A Powerful Way to Derive the Cosine and Sine Sum and Difference Identities

In this video I will show you how to derive the cosine and sine sum and difference identities. This is a really nice method because it only requires knowing Euler's formula. I hope this helps.

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Thank you:)

[damnyutoobe]

That Euler identity seems simple yet shows up so much and is so useful in many applications. I can see why it's held to such a high regard.

[jonko]

Very nice video! Love your content. Fun fact: Once you've used Euler's formula to derive the formulas for cos(x+y) and sin(x+y) you can get the formulas for cos(x-y) and sin(x-y) a little bit quicker by writing them as cos(x+(-y)) and sin(x+(-y)) and plug that into the sum formulas:
cos(x-y) = cos(x+(-y)) = cosx*cos(-y)-sinx*sin(-y) = cosxcosy+sinxsiny
sin(x-y) = sin(x+(-y)) = sinxcos(-y)+cosxsin(-y) = sinxcosy-cosxsiny

[ILoveMaths]

That's so neat! Perfect! I always had a hard time memorising these identities.
Notice that once you get the first two identities, you can just replace the y by -y and you can get the other two identities without having to start from scratch.

[ThomasLahn]

Neat. Note that once you have the identities for cos(x + y) and sin(x + y), you do not actually need to go back to the exponential function again because you can just replace y by −y, and then use the oddness of the sine function and the evenness of the cosine function as you did there.

Another neat derivation for these that I learned from my professor in Linear Algebra and can remember better because I can do it in my head, is to use 2D rotation matrices

R(φ) = [(cos(φ), −sin(φ)), (sin(φ), cos(φ))].

Their matrix multiplication is additive with respect to the parameter φ, and commutative, which is obvious if you consider how rotations work:

R(x + y) = R(y) R(x) = R(x) R(y).

Then you can identify the corresponding components. For example,

R₁₁(x + y)
= cos(x + y)
= R₁₁(x) R₁₁(y) + R₁₂(x) R₂₁(y)
= cos(x) cos(y) − sin(x) sin(y).

These two approaches combined show that multiplication of a vector by exp(𝚤 φ) is equivalent to a counter-clockwise rotation of it by φ around the axis perpendicular to the plane in which φ is measured, which helps to describe rotations in classical and quantum mechanics.

[bingeu]

Wow what a neat alternative way to derive it!

[JR-ucnk]

This video has perfect timing! Trying to sharpen my trig identity skills starting yesterday, excellent!

[telescopilan]

You're a blessing man. Thank you🙏
After doing math all day, your videos are therapy for me

[pavithran]

at 4:57 my mind got Goosebumps like the first time i got interested in math, thats so fascinating sir, thanks for making this video, one small request sir, can you tell is there any book for trigonometry to actually derive formula like this, because ever since i know this formula i have been memorizing that formula over there until i watch this video, thanks again and have a great week from india

[sergiolucas]

good video, man, never seen this way before :)

[Kapadozzz]

this reminds me about extracting roots of an imaginary number is pretty similar.

[NikhilSharma-jqsv]

You are lifesaver
I have been following you continuously.

[cardinalityofaset]

In my linear algebra class, the professor showed an easy geometric way to do it. It is really easy to find matrix that coresponds to the rotation in a plane through angle x. Now we notice that any mxn matrix A can be viewed as a function f_A, where f_A(x) = Ax, where x is n-dimentional vector. In our case, the matrix of a rotation through some angel is a 2x2 matrix. We also notice that multiplying two matrices (assuming the multiplication is defined) corresponds to the composition of the transformations (f_AB = f_A o f_B). And there you have it. On the left side of the equation (f_AB) will be matrix of a rotation through angle x+y (We can use the initial matrix, where we substitute x+y) and on the right side (f_A o f_B) we multipy matrix of a rotation through angle x (call it A) with matrix of a rotation through angle y (call it B) assuming that the composition of two rotations is a rotation. Because of the identity f_AB = f_A o f_B we know that the two matrices are equal. That means element in the position (i, j) of the left matrix is equal to the element (i, j) of the right matrix. This proves the identities of a conpound angle (sort of). The problem is that we dont know that the composition of two rotations is a rotation. That is not easy to show. You have to know bunch of stuff about singular decomposition of a matrix. However, it is a nice geometric way to do it :)

[UsernameDP]

Just straight up amazing, any formula that allows you to input arbitrary numbers for sine or cosine can be used to derive a formula consisting of an arbitrary input within sine and cosine. A cool way of thinking about algebraic proofs.

[ChechoColombia]

The only problem is that there is a circular argument because if you want to proof the Euler formula you must know the derivative of sine, which can only be proven if you know the identity of sin(x+y).

[NishantKumar-yhuw]

We could also do with series expansion.

[MeM_UK]

How do you choose which/how many problems to do as a self-learner? You said to do all the problems your teacher sets, but as a self learner should I just do all the problems? You clearly select... but on what basis?

[ShaolinMonkster]

Video of how euler formula is derived??

[nighatyasmin]

Why don’t you give a proof of Euler’s formula? It would be more interesting. The trig identities follow as corollaries.