One of my all-time favourite topics in Math is the Golden Ratio of (1 + √5)/2.
I happen to know that cos 36° = (1+√5)/4
Notice also that in a regular pentagon , we have interior angles of 108° , and if we draw in some diagonals, we have all kinds of 36, 72 and 108° angles.
And also note that 90-72 = 18, or 18 = 1/2 of 36
and WOW, all kinds of neat stuff here
Anyway, I will assume you don't know the Golden Ratio connection, and let's do it this way:
let x = 18°
then 5x = 90° , and then
2x = 90 - 3x
take sin of both sides
sin(2x) = sin(90 - 3x)
but on the right side I have a well-known trig identity
sin 2x = cos 3x
but we also know that cos 3x = 4cos^3 x - 3cos x , so
sin 2x = 4cos^3 x - 3cos x
2 sinx cosx = 4cos^3 x - 3cosx
divide by cosx
2 sinx = 4cos^2 x - 3
another ID ....
2 sinx = 4(1-sin^2 x) - 3
2 sinx = 4 - 4 sin^2 x - 3
4 sin^2 x + 2sinx - 1 = 0
sin x = (-2 ± √20)/8
= (-1 ± √5)/4 , but x was 18°, so we use the positive answer
sin 18° = (-1 + √5)/4 , I checked with my calculator, it is right!
recall that cos A = √(1 - sin^2 A)
so cos 18° = √(1 - (-1+√5)^2/16)
= √( (16 - (1 - 2√5 + 5)/16 )
= √(10 + 2√5)/4
YEAHHHHH
Ok, now we have sin18 and cos18,
tan18 and cot18 should be easy
I leave it up to you to work out sin9° and cos9°
hint: cos 2A = 1 - 2 sin^2 A , think 2A =18, then A = 9
Without using the Calculator, solve the ff: (show solution)
a. Cos 18°
b. Tan 18°
c. Cot 18°
d. Sin 9°
e. Cos 9°
[Hint: Using any of the formulas for Half- Angle, Double Angles, Reduction]
Good Day!!! Thank you for the time. :)
1 answer
1 answer