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The quantity \(\tan 7.5^\circ\) can be expressed in the form\(\tan 7.5^\circ = \sqrt{a} - \sqrt{b} + \sqrt{c} - d,\)
where \(a \ge b \ge c \ge d\) are positive integers. Find \(a + b + c + d.\)

 Apr 15, 2020
 #1
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a=3; b=2;c=1;d=0;p=1;m=0;n=frac(2#(a)-2#(b)+2#(c)-d);m=if(round(n*1E10)==round(tan(7.5)*1E10), goto loop, goto next);loop: printn,a,b,c,d;next:a++;if(a<35, goto5, 0);a=3;b++;if(b<35, goto5, 0);a=3;b=2;c++;if(c<35, goto5,0);a=3;b=3;c=1;d++;if(d<35, goto5, discard=0;

 

OUTPUT:  a = 6,   b = 3,   c = 2,   and   d = 2

Sqrt(6) - sqrt(3) + sqrt(2) - 2  =  tan(7.5 degrees).

 Apr 15, 2020
 #2
avatar+24995 
+2

The quantity \(\tan 7.5^\circ\) can be expressed in the form \(\tan 7.5^\circ = \sqrt{a} - \sqrt{b} + \sqrt{c} - d\),
where \(a \ge b \ge c \ge d\) are positive integers.
Find \(a + b + c + d\).


Formula: \(\boxed {\cos(2\varphi) = \cos^2(\varphi)-\sin^2(\varphi) \\ \Rightarrow \quad \sin^2(\varphi)=\dfrac{1-\cos(2\varphi)}{2},\quad \cos^2(\varphi)=\dfrac{1+\cos(2\varphi)}{2} }\)

 

\(\text{Let $\cos(30^\circ)=\dfrac{\sqrt{3}}{2} $}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\sin^2(15^\circ)} &=& \mathbf{\dfrac{1-\cos(30^\circ)}{2}} \\ \sin^2(15^\circ) &=& \dfrac{1-\dfrac{\sqrt{3}}{2}}{2} \\ \mathbf{ \sin^2(15^\circ) } &=& \mathbf{\dfrac{1}{2}-\dfrac{\sqrt{3}}{4}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\cos^2(15^\circ)} &=& \mathbf{\dfrac{1+\cos(30^\circ)}{2}} \\ \cos^2(15^\circ) &=& \dfrac{1+\dfrac{\sqrt{3}}{2}}{2} \\ \mathbf{ \cos^2(15^\circ) } &=& \mathbf{\dfrac{1}{2}+\dfrac{\sqrt{3}}{4}} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \mathbf{\tan^2(15^\circ)} &=& \mathbf{\dfrac{\sin^2(15^\circ)} {\cos^2(15^\circ)}} \\\\ \tan^2(15^\circ) &=& \dfrac{\left( \dfrac{1}{2}-\dfrac{\sqrt{3}}{4}\right) } { \left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{4} \right)} \\\\ \tan^2(15^\circ) &=& \dfrac{\left( \dfrac{1}{2}-\dfrac{\sqrt{3}}{4}\right) } { \left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{4} \right)}\times \dfrac{\left( \dfrac{1}{2}-\dfrac{\sqrt{3}}{4}\right) } { \left( \dfrac{1}{2}-\dfrac{\sqrt{3}}{4} \right)} \\\\ \tan^2(15^\circ) &=& \dfrac{\left( \dfrac{1}{4}-\dfrac{\sqrt{3}}{4}+\dfrac{3}{16} \right) } { \left( \dfrac{1}{4}-\dfrac{3}{16} \right)} \\\\ \tan^2(15^\circ) &=& 7 - 4\sqrt{3} \\ \hline \tan (15^\circ) &=& \sqrt{ 7 - 4\sqrt{3}} \\ \tan (15^\circ) &=& \sqrt{ 7 - \sqrt{16*3}} \\ \mathbf{\tan (15^\circ)} &=& \mathbf{\sqrt{ 7 - \sqrt{48}} } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{\tan (15^\circ)} &=& \mathbf{\sqrt{ 7 - \sqrt{48}} } \\\\ \sqrt{ 7 - \sqrt{48}} &=& \sqrt{u}-\sqrt{v} \\ \ 7 - \sqrt{48} &=& \left(\sqrt{u}-\sqrt{v}\right)^2 \\ \ 7 - \sqrt{48} &=& u+2\sqrt{uv}+v \\ \ 7 - \sqrt{48} &=& u+v+\sqrt{4uv} \quad \text{compare}\\ && \begin{array}{|rcll|} \hline u+v &=&7 \quad \text{ or }\quad v = 7-u \\ 4uv&=&48 \\ 4u(7-u) &=& 48 \\ u(7-u) &=& 12 \\ u^2 -7u+12 &=& 0 \\\\ u &=& \dfrac{7\pm \sqrt{49-4*12} }{2} \\ u &=& \dfrac{7\pm 1 }{2} \\ \mathbf{u} &=& \mathbf{4} \\\\ v &=& 7-u \\ v &=& 7-4 \\ \mathbf{v} &=& \mathbf{3} \\ \hline \end{array} \\ \mathbf{\tan (15^\circ)} &=& \mathbf{\sqrt{u}-\sqrt{v} } \\ \tan (15^\circ) &=& \sqrt{4}-\sqrt{3} \\ \mathbf{\tan (15^\circ)} &=& \mathbf{2-\sqrt{3}} \\ \hline \dfrac{1}{\tan (15^\circ)} &=& \left(\dfrac{1}{2-\sqrt{3}} \right) \times \left(\dfrac{2+\sqrt{3}}{2+\sqrt{3}} \right) \\\\ \dfrac{1}{\tan (15^\circ)} &=& \dfrac{2+\sqrt{3}}{4-3} \\\\ \mathbf{\dfrac{1}{\tan (15^\circ)}} &=& \mathbf{2+\sqrt{3}} \\ \hline \end{array}\)

 

Formula: \(\boxed {\tan(2\varphi) = \dfrac{2\tan(\varphi)}{1-\tan^2(\varphi)} }\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\tan(2\varphi)} &=& \mathbf{\dfrac{2\tan(\varphi)}{1-\tan^2(\varphi)}} \\\\ \tan(15^\circ) &=& \dfrac{2\tan(7.5^\circ)}{1-\tan^2(7.5^\circ)} \\\\ \tan(15^\circ)\left(1-\tan^2(7.5^\circ)\right) &=& 2\tan(7.5^\circ) \\ \tan(15^\circ) -\tan(15^\circ)\tan^2(7.5^\circ) &=& 2\tan(7.5^\circ) \\ \tan(15^\circ)\tan^2(7.5^\circ)+ 2\tan(7.5^\circ) -\tan(15^\circ) &=& 0 \quad | \quad : \tan(15^\circ) \\ \tan^2(7.5^\circ)+ 2\times\dfrac{1}{\tan(15^\circ)} \tan(7.5^\circ) - 1 &=& 0 \quad | \quad \mathbf{\dfrac{1}{\tan (15^\circ)}=2+\sqrt{3}} \\ \mathbf{\tan^2(7.5^\circ)+ 2(2+\sqrt{3}) \tan(7.5^\circ) - 1} &=& \mathbf{0} \\\\ \tan(7.5^\circ) &=& \dfrac{-2(2+\sqrt{3}) \pm \sqrt{4(2+\sqrt{3})^2-4(-1)} }{2} \\\\ \tan(7.5^\circ) &=& \dfrac{-2(2+\sqrt{3}) \pm \sqrt{4(2+\sqrt{3})^2+4} }{2} \\\\ \tan(7.5^\circ) &=& \dfrac{-2(2+\sqrt{3}) \pm \sqrt{4\left( (2+\sqrt{3})^2+1\right)} }{2} \\\\ \tan(7.5^\circ) &=& \dfrac{-2(2+\sqrt{3}) \pm 2\sqrt{(2+\sqrt{3})^2+1} }{2} \\\\ \tan(7.5^\circ) &=& -2-\sqrt{3} \pm \sqrt{(2+\sqrt{3})^2+1} \\\\ \tan(7.5^\circ) &=& -2-\sqrt{3} \pm \sqrt{8+4\sqrt{3}} \\ \tan(7.5^\circ) &=& -2-\sqrt{3} \pm \sqrt{8+\sqrt{16*3}} \\ \mathbf{\tan(7.5^\circ)} &=& \mathbf{-2-\sqrt{3} + \sqrt{8+\sqrt{48}}} \quad | \quad \tan(7.5^\circ) > 0\ ! \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{ \sqrt{8+\sqrt{48} } } &=& \mathbf{\sqrt{u}+\sqrt{v}} \\ \ 8+\sqrt{48} &=& \left(\sqrt{u}+\sqrt{v}\right)^2 \\ \ 8+\sqrt{48} &=& u+2\sqrt{uv}+v \\ \ 8+\sqrt{48} &=& u+v+\sqrt{4uv} \quad \text{compare}\\ && \begin{array}{|rcll|} \hline u+v &=&8 \quad \text{ or }\quad v = 8-u \\ 4uv&=&48 \\ 4u(8-u) &=& 48 \\ u(8-u) &=& 12 \\ u^2 -8u+12 &=& 0 \\\\ u &=& \dfrac{8\pm \sqrt{64-4*12} }{2} \\ u &=& \dfrac{8\pm 4 }{2} \\ \mathbf{u} &=& \mathbf{6} \\\\ v &=& 8-u \\ v &=& 8-6 \\ \mathbf{v} &=& \mathbf{2} \\ \hline \end{array} \\ \mathbf{ \sqrt{8+\sqrt{48}} }&=& \mathbf{ \sqrt{u}+\sqrt{v} } \\ \mathbf{ \sqrt{8+\sqrt{48}} } &=& \mathbf{ \sqrt{6}+\sqrt{2} } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\tan(7.5^\circ)} &=& \mathbf{-2-\sqrt{3} + \sqrt{8+\sqrt{48}}} \quad & | \quad \mathbf{ \sqrt{8+\sqrt{48}}=\sqrt{6}+\sqrt{2} } \\ \tan(7.5^\circ) &=& -2-\sqrt{3} + \sqrt{6}+\sqrt{2} \\ \mathbf{ \tan(7.5^\circ) } &=& \mathbf{\sqrt{6}-\sqrt{3}+\sqrt{2}-2} \\ \hline a+b+c+d &=& 6+3+2+2 \\ \mathbf{a+b+c+d} &=& \mathbf{13} \\ \hline \end{array}\)

 

laugh

 Apr 16, 2020

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