Question 1

I understand how this video proves the angle addition for sine, but not where this formula comes from to begin with, I feel like somewhere I missed a step.  It seems like a very complex proof for such a simple concept, why can't we just add sine a + sine b directly?  What is it about trig functions that makes angle additions so complicated?  I felt like I was grasping all the trig identities/unit circle definitions, etc up to this point but just crashed & burned here...  The video is very clear but it seems like there should be some sort of introductory video to the concept of adding angles & why we can't do it more easily...  Maybe it's just me?

Accepted Answer

The proof is where the formula comes from. We can't do sin(a + b) = sin(a) + sin(b) because sine does not distribute. It's similar to x^2: (a + b)^2 isn't a^2 + b^2, it's a^2 + 2ab + b^2. The same thing applies to sin(a + b): sin(a + b) = sin(a)cos(b) + cos(a)sin(b).

Question 2

It seems that this (very nice) proof only covers the case where x, y, and (x+y) are all acute angles.  How would one generalize this to cover *all* x and y?

Accepted Answer

This is a good question. Here's a proof I just came up with that the angle addition formula for sin() applies to angles in the second quadrant:

Given: pi/2 < a < pi and pi/2 < b < pi // a and b are obtuse angles less than 180°.
Define: c = a - pi/2 and d = b - pi/2 // c and d are acute angles.
Theorem: sin(c + d) = sin(c)*cos(d) + cos(c)*sin(d) // angle addition formula for sin().
Substitute: sin((a - pi/2) + (b - pi/2)) = sin(a - pi/2)*cos(b - pi/2) + cos(a - pi/2)*sin(b - pi/2)
Simplify: sin((a - pi/2) + (b - pi/2)) = sin(a + b - pi)
sin(a + b - pi) = -sin(a + b) // from unit circle
sin(a - pi/2) = -cos(a) and sin(b - pi/2) = -cos(b) // from unit circle
cos(a - pi/2) = sin(a) and cos(b - pi/2) = sin(b) // from unit circle
Substitute: -sin(a + b) = -cos(a)*sin(b) + sin(a)*(-cos(b))
Simplify and rearrange: sin(a + b) = sin(a)*cos(b) + cos(a)*sin(b) // Done

You can also use this method to generalize to angles greater than 180°, and it also works for the cos() addition formula.

Question 3

I understood all the parts of the proof, but is it okay to generalize this after proving it with only a right triangle that has a side of unit measure? Isn't there a way to prove it holds true for all angles, without taking the measure of a side as 1?

Accepted Answer

The trigonometric functions are all based on the ratios of the sides of right triangles and for a similar triangle (which is the case for triangles with the same angles) this ratio will remain constant as you scale up or down the length of the hypotenuse.

Question 4

Very clear and detailed proof, but where does this diagram even come from? I feel like I need to know where there diagram is derived from as well.

Accepted Answer

Hello Omar,

If you have a protractor I encourage you to construct the triangles to show the sum of SIN(30 degrees) plus SIN(45 degrees).

Tips: Let the length of the upper triangle's hypotenuse equal one. Observe that this is the only length that equals one. Take your time to do the procedure even if you need start over multiple times like I did. It's worth the effort and you will see how the triangles came to be. You will also have a nice drawing you can hang on you wall because you will "own" the identity.

Please leave a comment and let us know if you were successful.

Regards,

APD

Question 5

at 6:43 why is that angle labeled 90 - y?

Accepted Answer

We don't know what y is, but we know that those two angles add up to 90. That means the other one has to be y less than 90.

Think about it: 
y + (that angle) = 90
(that angle) = 90 - y

Question 6

Say we are using a unit circle. Say A is the center of that unit circle. And we want the proof related that senario. In that case AB is as above 1 but AC would also be 1.... So the above proof would need to be thought of in a different way. Is there any proof of the sine addition identity using the unit circle?

Accepted Answer

𝐴𝐶 wouldn't be equal to 1, because point 𝐶 wouldn't be on the unit circle.

Question 7

sin(∠ABC−60∘) , im confused whether I should be subtracting or adding the two because I'm subtracting -60∘ instead of adding 60∘ : sin(theta)*cos(60) + cos(theta)*sin(60)
                                                                      or: sin(theta)*cos(60) - cos(theta)*sin(60)
Do i always add sinacosb and cosasinb regardless of positive or negative value of 'b' in sin(a+b)??
Also what about for cos(a+b)? ( Where the value of b is negative)
Sorry if this doesn't make any sense...

Accepted Answer

*Addition Formulas*:
Sin(a+b)=sin(a)cos(b)+cos(a)sin(b).
Cos(a+b)=cos(a)cos(b) - sin(a)sin(b).
Tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b)).
*Subtraction Formulas*:
Sin(a-b)=sin(a)cos(b) - cos(a)sin(b).
Cos(a-b)=cos(a)cos(b)+sin(a)sin(b).
Tan(a-b)=(tan(a)-tan(b))/(1+tan(a)tan(b)).

Question 8

How to prove sin0/cos0+cos0/sin0=sec0 cos ec0?
I mean i'm putting 0 because i couldnt type theater sign..

Accepted Answer

Ok, first the variable is called theta (θ).

https://en.wikipedia.org/wiki/Theta

As for your problem:

```sin(θ)/cos(θ) + cos(θ)/sin(θ) = sec(θ)•csc(θ)

sin(θ)/cos(θ) + cos(θ)/sin(θ) = 1/cos(θ)•1/sin(θ)

sin^2(θ)/cos(θ) + cos(θ) = 1/cos(θ)•sin(θ)/sin(θ)

sin^2(θ) + cos^2(θ) = cos(θ)/cos(θ)•sin(θ)/sin(θ)

sin^2(θ) + cos^2(θ) = 1•1

sin^2(θ) + cos^2(θ) = 1 QED```

Precalculus

Course: Precalculus > Unit 2

Proof of the sine angle addition identity

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