The Best Diagram for Angle Sum Identities

Рет қаралды 3,173

Күн бұрын

Пікірлер: 33

@mcrow312166 2 ай бұрын

I love how simple this proof is. I also like the other proof with the constructed perpendicular lines and adding the lengths but the elegance of this proof is charming

@oxbmaths 2 ай бұрын

Nice. Very elegant elementary proof and one that is easy to remember.

@skinnykevin3998 2 ай бұрын

I've never seen this argument before. It's incredible!

@iMíccoli 2 ай бұрын

Beautiful ❤

@cauchym9883 2 ай бұрын

I could derive the first result from your figure without resorting to the formula 1/2*a*b*sin(\theta+\phi) by using the law of sines instead: sin(\theta+\phi)/(c+d)=sin(90°-\theta)/a=h/(a*b) Multiplying both sides with (c+d) yields: sin(\theta+\phi)/(c+d)=h*c/(a*b) + h*d/(a*b)=sin(\theta)*cos(\phi)+cos(\theta)*sin(\phi).

@DrBarker 2 ай бұрын

This is very neat, I wish I'd thought of that!

@ojas3464 2 ай бұрын

@@DrBarker Having proved the first, and using the fact that θ is acute iff 90° - θ so, replace every occurrence of one of them, say θ by 90° - θ' on both sides to obtain the cosine of sums

@gibbogle 2 ай бұрын

That works for me, since I know the law of sines.

@guslackner9270 29 күн бұрын

You don’t have to use Dr. Barker’s identity formulaically though, you can just drop a perpendicular to line b from the angle between a and d. Then the length of that dropped line is a*sin(theta+phi) by the definition of sine. It doesn’t matter that the bases which sum to b are different lengths because the area formula requires both triangles to be multiplied by one half. So the area is just the shared side a*sin(theta+phi) times b times 1/2.

@KSignalEingang 2 ай бұрын

If you're just looking for a way to derive the angle sum formulas for sin and cos, Euler's e^iθ = cosθ + i*sinθ provides a much more compact way of doing so, with fairly straightforward algebra.

@bpark10001 2 ай бұрын

This is much messier than stacking 2 right triangles, having hypotenuse of the first be also the long leg of the second. The stacking scheme relies only upon the definition of sine is opposite over hypotenuse, cosine is adjacent over hypotenuse. Other than that only simple geometry (similar triangles) is required. Using the sine triangle area rule is itself more complicated than the sine of angle sum. It also can't operate in other than the 1st quadrant.

@ojas3464 2 ай бұрын

My memory may be vague, I thing S L Loney uses this suggestion in Volume 1, Plane Trigonometry

@kikilolo6771 2 ай бұрын

Nice I didn't want to learn all these (if you have other techniques for the other formulas I'm interested)

@mekbebtamrat817 2 ай бұрын

Great pedagogical tool

@gibbogle 2 ай бұрын

That's good, but I didn't know the starting formula for the area: half the sine of the angle times the product of the two adjacent sides. That's news to me. Where does that come from?

@jonahansen 20 күн бұрын

Took me for a pause, but it's pretty easy to derive. Just draw out a triangle with two known sides and an included angle. Actually handy to remember..

@ojas3464 2 ай бұрын

👍

@anandarunakumar6819 2 ай бұрын

Lovely proof. Good approach to those who find it hard to remember the traditional methods.

@dakcom-mk6mp 2 ай бұрын

Nice

@bobbybannerjee5156 2 ай бұрын

Your proofs are (of course) valid, but the wrong way round. We prove the area of a triangle 📐 theorem using the sin(a+b) rule. Likewise, the cosine rule comes way later in the study of trigonometry. Though far more difficult, the standard way of proving them using the "double triangle" and distance methods from first principles is more satisfying.

@codatheseus5060 2 ай бұрын

I love your videos, I'd love to see you do a video where you do something like recreating trig functions while trying to avoid standard methods, or, making analogous functions out of the remainder mod function (just distinguishing it from the absolute value "mod") I also think it may be fun to do something like this: I noticed the other day that for n-dimensional-squares you can count in binary up to the number of places for dimensions for vertices locations. I'd love to see some way of mathing stuff so that works with other shapes, and then trying to find the underlying pattern in how that changes based on the shape

@DrBarker 2 ай бұрын

Thank you! I actually have a few old videos on hypercubes which might be related to what you're looking for.

@robertsandy3794 2 ай бұрын

This is a nice proof for small angles as you said

@nixonchan1383 2 ай бұрын

N e a t !

@guigsbi7979 2 ай бұрын

Why is area for triangle 1/2*ab*sin(theta+phi) ?

@Happy_Abe 2 ай бұрын

From 1/2*bh where b is base and h is height If you view a as a base then using trigonometry you can see that the height from that base will be b*sin(theta+phi) which comes from the sin being h/b since that angle is opposite to the height and b will be the hypotenuse of that half triangle.

@anandharamang3289 2 ай бұрын

In vector maths, cross product of b & c vector, bc sinA is the area of parallelogram. Half of the area is area of the triangle . 1/2 bc sin(theta +chi)

@guigsbi7979 2 ай бұрын

@@Happy_Abe thanks

@atzuras 2 ай бұрын

@anandharamang3289 you are right but that is a circular answer. The cross product is the definition of a vector operation. The property of being the equal to the area has to be proven by geometry. As shown in answers below.