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:

\(\cos(-\theta) = \cos\theta\)

\(\sin(-\theta)=-\sin\theta\)

\(\sin\theta = \cos(90-\theta)\)

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: \(\sin(\alpha+\beta)=\sin\alpha \cos\beta+\sin\beta\cos\alpha\). 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 \(\sin(\alpha-\beta)\). To do this, let's refer back to the one formula we do know: \(\sin(\alpha+\beta)=\sin\alpha \cos\beta+\sin\beta\cos\alpha\). Note that if we replace \(\beta\) with \(-\beta\), then we can derive a formula for \(\sin(\alpha-\beta)\). Let's do that: \[\begin{align*} \sin(\alpha-\beta)&=\sin(\alpha+(-\beta))\\ &=\sin\alpha \cos(-\beta)+\sin(-\beta)\cos\alpha\\ &= \sin\alpha \cos\beta + (-\sin\beta)\cos\alpha\\ &= \boxed{\sin\alpha\cos\beta-\sin\beta\cos\alpha} \end{align*}\] and we are done. \(\Box\)

**Proving the Cosine Angle Sum formula**

We wish to find a formula for \(\cos(\alpha+\beta)\). 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 \(\cos(\theta)=\sin(90-\theta)\) for that.

\[\begin{align*} \cos(\alpha+\beta)&= \sin(90-(\alpha+\beta))\\ &= \sin((90-\alpha)-\beta)\\ &\text{Now we can use Sine Angle Difference formula:}\\ &= \sin(90-\alpha)\cos(\beta)-\sin\beta\cos(90-\alpha)\\ &= \cos\alpha\cos\beta-\sin\beta\sin\alpha\\ &= \boxed{\cos\alpha\cos\beta-\sin\alpha\sin\beta} \end{align*}\]

**Proving the Cosine Angle Difference formula**

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

\[\begin{align*} \cos(\alpha-\beta)&= \cos(\alpha+(-\beta))\\ &= \cos\alpha\cos(-\beta)-\sin\alpha\sin(-\beta)\\ &=\cos\alpha\cos\beta-\sin\alpha(-\sin\beta)\\ &= \boxed{\cos\alpha\cos\beta+\sin\alpha\sin\beta} \end{align*}\] 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 \(\sin(2\theta)\). This is simple using the Sine Angle Sum formula; simply plug in \(\alpha=\theta\) and \(\beta=\theta\).

\[\begin{align*} \sin(2\theta)&= \sin(\theta + \theta)\\ &= \sin\theta\cos\theta+\sin\theta\cos\theta\\ &=\boxed{2\sin\theta\cos\theta} \end{align*}\] and we are done. \(\Box\)

**Proving the Cosine Double-angle formula**

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

\[\begin{align*} \cos(2\theta)&=\cos(\theta+\theta)\\ &= \cos\theta\cos\theta-\sin\theta\sin\theta\\ &= \cos^2\theta-\sin^2\theta\\ &\text{We now can use }\sin^2\theta+\cos^2\theta=1\text{ to simplify this:}\\ &=\boxed{1-2\sin^2\theta}\\ &= \boxed{2\cos^2\theta-1} \end{align*}\] 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 \(\sin\left(\frac{theta}{2}\right)\). 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 \(\frac{\theta}{2}\) into a double-angle formula, then solving for \(\sin\left(\frac{\theta}{2}\right)\) will give us a formula.

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

\[\begin{align*} \cos\left(2\left(\frac{\theta}{2}\right)\right)&= \cos(\theta)\\ &= 1-2\sin^2\left(\frac{\theta}{2}\right)\\ &\implies 2\sin^2\left(\frac{\theta}{2}\right)=1-\cos\theta\\ &\implies \sin^2\left(\frac{\theta}{2}\right)=\dfrac{1-\cos\theta}{2}\\ &\implies \sin\left(\frac{\theta}{2}\right)=\boxed{\pm\sqrt{\dfrac{1-\cos\theta}{2}}} \end{align*}\]

Note that in this case, the \(\pm\) doesn't mean \(\sin\left(\frac{\theta}{2}\right)\) has two values; it simply means that if \(\frac{\theta}{2}\) 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 \(\cos\frac{\theta}{2}\). We can use a similar strategy for this as the previous proof, but instead we use \(\cos(2\theta)=2\cos^2\theta-1\). \[\begin{align*} \cos\left(2\left(\frac{\theta}{2}\right)\right)&= \cos(\theta)\\ &= 2\cos^2\left(\frac{\theta}{2}\right)-1\\ &\implies 2\cos^2\left(\frac{\theta}{2}\right)=\cos\theta+1\\ &\implies \cos^2\left(\frac{\theta}{2}\right)=\dfrac{\cos\theta+1}{2}\\ &\implies \cos\left(\frac{\theta}{2}\right)=\boxed{\pm\sqrt{\dfrac{\cos\theta+1}{2}}} \end{align*}\] Again, the \(\pm\) sign doesn't mean \(\cos\dfrac{\cos\theta+1}{2}\) has two values, it simply means take the appropriate sign based on what quadrant \(\frac{\theta}{2}\) 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 \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) and \(\tan(-\theta)=-\tan(\theta)\). 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 \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) will give us the formula for tangent.

\[\begin{align*} \tan(\alpha+\beta)&= \dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\\ &= \dfrac{\sin\alpha\cos\beta+\sin\beta\cos\alpha}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}\\ &\text{This is where the trick comes: }\\ &\text{multiply the top and bottom by }\dfrac{\frac{1}{\cos\alpha\cos\beta}}{\frac{1}{\cos\alpha\cos\beta}}\\ &=\dfrac{\frac{\sin\alpha}{\cos\alpha}+\frac{\sin\beta}{\cos\beta}}{1-\frac{\sin\alpha}{\cos\alpha}\cdot\frac{\sin\beta}{\cos\beta}}\\ &=\boxed{\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}} \end{align*}\] and we are done. \(\Box\)

**Proving Tangent Angle Difference formula**

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

\[\begin{align*} \tan(\alpha-\beta)&= \tan(\alpha+(-\beta))\\ &= \dfrac{\tan\alpha+\tan(-\beta)}{1-\tan\alpha\tan(-\beta)}\\ &=\boxed{\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}} \end{align*}\] and we are done. \(\Box\).

**Proving Tangent Double-angle formula**

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

\[\begin{align*} \tan(2\theta)&=\tan(\theta+\theta)\\ &= \dfrac{\tan\theta+\tan\theta}{1-\tan\theta\tan\theta}\\ &=\boxed{\dfrac{2\tan\theta}{1-\tan^2\theta}} \end{align*}\] 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.

\[\begin{align*} \tan\left(\frac{\theta}{2}\right)&= \dfrac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)}\\ &= \dfrac{\pm\sqrt{\dfrac{1-\cos\theta}{2}}}{\pm\sqrt{\dfrac{\cos\theta+1}{2}}}\\ &= \pm\dfrac{\sqrt{1-\cos\theta}}{\sqrt{\cos\theta+1}}\\ &\text{Rationalizing the denominator:}\\ &= \pm \dfrac{\sqrt{(1-\cos\theta)(\cos\theta+1)}}{\cos\theta+1}\\ &= \pm\dfrac{\sqrt{1-\cos^2\theta}}{\cos\theta+1}\\ &\text{Using }\sin^2\theta+\cos^2\theta=1\text{:}\\ &= \pm\dfrac{\sin\theta}{\cos\theta+1} \end{align*}\] and we are done. \(\Box\).

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

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

## Comments

Sort by:

TopNewestHi 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 · 3 years, 1 month ago

Log in to reply

– Josh Rowley · 3 years, 1 month ago

Really nice post but you've written \( Sin(\alpha + \beta) = Sin(\alpha)Cos(\beta) + Sin(\beta) + Cos(\alpha) \) twice at the start instead of \( Sin(\alpha + \beta) = Sin(\alpha)Cos(\beta) + Cos(\alpha)Sin(\beta) \)Log in to reply

Daniel – Daniel Liu · 3 years, 1 month ago

Log in to reply

– Josh Rowley · 3 years, 1 month ago

The second one is in Proving the Sine Angle Difference formulaLog in to reply

this stuff is very useful really!! – Will Jain · 2 years, 9 months ago

Log in to reply

what's all this for? Complex numbers are the best! :P – Vincent Huang · 3 years ago

Log in to reply

its really good but may not be easy for every one .not for me too – Siddharth Singh · 3 years ago

Log in to reply

– Daniel Liu · 3 years ago

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.Log in to reply

– Shaan Ragib · 2 years, 9 months ago

me too.Log in to reply

Are there any exercises accompanying this post? Thanks in advance for any information!! – Shabarish Ch · 3 years, 1 month ago

Log in to reply

Hint: somehow use \(\cos(\alpha+\beta)\) and \(\cos(\alpha-\beta)\). – Daniel Liu · 3 years ago

Log in to reply

– Shabarish Ch · 3 years ago

Will definitely try, thanks!!Log in to reply

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 · 2 years, 5 months ago

Log in to reply

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 · 2 years, 9 months ago

Log in to reply

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 · 2 years, 3 months ago

Log in to reply