Prove that tan50° + tan60° + tan70° = tan80°.
Note that we require a proof. Showing that the equation holds for however many deicmal places is not a proof. The numbers we are concerned with have infinite decimals, and so you'll have to do some trigonometric algebra to solve this one.
Hint: the first step is to get both sides in terms of the same type of value. Try rewriting them both in terms of tan10°.
sin60° = ½√3
cos60° = ½
tan60° = √3
Let C = cos10°
Let S = sin10°
tan50° + tan60° + tan70°
= sin(60°−10°)/cos(60°−10°) + √3 + sin(60°+10°)/cos(60°+10°)
= (½√3·C − ½·S)/(½·C + ½√3·S) + √3 + (½√3·C + ½·S)/(½·C − ½√3·S)
= [ (√3·C − S)·(C − √3·S) + √3·(C + √3·S)·(C − √3·S) + (√3·C + S)·(C + √3·S) ] ÷ [ (C + √3·S)·(C − √3·S) ]
= (3√3·C² − √3·S²) / (C² − 3·S²)
Now, tan80° = 1/tan10° = C/S
So the initial hypothesis is true if and only if:
(3√3·C² − √3·S²) / (C² − 3S²) = C/S
⇒ 3√3·C²S − √3·S³ = C³ − 3CS²
⇒ C³ − 3√3·C²S − 3CS² + √3·S³ = 0
By de Moivre's theorem:
(cos10° +i·sin10°)³ = cos30° + i·sin30°
⇒ C³ − 3CS² = cos30° = ½√3
⇒ 3C²S − S³ = sin30° = ½
So,
C³ − 3√3·C²S − 3CS² + √3·S³
= (C³ − 3CS²) − √3·(3C²S − S³)
= (½√3) − √3·(½)
= 0
QED