Geometry Arc Circle Area – PUTNAM Problem A2 1998 – Trigonometry Identity Proof
Geometry Arc Circle Area
I Found this Interesting Math Olympiad PUTNAM Problem and Its Solution (not mine) whilst browsing.

.
.
.
A Comment for edification
A Very Clever Solution you have there.
Comment: Why Multiply by 2?
(i) Twice the First Picture in Addend
Note that
“First Picture in Addend” = (Area of Triangle) = (half of area of Top Left Rectangle)
Thus doubling it,
“Twice the First Picture in Addend” = Twice (half of area of Top Left Rectangle) = (Area of Top Left Rectangle)
(ii) Twice the Second Picture in Addend
Note that
“Second Picture in Addend” = (Area of Triangle) = (half of area of Bottom Right Rectangle)
Thus doubling it,
“Twice the Second Picture in Addend” = Twice (half of area of Bottom Right Rectangle) = (Area of Bottom Right Rectangle)
(iii) Twice the Third Picture in Addend
Note that
“Third Picture in Addend” = (Area of part-Arc)
Thus doubling it,
“Twice the Third Picture in Addend” = Twice (Area of part-Arc) = (Area of part-Arc COUNTED TWICE, one for its calculation in A and the other for its calculation in B.)
Yes only this area is counted “twice” in the Overall Sum because it occurs in both “A and B respectively” .
(iv) Total Shaded Area
(Total Shaded Area) or (A + B.)
= 2 x [ (half of area of Top Left Rectangle) + (half of area of Bottom Right Rectangle) + (Area of part-Arc)]
= 2 x [ Area of Arc with angle “theta” & radius “1”]
= 2 x [ (1/2) x r^2 x (theta) ], radius given as 1
= 2 x [ (1/2) x 1^2 x (theta) ]
= 2 x (1/2) x (theta)
= (theta)
= s
Depending only on the arc length s, when radius is given as claimed.
QED
Note:
Using (theta) in Radians,
r = radius = 1
(theta) = angle in Radians
s = arc length
Then,
s = r (theta)
s = (1) (theta)
s = (theta) which was substituted in the last line above.
P/S: A Visual Geometrical Proof for the infamous elementary trigonometric identities

How to understand the proof above?
It’s difficult to explain here. Here are some useful pointers:
Example: Look at shaded triangle in Second Picture above.
Sin (a + b)
= (Vertical dotted line)/ (Hypotenuse)
= (Cos a Sin b + Sin a Cos b) / 1
= (Cos a Sin b + Sin a Cos b)
Can you see it?
Note:
(Vertical dotted line) = (Cos a Sin b + Sin a Cos b)
By diagram construction.
Example: In the right triangle with angle “a” in the Top Right Corner,
It has (hypotenuse) = Sin b
and an ‘angle a’.
So, its
Sin a = (‘horizontal length’)/(hypotenuse)
Sin a = (‘horizontal length’)/(Sin b)
So, rewriting this we obtain:
(‘horizontal length’) = (Sin a) x (Sin b) as shown in Pic etc.
STAY CONNECTED