What a beautiful identity

Geometry Level 2

$\large \tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ$

What is the value of the expression above?

5 4 3 2 1

1 solution

Vineet Golcha
May 24, 2015

tan 9 - tan 27 - tan 63 + tan 81,

= tan 9 - tan 27 - tan (90-27) + tan (90-9),

= tan 9 - tan 27 - cot 27 + cot 9,

=[tan 9 + cot 9] - [ tan 27 + cot 27],

=[sin 9/ cos 9 + cos 9/ sin 9] - [ sin 27/cos 27 + cos 27/sin 27],

= 2/sin 18 - 2/sin 54,

= 2[ 1/sin 18 - 1/ sin 54],

=2 [ (sin 54 - sin 18)/ ( sin 18 sin 54) ],

= 2 [ ( 2 cos 36 sin 18)/( sin 18 sin 54)],

= 4 cos 36 / sin 54,

= 4 sin 54 / sin 54,

= 4

Moderator note:

Nice work! Bonus question: Can you find the exact form of $\sin(54^\circ)$ ?

Let A=18°. Then 3A = 90- 2A, so cos(3A) = sin(2A). Expanding each of these trigonometric functions gives

4cos^3(A)−3cos(A)=2sin(A)cos(A) ⇒4cos^2(A)−3=2sin(A) ⇒ 4−4sin^2(A)−3=2sin(A) ⇒4sin^2(A)+2sin(A)−1=0, Solving for sin(A) ⇒sin(A) = (−2+20)/√8=(−1+5)/√4 ⇒sin54∘ = sin(3A) = 3sin(A)−4sin^3(A)=(1+5)/√4

Vijay Simha - 6 years ago

sin 54 = cos (90-36) = cos 36 (because it's on the first quadrant, the result is not negative.)

this is my very first post i hope this was right

Ardianto Kurniawan - 6 years ago

That's right.

Chung Kevin - 6 years ago

how did you conclude the 4th last line from the previous one...i am a big novice to trig...

Ritwik Jain - 6 years ago

we have this trigonometry equation where

sin A - sin B = 2 cos (A+B)/2 sin(A-B)/2

sin A + sin B = 2 sin (A+B)/2 x cos(A-B)/2

sin A - sin B = 2 cos (A+B)/2 x sin(A-B)/2

cos A + cos B = 2 cos (A+B)/2 x cos (A-B)/2

cos A - cos B = - 2 sin (A+B)/2 x sin (A-B)/2

remember this structure as a letter like this so it could be easy to memorize :

S+ S = 2SS

S - S = 2CS

C + C = 2CC

C - C = -2SS

if you wants to reverse from the left to the right then you gotta divide the angle inside like what i wrote on the top. but when you want to reverse from the right to the left for example:

2Sin(A)CosB = Sin (A+B) + Sin (A-B)

and so oon..

Ardianto Kurniawan - 6 years ago

vineet golcha-i did not understand the sixth line of your solution.

Ayush Singh - 6 years ago

Vinceet added each set of fractions. The result for both in the numerator would be a sin^2 + cos^2 situation, which equals 1. The result for both in the denominator would involved using the sine double angle formula to get 0.5(sin 2x). Since the 0.5 is in the denominator, you get a 2 in the numerator.

Louis W - 6 years ago

What a beautiful identity

1 solution

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