Need some tanning

Geometry Level 4

tan ( 6 3 ) = a b + c b \large \tan(63^\circ) = \sqrt{\sqrt a-\sqrt b} + \sqrt{\sqrt c-\sqrt b}

The equation above is true for positive integers a , b , a,b, and c . c. What is the value of a + b + c ? a+b+c?


The answer is 81.

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3 solutions

tan 6 3 = sin 6 3 cos 6 3 = cos ( 9 0 6 3 ) cos 6 3 = cos 2 7 cos 6 3 = cos ( 4 5 1 8 ) cos ( 4 5 + 1 8 ) = 1 2 cos 1 8 + 1 2 sin 1 8 1 2 cos 1 8 1 2 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 + 2 sin 1 8 cos 1 8 + sin 2 1 8 cos 2 1 8 sin 2 1 8 = 1 + sin 3 6 cos 3 6 \begin{aligned} \tan{63^\circ} & = \frac{\sin{63^\circ}}{\cos{63^\circ}} = \frac{\cos{(90^\circ - 63^\circ)}}{\cos{63^\circ}} = \frac{\cos{27^\circ}}{\cos{63^\circ}} = \frac{\cos{(45^\circ - 18^\circ)}}{\cos{(45^\circ + 18^\circ)}} \\ & = \frac{\frac{1}{\sqrt{2}}\cos{18^\circ}+ \frac{1}{\sqrt{2}}\sin{18^\circ}}{\frac{1}{\sqrt{2}}\cos{18^\circ}- \frac{1}{\sqrt{2}}\sin{18^\circ}} = \frac{\cos{18^\circ}+\sin{18^\circ}}{\cos{18^\circ}-\sin{18^\circ}} \\ & = \frac{(\cos{18^\circ}+\sin{18^\circ})(\cos{18^\circ} + \sin{18^\circ})} {(\cos{18^\circ}-\sin{18^\circ})(\cos{18^\circ}+\sin{18^\circ})} \\ & = \frac{\cos^2{18^\circ} + 2 \sin{18^\circ} \cos{18^\circ} + \sin^2{18^\circ}}{\cos^2{18^\circ}-\sin^2{18^\circ}} \\ & = \frac {1+\sin{36^\circ}}{\cos{36^\circ}} \end{aligned}

Now we know that:

sin 2 3 6 + cos 2 3 6 = 1 cos 2 5 4 + cos 2 3 6 = 1 Let θ = 1 8 cos 2 3 θ + cos 2 2 θ = 1 Let cos θ = cos 1 8 = x ( 4 x 3 3 x ) 2 + ( 2 x 2 1 ) 2 = 1 16 x 6 24 x 4 + 9 x 2 + 4 x 4 4 x 2 + 1 = 1 16 x 6 20 x 4 + 5 x 2 = 0 Since x 0 16 x 4 20 x 2 + 5 = 0 x 2 = 20 ± 400 320 32 cos 2 1 8 = 5 + 5 8 5 5 8 θ > 1 8 , rejected \begin{aligned} \sin^2{36^\circ} + \cos^2{36^\circ} & = 1 \\ \Rightarrow \cos^2{54^\circ} + \cos^2{36^\circ} & = 1 \quad \quad \color{#3D99F6} {\text{Let }\theta = 18^\circ} \\ \cos^2{3\theta} + \cos^2{2\theta} & = 1 \quad \quad \color{#3D99F6} {\text{Let } \cos{\theta} = \cos{18^\circ} = x } \\ \Rightarrow (4x^3-3x)^2 + (2x^2-1)^2 & = 1 \\ 16x^6 - 24x^4 + 9x^2 + 4x^4 - 4x^2 + 1 & = 1 \\ 16x^6 - 20x^4 + 5x^2 & = 0 \quad \quad \color{#3D99F6} {\text{Since } x \ne 0} \\ 16x^4 - 20x^2 + 5 & = 0 \\ \Rightarrow x^2 & = \frac{20 \pm \sqrt{400-320}}{32} \\ \Rightarrow \cos^2{18^\circ} & = \frac{5 + \sqrt{5}}{8} \\ & \quad \color{#3D99F6}{ \frac{5 - \sqrt{5}}{8} \Rightarrow \theta > 18^\circ, \text{rejected}}\end{aligned}

Now, we have:

cos 3 6 = 2 cos 2 1 8 1 = 5 + 5 4 1 = 1 + 5 4 sin 3 6 = 1 cos 2 3 6 = 16 ( 6 + 2 5 ) 4 = 10 20 4 tan 6 3 = 1 + sin 3 6 cos 3 6 = 4 + 10 20 1 + 5 = ( 4 + 10 20 ) ( 5 1 ) ( 5 + 1 ) ( 5 1 ) = 4 ( 5 1 ) + ( 10 20 ) ( 6 20 ) 4 = 4 ( 5 1 ) + 60 + 20 16 20 4 = 5 1 + 5 20 = ( 5 1 ) 2 + 25 20 = 6 2 5 + 25 20 = 36 20 + 25 20 \begin{aligned} \cos{36^\circ} & = 2 \cos^2{18^\circ} - 1 =\frac{5 + \sqrt{5}}{4} - 1 =\frac{1 + \sqrt{5}}{4} \\ \sin{36^\circ} & = \sqrt{1-\cos^2{36^\circ}} = \frac{\sqrt{16-(6+2\sqrt{5})}}{4} = \frac{\sqrt{10-\sqrt{20}}}{4} \\ \Rightarrow \tan{63^\circ} & = \frac {1+\sin{36^\circ}}{\cos{36^\circ}} = \frac{4+\sqrt{10-\sqrt{20}}}{1+\sqrt{5}} \\ & = \frac{(4+\sqrt{10-\sqrt{20}})(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)} \\ &= \frac{4(\sqrt{5}- 1) + \sqrt{(10-\sqrt{20})(6-\sqrt{20})}}{4} \\ &= \frac{4(\sqrt{5}- 1) + \sqrt{60+20-16\sqrt{20}}}{4} \\ & = \sqrt{5}- 1 + \sqrt{5-\sqrt{20}} \\ & = \sqrt{(\sqrt{5}- 1)^2} + \sqrt{ \sqrt{25}-\sqrt{20}} \\ & = \sqrt{6-2\sqrt{5}} + \sqrt{\sqrt{25}-\sqrt{20}} \\ & = \sqrt{\sqrt{36}-\sqrt{20}} + \sqrt{\sqrt{25}-\sqrt{20}} \end{aligned}

a + b + c = 36 + 20 + 25 = 81 \Rightarrow a + b + c = 36+20+25 = \boxed{81}

Moderator note:

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 \sin 18^\circ, \cos 18^ \circ is to look at the regular pentagon, since cos 1 8 = sin 10 8 \cos 18 ^\circ = \sin 108 ^ \circ , which is the interior angle.

Awesome solution sir

Sumukh Bansal - 3 years, 6 months ago

tan ( 6 3 ) = cot ( 5 4 2 ) \tan(63^\circ)=\cot\left(\dfrac{54^\circ}{2}\right)

Now, let's use: cot ( θ 2 ) = cot ( θ ) + csc ( θ ) \cot\left(\dfrac{\theta}{2}\right)=\cot(\theta)+\csc(\theta) .

So, the value we want is cot ( 5 4 ) + csc ( 5 4 ) \cot(54^\circ)+\csc(54^\circ)

We know that sin ( 5 4 ) = 1 + 5 4 \sin(54^\circ)=\dfrac{1+\sqrt{5}}{4} , so csc ( 5 4 ) = 4 1 + 5 = 5 1 \csc(54^\circ)=\dfrac{4}{1+\sqrt{5}}=\sqrt{5}-1 .

Also, cot 2 ( 5 4 ) + 1 = csc 2 ( 5 4 ) \cot^2(54^\circ)+1=\csc^2(54^\circ) , so cot ( 5 4 ) = ( 5 1 ) 2 1 = 5 2 5 \cot(54^\circ)=\sqrt{(\sqrt{5}-1)^2-1}=\sqrt{5-2\sqrt{5}} .

So, tan ( 6 3 ) = 5 2 5 + 5 1 \tan(63^\circ)=\sqrt{5-2\sqrt{5}}+\sqrt{5}-1 .

Finally, manipulate it so it match with the required form:

tan ( 6 3 ) = 5 2 2 2 5 + ( 5 1 ) 2 \tan(63^\circ)=\sqrt{\sqrt{5^2}-\sqrt{2^2 \cdot 5}}+\sqrt{(\sqrt{5}-1)^2}

tan ( 6 3 ) = 25 20 + 36 20 \tan(63^\circ)=\sqrt{\sqrt{25}-\sqrt{20}}+\sqrt{\sqrt{36}-\sqrt{20}}

Hence, a = 25 a=25 , b = 20 b=20 , c = 36 c=36 and a + b + c = 81 a+b+c=\boxed{81} .

Note: to obtatin sin ( 5 4 ) \sin(54^\circ) 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 \cos(2 \times 72^\circ)=\cos(3\times 72^\circ) \\ 2\cos^2(72^\circ)-1=4\cos^3(72^\circ)-3\cos(72^\circ) \\ 4\cos^3(72^\circ)-2\cos^2(72^\circ)-3\cos(72^\circ)+1=0 \\ (\cos(72^\circ)-1)(4\cos^2(72^\circ)+2\cos(72^\circ)-1)=0

Obviously cos ( 7 2 ) 1 \cos(72^\circ) \neq 1 , and 0 < cos ( 7 2 ) < 1 0 < \cos(72^\circ) < 1 , so let's find the positive root of the quadratic factor:

cos ( 7 2 ) = 1 + 5 4 \cos(72^\circ)=\dfrac{-1+\sqrt{5}}{4}

So, cos ( 14 4 ) = 2 cos 2 ( 7 2 ) 1 = 1 + 5 4 \cos(144^\circ)=2\cos^2(72^\circ)-1=-\dfrac{1+\sqrt{5}}{4}

And sin ( 5 4 ) = cos ( 3 6 ) = cos ( 14 4 ) = 1 + 5 4 \sin(54^\circ)=\cos(36^\circ)=-\cos(144^\circ)=\dfrac{1+\sqrt{5}}{4}

Mas Mus
Oct 10, 2015

tan 6 3 = sin ( 4 5 + 1 8 ) cos ( 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 ) = 1 + 2 sin 1 8 cos 1 8 cos 3 6 \tan63^{\circ}=\dfrac{\sin(45^{\circ}+18^{\circ})}{\cos(45^{\circ}+18^{\circ})}=\dfrac{\left(\cos18^{\circ}+\sin18^{\circ}\right)\left(\cos18^{\circ}+\sin18^{\circ}\right)}{\left(\cos18^{\circ}-\sin18^{\circ}\right)\left(\cos18^{\circ}+\sin18^{\circ}\right)}=\dfrac{1+2\sin18^{\circ}\cos18^{\circ}}{\cos36^{\circ}}

Note that sin 1 8 = 5 1 4 \sin18^{\circ}=\dfrac{\sqrt{5}-1}{4} (proof is left for the reader), then cos 1 8 = 10 + 2 5 4 \cos18^{\circ}=\dfrac{\sqrt{10+2\sqrt{5}}}{4} and cos 3 6 = 5 + 1 4 \cos36^{\circ}=\dfrac{\sqrt{5}+1}{4} .

Now, we have

tan 6 3 = 1 + 2 ( 5 1 4 ) ( 10 + 2 5 4 ) 5 + 1 4 = ( 5 1 ) + ( 3 5 ) ( 10 + 2 5 ) 4 = ( ( 5 + 1 ) 2 5 × 1 ) + ( ( 9 + 5 ) 2 9 × 5 ) ( 10 + 2 5 ) 4 = 36 20 + 80 32 5 4 = 36 20 + 25 20 \begin{array}{c}&\tan63^{\circ}&=\dfrac{1+2\left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{10+2\sqrt{5}}}{4}\right)}{\frac{\sqrt{5}+1}{4}}\\&=\left(\color{#D61F06}{\sqrt{5}-1}\right)+\dfrac{\left(\color{#D61F06}{3-\sqrt{5}}\right)\left(\sqrt{10+2\sqrt{5}}\right)}{4}\\&=\left(\color{#D61F06}{\sqrt{(5+1)-2\sqrt{5\times{1}}}}\right)+\dfrac{\left(\color{#D61F06}{\sqrt{(9+5)-2\sqrt{9\times{5}}}}\right)\left(\sqrt{10+2\sqrt{5}}\right)}{4}\\&=\sqrt{\sqrt{36}-\sqrt{20}}+\dfrac{\sqrt{80-32\sqrt{5}}}{4}\\&=\sqrt{\sqrt{36}-\sqrt{20}}+\sqrt{\sqrt{25}-\sqrt{20}}\end{array} .

Hence, a = 36 , b = 25 , c = 20 a=36,~ b=25,~ c=20 , and the required answer is a + b + c = 36 + 25 + 20 = 81 a+b+c=36+25+20=81

Note : ( x + y ) 2 x × y = x y , x > y \color{#D61F06}{\boxed{\textbf{Note}:~\sqrt{(x+y)-2\sqrt{x\times{y}}}=\sqrt{x}-\sqrt{y},x>y}}

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