\theta)&=\tan(\theta+\theta)\\ &= \dfrac{\tan\theta+\tan\theta}{1-\tan\theta\tan\theta}\\ &=\boxed{\dfrac{2\tan\theta}{1-\tan^2\theta}} \end{aligned} and we are done. □\Box

Proving Tangent Half-angle formula

We can prove the tangent half-angle formula much like we did for the tangent angle sum formula: by using both the sine and cosine half-angle formulas.

tan⁡(θ2)=sin⁡(θ2)cos⁡(θ2)=±1−cos⁡θ2±cos⁡θ+12=±1−cos⁡θcos⁡θ+1Rationalizing the denominator:=±(1−cos⁡θ)(cos⁡θ+1)cos⁡θ+1=±1−cos⁡2θcos⁡θ+1Using sin⁡2θ+cos⁡2θ=1:=±sin⁡θcos⁡θ+1\begin{aligned} \tan\left(\frac{\theta}{2}\right)&= \dfrac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)}\\ &= \dfrac{\pm\sqrt{\dfrac{1-\cos\theta}{2}}}{\pm\sqrt{\dfrac{\cos\theta+1}{2}}}\\ &= \pm\dfrac{\sqrt{1-\cos\theta}}{\sqrt{\cos\theta+1}}\\ &\text{Rationalizing the denominator:}\\ &= \pm \dfrac{\sqrt{(1-\cos\theta)(\cos\theta+1)}}{\cos\theta+1}\\ &= \pm\dfrac{\sqrt{1-\cos^2\theta}}{\cos\theta+1}\\ &\text{Using }\sin^2\theta+\cos^2\theta=1\text{:}\\ &= \pm\dfrac{\sin\theta}{\cos\theta+1} \end{aligned} and we are done. □\Box.

As an exercise, prove that the ±\pm sign can be entirely omitted; in other words, the formula is simply tan⁡(θ2)=sin⁡θcos⁡θ+1\boxed{\tan\left(\frac{\theta}{2}\right)=\dfrac{\sin\theta}{\cos\theta+1}}

Hope this post helped you guys memorize the multitude of Trigonometric Identities. And if you ever forget, you know you can prove it again!

Daniel

#Geometry #Trigonometry #TrigonometricIdentities #CosinesGroup #Proofs #Guide

Note by Daniel Liu
7 years, 3 months ago

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Comments

Hi all,

Sorry for the really long post this time. I think I should have divided it into three parts instead of two, but I already wrote it now and so here it is! Sorry if it is too much to digest :P

Yet another #CosinesGroup post, this time after a month of not posting (Oops, sorry!) Hope you guys enjoy it!

Daniel

Daniel Liu - 7 years, 3 months ago

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Really nice post but you've written Sin(α+β)=Sin(α)Cos(β)+Sin(β)+Cos(α) Sin(\alpha + \beta) = Sin(\alpha)Cos(\beta) + Sin(\beta) + Cos(\alpha) twice at the start instead of Sin(α+β)=Sin(α)Cos(β)+Cos(α)Sin(β) Sin(\alpha + \beta) = Sin(\alpha)Cos(\beta) + Cos(\alpha)Sin(\beta)

Josh Rowley - 7 years, 3 months ago

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Thanks for proofreading! However, I could only find that mistake once.

Daniel

Daniel Liu - 7 years, 3 months ago

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@Daniel Liu – The second one is in Proving the Sine Angle Difference formula

Josh Rowley - 7 years, 3 months ago

this stuff is very useful really!!

will jain - 6 years, 11 months ago

its really good but may not be easy for every one .not for me too

Siddharth Singh - 7 years, 3 months ago

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Yea, I agree. Some of the later proofs get harder to remember. However, try to remember the proof; this is not only better than memorizing the formulas, but you will gain problem-solving skills. You can more easily recognize when to apply these things to situations.

Daniel Liu - 7 years, 3 months ago

me too.

shaan ragib - 7 years ago

what's all this for? Complex numbers are the best! :P

Vincent Huang - 7 years, 3 months ago

Are there any exercises accompanying this post? Thanks in advance for any information!!

Shabarish Ch - 7 years, 3 months ago

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As an exercise, I suppose you can try to prove the formulas for sin⁡αsin⁡β\sin\alpha\sin\beta and cos⁡αcos⁡β\cos\alpha\cos\beta. Good luck!

Hint: somehow use cos⁡(α+β)\cos(\alpha+\beta) and cos⁡(α−β)\cos(\alpha-\beta).

Daniel Liu - 7 years, 3 months ago

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Will definitely try, thanks!!

Shabarish Ch - 7 years, 3 months ago

Thank you Daniel. By the way, in my country, I have to learn more and more Trigonometric formulas! But once you learn about them, I think you can't easily forget them, just like me!

Dang Anh Tu - 7 years ago

Can you add this set of notes into the Wiki pages for Trigonometric Identities?

Copy and paste the appropriate parts into the summary write-up.

Thanks!

Calvin Lin Staff - 6 years, 7 months ago

is'nt there any easy way t keep thm in mind 4ever.........i do remember bt later i 4get.....!!....thn again i hv too....!!

Bulbul Shekhu - 6 years, 6 months ago
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