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Geometry Level 3

If A = cot 0. 5 tan 0. 5 2 tan 1 4 tan 2 8 cot 4 A= \cot 0.5^{\circ} - \tan 0.5^{\circ} - 2 \tan 1^{\circ} - 4 \tan 2^{\circ} - 8 \cot 4^{\circ} , evaluate A + 20 A+20 .


The answer is 20.00.

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2 solutions

Let x = 0.5 x=0.5 . Then, cot x tan x = cos x sin x sin x cos x = 2 ( cos 2 x sin 2 x ) 2 sin x cos x = 2 cos 2 x sin 2 x = 2 cot 2 x \cot x - \tan x = \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x} = \frac{2\left(\cos ^2x-\sin ^2x\right)}{2\sin x\cos x} = \frac{2\cos 2x}{\sin 2x} = 2\cot 2x

Now,

A = cot x tan x 2 tan 2 x 4 tan 4 x 8 cot 8 x A = \cot x - \tan x - 2 \tan 2x - 4 \tan 4x - 8 \cot 8x

= 2 cot 2 x 2 tan 2 x 4 tan 4 x 8 cot 8 x = 2\cot 2x - 2 \tan 2x - 4 \tan 4x - 8 \cot 8x

= 2 ( cot 2 x tan 2 x ) 4 tan 4 x 8 cot 8 x = 2(\cot 2x - \tan 2x) - 4 \tan 4x - 8 \cot 8x

= 2 ( 2 cot 4 x ) 4 tan 4 x 8 cot 8 x = 2(2 \cot 4x) - 4 \tan 4x - 8 \cot 8x

= 4 ( cot 4 x tan 4 x ) 8 cot 8 x = 4( \cot 4x - \tan 4x) - 8 \cot 8x

= 4 ( 2 cot 8 x ) 8 cot 8 x = 4(2 \cot 8x) - 8 \cot 8x

= 8 cot 8 x 8 cot 8 x = 0 = 8 \cot 8x - 8 \cot 8x = 0

Hence, A + 20 = 0 + 20 = 20 A+20=0+20=\boxed{20}

Let x = 0. 5 x = 0.5^\circ . Then we have:

A = cot x tan x 2 tan 2 x 4 tan 4 x 8 cot 8 x = 1 tan x tan x 2 tan 2 x 4 tan 4 x 8 tan 8 x = 1 tan 2 x tan x 2 tan 2 x 4 tan 4 x 8 cot 8 x = 2 cot 2 x 2 tan 2 x 4 tan 4 x 8 cot 8 x cot θ tan θ = 2 cot 2 θ = 4 cot 4 x 4 tan 4 x 8 cot 8 x = 8 cot 8 x 8 cot 8 x = 0 \begin{aligned} A & = \color{#3D99F6}{\cot x - \tan x} - 2 \tan 2x - 4 \tan 4x - 8 \cot 8x \\ & = \color{#3D99F6}{\frac 1{\tan x} - \tan x} - 2 \tan 2x - 4 \tan 4x - \frac 8{\tan 8x} \\ & = \color{#3D99F6}{\frac {1-\tan^2 x}{\tan x}} - 2 \tan 2x - 4 \tan 4x - 8 \cot 8x \\ & = \color{#3D99F6}{2\cot 2x} - 2 \tan 2x - 4 \tan 4x - 8 \cot 8x & \small \color{#3D99F6}{\implies \cot \theta - \tan \theta = 2 \cot 2 \theta} \\ & = 4 \cot 4x - 4 \tan 4x - 8 \cot 8x \\ & = 8 \cot 8x - 8 \cot 8x \\ & = 0 \end{aligned}

A + 20 = 0 + 20 = 20 \implies A + 20 = 0 + 20 = \boxed{20}

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