tan ( 6 3 ∘ ) = a − b + c − b
The equation above is true for positive integers a , b , and c . What is the value of a + b + c ?
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Great way to reduce it to values that we know how to get to. Another way to remember how to calculate sin 1 8 ∘ , cos 1 8 ∘ is to look at the regular pentagon, since cos 1 8 ∘ = sin 1 0 8 ∘ , which is the interior angle.
Awesome solution sir
tan ( 6 3 ∘ ) = cot ( 2 5 4 ∘ )
Now, let's use: cot ( 2 θ ) = cot ( θ ) + csc ( θ ) .
So, the value we want is cot ( 5 4 ∘ ) + csc ( 5 4 ∘ )
We know that sin ( 5 4 ∘ ) = 4 1 + 5 , so csc ( 5 4 ∘ ) = 1 + 5 4 = 5 − 1 .
Also, cot 2 ( 5 4 ∘ ) + 1 = csc 2 ( 5 4 ∘ ) , so cot ( 5 4 ∘ ) = ( 5 − 1 ) 2 − 1 = 5 − 2 5 .
So, tan ( 6 3 ∘ ) = 5 − 2 5 + 5 − 1 .
Finally, manipulate it so it match with the required form:
tan ( 6 3 ∘ ) = 5 2 − 2 2 ⋅ 5 + ( 5 − 1 ) 2
tan ( 6 3 ∘ ) = 2 5 − 2 0 + 3 6 − 2 0
Hence, a = 2 5 , b = 2 0 , c = 3 6 and a + b + c = 8 1 .
Note: to obtatin sin ( 5 4 ∘ ) we can consider that:
cos ( 2 × 7 2 ∘ ) = cos ( 3 × 7 2 ∘ ) 2 cos 2 ( 7 2 ∘ ) − 1 = 4 cos 3 ( 7 2 ∘ ) − 3 cos ( 7 2 ∘ ) 4 cos 3 ( 7 2 ∘ ) − 2 cos 2 ( 7 2 ∘ ) − 3 cos ( 7 2 ∘ ) + 1 = 0 ( cos ( 7 2 ∘ ) − 1 ) ( 4 cos 2 ( 7 2 ∘ ) + 2 cos ( 7 2 ∘ ) − 1 ) = 0
Obviously cos ( 7 2 ∘ ) = 1 , and 0 < cos ( 7 2 ∘ ) < 1 , so let's find the positive root of the quadratic factor:
cos ( 7 2 ∘ ) = 4 − 1 + 5
So, cos ( 1 4 4 ∘ ) = 2 cos 2 ( 7 2 ∘ ) − 1 = − 4 1 + 5
And sin ( 5 4 ∘ ) = cos ( 3 6 ∘ ) = − cos ( 1 4 4 ∘ ) = 4 1 + 5
tan 6 3 ∘ = cos ( 4 5 ∘ + 1 8 ∘ ) sin ( 4 5 ∘ + 1 8 ∘ ) = ( cos 1 8 ∘ − sin 1 8 ∘ ) ( cos 1 8 ∘ + sin 1 8 ∘ ) ( cos 1 8 ∘ + sin 1 8 ∘ ) ( cos 1 8 ∘ + sin 1 8 ∘ ) = cos 3 6 ∘ 1 + 2 sin 1 8 ∘ cos 1 8 ∘
Note that sin 1 8 ∘ = 4 5 − 1 (proof is left for the reader), then cos 1 8 ∘ = 4 1 0 + 2 5 and cos 3 6 ∘ = 4 5 + 1 .
Now, we have
tan 6 3 ∘ = ( 5 − 1 ) + 4 ( 3 − 5 ) ( 1 0 + 2 5 ) = ( ( 5 + 1 ) − 2 5 × 1 ) + 4 ( ( 9 + 5 ) − 2 9 × 5 ) ( 1 0 + 2 5 ) = 3 6 − 2 0 + 4 8 0 − 3 2 5 = 3 6 − 2 0 + 2 5 − 2 0 = 4 5 + 1 1 + 2 ( 4 5 − 1 ) ( 4 1 0 + 2 5 ) .
Hence, a = 3 6 , b = 2 5 , c = 2 0 , and the required answer is a + b + c = 3 6 + 2 5 + 2 0 = 8 1
Note : ( x + y ) − 2 x × y = x − y , x > y
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tan 6 3 ∘ = cos 6 3 ∘ sin 6 3 ∘ = cos 6 3 ∘ cos ( 9 0 ∘ − 6 3 ∘ ) = cos 6 3 ∘ cos 2 7 ∘ = cos ( 4 5 ∘ + 1 8 ∘ ) cos ( 4 5 ∘ − 1 8 ∘ ) = 2 1 cos 1 8 ∘ − 2 1 sin 1 8 ∘ 2 1 cos 1 8 ∘ + 2 1 sin 1 8 ∘ = cos 1 8 ∘ − sin 1 8 ∘ cos 1 8 ∘ + sin 1 8 ∘ = ( cos 1 8 ∘ − sin 1 8 ∘ ) ( cos 1 8 ∘ + sin 1 8 ∘ ) ( cos 1 8 ∘ + sin 1 8 ∘ ) ( cos 1 8 ∘ + sin 1 8 ∘ ) = cos 2 1 8 ∘ − sin 2 1 8 ∘ cos 2 1 8 ∘ + 2 sin 1 8 ∘ cos 1 8 ∘ + sin 2 1 8 ∘ = cos 3 6 ∘ 1 + sin 3 6 ∘
Now we know that:
sin 2 3 6 ∘ + cos 2 3 6 ∘ ⇒ cos 2 5 4 ∘ + cos 2 3 6 ∘ cos 2 3 θ + cos 2 2 θ ⇒ ( 4 x 3 − 3 x ) 2 + ( 2 x 2 − 1 ) 2 1 6 x 6 − 2 4 x 4 + 9 x 2 + 4 x 4 − 4 x 2 + 1 1 6 x 6 − 2 0 x 4 + 5 x 2 1 6 x 4 − 2 0 x 2 + 5 ⇒ x 2 ⇒ cos 2 1 8 ∘ = 1 = 1 Let θ = 1 8 ∘ = 1 Let cos θ = cos 1 8 ∘ = x = 1 = 1 = 0 Since x = 0 = 0 = 3 2 2 0 ± 4 0 0 − 3 2 0 = 8 5 + 5 8 5 − 5 ⇒ θ > 1 8 ∘ , rejected
Now, we have:
cos 3 6 ∘ sin 3 6 ∘ ⇒ tan 6 3 ∘ = 2 cos 2 1 8 ∘ − 1 = 4 5 + 5 − 1 = 4 1 + 5 = 1 − cos 2 3 6 ∘ = 4 1 6 − ( 6 + 2 5 ) = 4 1 0 − 2 0 = cos 3 6 ∘ 1 + sin 3 6 ∘ = 1 + 5 4 + 1 0 − 2 0 = ( 5 + 1 ) ( 5 − 1 ) ( 4 + 1 0 − 2 0 ) ( 5 − 1 ) = 4 4 ( 5 − 1 ) + ( 1 0 − 2 0 ) ( 6 − 2 0 ) = 4 4 ( 5 − 1 ) + 6 0 + 2 0 − 1 6 2 0 = 5 − 1 + 5 − 2 0 = ( 5 − 1 ) 2 + 2 5 − 2 0 = 6 − 2 5 + 2 5 − 2 0 = 3 6 − 2 0 + 2 5 − 2 0
⇒ a + b + c = 3 6 + 2 0 + 2 5 = 8 1