Notorious Trigonometry

Geometry Level 3

tan ⁡ 9 ∘ − tan ⁡ 2 7 ∘ − tan ⁡ 6 3 ∘ + tan ⁡ 8 1 ∘ = ? \tan 9^{\circ} -\tan 27^{\circ} -\tan 63^{\circ} +\tan 81^{\circ} = ?

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  • This should be solved without a calculator.
  • This problem is not original.

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The answer is 4.

Donglin Loo by donglin loo , 22, Singapore

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2 solutions

Donglin Loo
Jan 26, 2018

tan ⁡ 9 ∘ − tan ⁡ 2 7 ∘ − tan ⁡ 6 3 ∘ + tan ⁡ 8 1 ∘ \tan 9^{\circ} -\tan 27^{\circ} -\tan63^{\circ} +\tan 81^{\circ}

= tan ⁡ 9 ∘ − tan ⁡ 2 7 ∘ − cot ⁡ 2 7 ∘ + tan ⁡ 9 ∘ =\tan 9^{\circ} -\tan 27^{\circ} -\cot 27^{\circ} +\tan 9^{\circ}

= ( tan ⁡ 9 ∘ + cot ⁡ 9 ∘ ) − ( tan ⁡ 2 7 ∘ − cot ⁡ 2 7 ∘ ) =(\tan 9^{\circ} +\cot 9^{\circ}) -(\tan 27^{\circ} -\cot 27^{\circ})

= ( sin ⁡ 9 ∘ cos ⁡ 9 ∘ + cos ⁡ 9 ∘ sin ⁡ 9 ∘ ) − ( sin ⁡ 2 7 ∘ cos ⁡ 2 7 ∘ + cos ⁡ 2 7 ∘ sin ⁡ 2 7 ∘ ) =(\cfrac{\sin 9^{\circ}}{\cos 9^{\circ}} +\cfrac{\cos 9^{\circ}}{\sin 9^{\circ}}) -(\cfrac{\sin 27^{\circ}}{\cos 27^{\circ}} +\cfrac{\cos 27^{\circ}}{\sin 27^{\circ}})

= ( sin ⁡ 9 ∘ ) 2 + ( cos ⁡ 9 ∘ ) 2 sin ⁡ 9 ∘ cos ⁡ 9 ∘ − ( sin ⁡ 2 7 ∘ ) 2 + ( cos ⁡ 2 7 ∘ ) 2 sin ⁡ 2 7 ∘ cos ⁡ 2 7 ∘ =\cfrac{(\sin 9^{\circ})^{2}+(\cos 9^{\circ})^{2}}{\sin 9^{\circ}\cos 9^{\circ}}-\cfrac{(\sin 27^{\circ})^{2}+(\cos 27^{\circ})^{2}}{\sin 27^{\circ}\cos 27^{\circ}}

= 1 sin ⁡ 9 ∘ cos ⁡ 9 ∘ − 1 sin ⁡ 2 7 ∘ cos ⁡ 2 7 ∘ =\cfrac{1}{\sin 9^{\circ}\cos 9^{\circ}}-\cfrac{1}{\sin 27^{\circ}\cos 27^{\circ}}

= 2 2 sin ⁡ 9 ∘ cos ⁡ 9 ∘ − 2 2 s i n 2 7 ∘ cos ⁡ 2 7 ∘ =\cfrac{2}{2\sin 9^{\circ}\cos 9^{\circ}}-\cfrac{2}{2sin 27^{\circ}\cos 27^{\circ}}

= 2 sin ⁡ 1 8 ∘ − 2 sin ⁡ 5 4 ∘ =\cfrac{2}{\sin 18^{\circ}}-\cfrac{2}{\sin 54^{\circ}}

= 2 ( sin ⁡ 5 4 ∘ − sin ⁡ 1 8 ∘ ) sin ⁡ 5 4 ∘ sin ⁡ 1 8 ∘ =\cfrac{2(\sin 54^{\circ}-\sin 18^{\circ})}{\sin 54^{\circ}\sin 18^{\circ}}

= 2 ( 2 cos ⁡ 3 6 ∘ sin ⁡ 1 8 ∘ ) sin ⁡ 5 4 ∘ sin ⁡ 1 8 ∘ =\cfrac{2(2\cos 36^{\circ} \sin 18^{\circ})}{\sin 54^{\circ}\sin 18^{\circ}}

= 4 cos ⁡ 3 6 ∘ sin ⁡ 5 4 ∘ =\cfrac{4\cos 36^{\circ}}{\sin 54^{\circ}}

cos ⁡ 3 6 ∘ = sin ⁡ ( 9 0 ∘ − 3 6 ∘ ) = sin ⁡ 5 4 ∘ \cos 36^{\circ}=\sin (90^{\circ}-36^{\circ})=\sin 54^{\circ}

cos ⁡ 3 6 ∘ sin ⁡ 5 4 ∘ = 1 \cfrac{\cos 36^{\circ}}{\sin 54^{\circ}}=1

So, tan ⁡ 9 ∘ − tan ⁡ 2 7 ∘ − tan ⁡ 6 3 ∘ + tan ⁡ 8 1 ∘ = 4 ( 1 ) = 4 \tan 9^{\circ} -\tan 27^{\circ} -\tan63^{\circ} +\tan 81^{\circ}=4(1)=4

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Chew-Seong Cheong
Jun 25, 2018

X = tan ⁡ 9 ∘ − tan ⁡ 2 7 ∘ − tan ⁡ 6 3 ∘ + tan ⁡ 8 1 ∘ Since tan ⁡ ( 9 0 ∘ − θ ) = cot ⁡ θ = 1 tan ⁡ θ = tan ⁡ 9 ∘ + 1 tan ⁡ 9 ∘ − tan ⁡ 2 7 ∘ − 1 tan ⁡ 2 7 ∘ = sin ⁡ 9 ∘ cos ⁡ 9 ∘ + cos ⁡ 9 ∘ sin ⁡ 9 ∘ − sin ⁡ 2 7 ∘ cos ⁡ 2 7 ∘ − cos ⁡ 2 7 ∘ sin ⁡ 2 7 ∘ = sin ⁡ 2 9 ∘ + cos ⁡ 2 9 ∘ sin ⁡ 9 ∘ cos ⁡ 9 ∘ − sin ⁡ 2 2 7 ∘ + cos ⁡ 2 2 7 ∘ sin ⁡ 2 7 ∘ cos ⁡ 2 7 ∘ As sin ⁡ 2 θ + cos ⁡ 2 θ = 1 = 1 sin ⁡ 9 ∘ cos ⁡ 9 ∘ − 1 sin ⁡ 2 7 ∘ cos ⁡ 2 7 ∘ Also sin ⁡ 2 θ = 2 sin ⁡ θ cos ⁡ θ = 2 sin ⁡ 1 8 ∘ − 2 sin ⁡ 5 4 ∘ = 2 ( sin ⁡ 5 4 ∘ − sin ⁡ 1 8 ∘ ) sin ⁡ 1 8 ∘ sin ⁡ 5 4 ∘ By sin ⁡ 3 θ = 3 sin ⁡ θ − 4 sin ⁡ 3 θ = 2 ( 3 sin ⁡ 1 8 ∘ − 4 sin ⁡ 3 1 8 ∘ − sin ⁡ 1 8 ∘ ) sin ⁡ 1 8 ∘ sin ⁡ 5 4 ∘ = 4 ( 1 − 2 sin ⁡ 2 1 8 ∘ ) sin ⁡ 5 4 ∘ By cos ⁡ 2 θ = 1 − 2 sin ⁡ 2 θ = 4 cos ⁡ 3 6 ∘ cos ⁡ 3 6 ∘ Since sin ⁡ ( 9 0 ∘ − θ ) = cos ⁡ θ = 4 \begin{aligned} X & = \tan 9^\circ - \tan 27^\circ - {\color{#3D99F6}\tan 63^\circ} + \color{#3D99F6} \tan 81^\circ & \small \color{#3D99F6} \text{Since }\tan (90^\circ - \theta) = \cot \theta = \frac 1{\tan \theta} \\ & = \tan 9^\circ + {\color{#3D99F6} \frac 1{\tan 9^\circ}} - \tan 27^\circ - \color{#3D99F6}\frac 1{\tan 27^\circ} \\ & = \frac {\sin 9^\circ}{\cos 9^\circ} + \frac {\cos 9^\circ}{\sin 9^\circ} - \frac {\sin 27^\circ}{\cos 27^\circ} - \frac {\cos 27^\circ}{\sin 27^\circ} \\ & = \frac {\sin^2 9^\circ + \cos^2 9^\circ}{\sin 9^\circ \cos 9^\circ} - \frac {\sin^2 27^\circ + \cos^2 27^\circ}{\sin 27^\circ \cos 27^\circ} & \small \color{#3D99F6} \text{As } \sin^2 \theta + \cos^2 \theta = 1 \\ & = \frac 1{\sin 9^\circ \cos 9^\circ} - \frac 1{\sin 27^\circ \cos 27^\circ} & \small \color{#3D99F6} \text{Also } \sin 2 \theta = 2\sin \theta \cos \theta \\ & = \frac 2{\sin 18^\circ} - \frac 2{\sin 54^\circ} \\ & = \frac {2({\color{#3D99F6}\sin 54^\circ} - \sin 18^\circ)}{\sin 18^\circ \sin 54^\circ} & \small \color{#3D99F6} \text{By } \sin 3 \theta = 3\sin \theta - 4 \sin^3 \theta \\ & = \frac {2({\color{#3D99F6}3\sin 18^\circ - 4\sin^3 18^\circ} - \sin 18^\circ)}{\sin 18^\circ \sin 54^\circ} \\ & = \frac {4\color{#3D99F6}(1 - 2\sin^2 18^\circ)}{\color{#D61F06}\sin 54^\circ} & \small \color{#3D99F6} \text{By } \cos 2 \theta = 1-2\sin^2 \theta \\ & = \frac {4\color{#3D99F6}\cos 36^\circ}{\color{#D61F06}\cos 36^\circ} & \small \color{#D61F06} \text{Since }\sin (90^\circ - \theta) = \cos \theta \\ & = \boxed{4} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Nicely done👍

donglin loo - 2 years, 11 months ago

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