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My trigonometric problem is:

If $A = \tan6^{\circ} \tan42^{\circ}$ B = cot$66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$.

Working :

$$B = \cot 66^{\circ} \cot78^{\circ} = 1- \frac{\tan24^{\circ}+\tan18^{\circ}}{\tan42^{\circ}}$$

$$A= \tan6^{\circ} \tan42^{\circ} = 1- \frac{\tan6^{\circ} +\tan42^{\circ}}{\tan48^{\circ}}$$

but it seems this is the wrong way of doing this...please suggest. Thanks!

Sachin
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3 Answers3

4

Using

$2\cos A\cos B=\cos(A-B)+\cos(A+B)$ and $2\sin A\sin B=\cos(A-B)-\cos(A+B),$

$$A=\frac{\sin 6^\circ\cdot \sin 42^\circ}{\cos 6^\circ\cdot \cos 42^\circ}=\frac{\cos36^\circ-\cos48^\circ}{\cos36^\circ+\cos48^\circ}$$

Applying Componendo and dividendo, $$\frac{1+A}{1-A}=\frac{\cos36^\circ}{\cos48^\circ}$$

Similarly, Using

$2\sin A\cos B=\sin(A+B)+\sin(A-B)$ and $2\cos A\sin B=\sin(A+B)-\sin(A-B),$

$$B=\frac{\cos66^\circ \cos72^\circ}{\sin66^\circ \sin72^\circ}=\frac{\cos66^\circ \sin18^\circ}{\sin66^\circ \cos18^\circ}=\frac{\sin84^\circ-\sin48^\circ}{\sin84^\circ+\sin48^\circ}$$

Applying Componendo and dividendo, $$\frac{1+B}{1-B}=\frac{\sin84^\circ}{\sin48^\circ}$$

$$\implies \frac{1+A}{1-A}\cdot\frac{1+B}{1-B}=\frac{\sin84^\circ\cdot \cos36^\circ}{\sin48^\circ\cdot \cos48^\circ}=\frac{2\sin84^\circ\cdot \cos36^\circ}{\sin(2\cdot48)^\circ}=2\cos36^\circ$$ as $\sin96^\circ=\sin(180-96)^\circ=\sin84^\circ$

Now $\cos36^\circ$ can be found here

lab bhattacharjee
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  • Hi,Still unable to find relation between A and B ..please suggest further.. – Sachin Jun 29 '13 at 14:58
  • @sultan, so we have $$\frac{1+AB+A+B}{1+AB-(A+B)}=\frac{\sqrt5+1}2$$

    Applying Componendo and dividendo, $$\frac{1+AB}{A+B}=\frac{\sqrt5+3}{\sqrt5-1}=\frac{(\sqrt5+3)(\sqrt5+1)}{(\sqrt5-1)(\sqrt5+1)}=\sqrt5+2$$ Now can you express $A$ in terms of $B$ or the opposite if need be?

     – lab bhattacharjee Jun 29 '13 at 15:03
  • @sultan,can you please re-confirm the last angle is it $72$ or $78?$ – lab bhattacharjee Jun 29 '13 at 15:07
  • @sultan, If the last angle is $78^\circ$ instead of $72^\circ$

    As $\tan x\tan(60^\circ-x)\tan(60^\circ-x)=\tan3x$

    $x=6\implies \tan 6^\circ\tan(54^\circ)\tan(66^\circ)=\tan 18^\circ\ \ \ \ \ (1)$

    and $x=18\implies \tan 18^\circ\tan(42^\circ)\tan(78^\circ)=\tan54^\circ\ \ \ \ (2)$

    $$(1)\cdot(2)\implies \tan 6^\circ \tan(42^\circ)\tan(66^\circ)\tan(78^\circ)=1$$

    $$\implies \tan 6^\circ \tan(42^\circ)=\cot(66^\circ)\cot(78^\circ)\implies A=B$$

     – lab bhattacharjee Jun 29 '13 at 15:20
  • is it 60+x instead of 60-x – Sachin Jun 29 '13 at 16:44
  • @sultan, sorry its $\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan 3x$. Btw, have you re-confirmed & could you follow my method in the comment – lab bhattacharjee Jun 29 '13 at 16:46
2

First, note that $A \approx 0.0946362785$, $\ \ B \approx 0.0946362785$.

Now, we will prove that $\ \ \ \Large{A=B.}$

a).

$$ \dfrac{A}{B} = \dfrac {\sin 6^\circ \sin 42^\circ} {\cos 6^\circ \cos 42^\circ} \cdot \dfrac {\sin 66^\circ \sin 78^\circ} {\cos 66^\circ \cos 78^\circ} = \dfrac {\bigl( 2 \sin 6^\circ \sin 66^\circ \bigr) \cdot \bigl( 2 \sin 42^\circ \sin 78^\circ \bigr)} {\bigl( 2 \cos 6^\circ \cos 42^\circ \bigr) \cdot \bigl( 2 \cos 66^\circ \cos 78^\circ \bigr)}. \tag{1} $$

b). Applying formulas

$\ \ \ \ 2\sin\alpha\sin\beta = \cos(\alpha-\beta) - \cos(\alpha+\beta)$, $\ \ \ $

$\ \ \ \ 2\cos\alpha\sin\beta = \cos(\alpha-\beta) + \cos(\alpha+\beta)$,

$ \ \ \ \ \ (1) \implies$ $$ \dfrac{A}{B} = \dfrac {\bigl( \cos 60^\circ - \cos72^\circ \bigr) \cdot \bigl( \cos 36^\circ - \cos 120^\circ \bigr)} {\bigl( \cos 60^\circ + \cos72^\circ \bigr) \cdot \bigl( \cos 36^\circ + \cos 120^\circ \bigr)} = \dfrac {\bigl( \frac{1}{2} - \cos72^\circ \bigr) \cdot \bigl( \cos 36^\circ + \frac{1}{2} \bigr)} {\bigl( \frac{1}{2} + \cos72^\circ \bigr) \cdot \bigl( \cos 36^\circ - \frac{1}{2} \bigr)}. \tag{2} $$

c). It is known, that $\ \ \ \ \cos72^\circ = \sin 18^\circ = \frac{1}{4}(\sqrt{5}-1)$,

so $\ \ \ \ \ \ \cos 36^\circ = 1 - 2(\sin 18^\circ)^2 = \frac{8}{8} - \frac{1}{8}(5-2\sqrt{5}+1) = \frac{1}{4}(\sqrt{5}+1)$,

and $\ \ \ \ \cos 72^\circ \cos 36^\circ = \frac{1}{16}(\sqrt{5}-1)(\sqrt{5}+1) = \frac{4}{16}=\frac{1}{4}$.

Hence (2) $\implies$

$$ \dfrac{A}{B} = \dfrac {\frac{1}{2}\cos 36^\circ - \cos 72^\circ \cos 36^\circ +\frac{1}{4}-\frac{1}{2}\cos 72^\circ} {\frac{1}{2}\cos 36^\circ +\cos 72^\circ \cos 36^\circ -\frac{1}{4}-\frac{1}{2}\cos 72^\circ } = \dfrac {\frac{1}{2}(\cos 36^\circ - \cos 72^\circ)} {\frac{1}{2}(\cos 36^\circ - \cos 72^\circ) } =\Large{1}. \tag{3} $$

Proved.

Oleg567
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$$A=Tan(6^\circ)\,Tan(42^\circ)=Tan(2.3^\circ)Tan(45^\circ-3^\circ)=\frac{2Tan(3^\circ)}{1-Tan^2(3^\circ)}\frac{1-Tan(3^\circ)}{1+Tan(3^\circ)}=\frac{2Tan(3^\circ)}{\left(1+Tan(3^\circ)\right)^2}$$ $\:$

$$ B=Tan(24^\circ)\,Tan(12^\circ)=Tan(2.12^\circ)Tan(12^\circ)=\frac{2Tan^2(12^\circ)}{1-Tan^2(12^\circ)} \implies$$

$\:$ $$Tan(12^\circ)=\sqrt{\frac{B}{B+2}}$$ But $$Tan(12^\circ)=Tan(15^\circ-3^\circ)=\frac{(2-\sqrt{3})-Tan(3^\circ)}{(2-\sqrt{3})+Tan(3^\circ)}\implies$$

$$Tan(3^\circ)=\left(\frac{\sqrt{B+2}-\sqrt{B}}{\sqrt{B+2}+\sqrt{B}}\right)(2-\sqrt{3})$$ Substitute $\,$$Tan(3^\circ)$ above in the Expression of $A$ to get the Required Relation

Ekaveera Gouribhatla
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