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# (tan15∘)x(tan25∘)x(tan35∘)x(tan85∘) I know the answer but I do not know the solution answer: 1 help me pls

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(tan15∘)x(tan25∘)x(tan35∘)x(tan85∘)
I know the answer but I do not know the solution
help me pls

Guest Mar 28, 2017
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#1
+18777
+1

(tan15∘)x(tan25∘)x(tan35∘)x(tan85∘)

Formula:

$$\begin{array}{|lcll|} \hline \sin (x) \; \sin (y) = \frac{1}{2}\Big(\cos (x-y) - \cos (x+y)\Big) \\ \cos (x) \; \cos (y) = \frac{1}{2}\Big(\cos (x-y) + \cos (x+y)\Big) \\ \sin (2x) = 2\cdot \sin(x)\cdot \cos(x) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && \tan(15^{\circ})\cdot \tan(25^{\circ})\cdot \tan(35^{\circ})\cdot \tan(85^{\circ}) \\ &=& \frac{\sin(15^{\circ})\cdot \sin(25^{\circ})\cdot \sin(35^{\circ})\cdot \sin(85^{\circ})} {\cos(15^{\circ})\cdot \cos(25^{\circ})\cdot \cos(35^{\circ})\cdot \cos(85^{\circ})} \\ &=& \frac{ [\sin(85^{\circ})\cdot\sin(15^{\circ})]\cdot [\sin(35^{\circ})\cdot \sin(25^{\circ})] } { [\cos(85^{\circ})\cdot\cos(15^{\circ})]\cdot [\cos(35^{\circ})\cdot \cos(25^{\circ})] } \\\\ && \sin(85^{\circ})\cdot\sin(15^{\circ}) = \frac{1}{2}\Big(\cos (85^{\circ}-15^{\circ}) - \cos (85^{\circ}+15^{\circ})\Big) \\ && \mathbf{ \sin(85^{\circ})\cdot\sin(15^{\circ}) = \frac{1}{2}\Big(\cos (70^{\circ}) - \cos (100^{\circ})\Big) } \\\\ && \sin(35^{\circ})\cdot\sin(25^{\circ}) = \frac{1}{2}\Big(\cos (35^{\circ}-25^{\circ}) - \cos (35^{\circ}+25^{\circ})\Big) \\ && \mathbf{ \sin(35^{\circ})\cdot\sin(25^{\circ}) = \frac{1}{2}\Big(\cos (10^{\circ}) - \cos (60^{\circ})\Big) } \\\\ && \cos(85^{\circ})\cdot\cos(15^{\circ}) = \frac{1}{2}\Big(\cos (85^{\circ}-15^{\circ}) + \cos (85^{\circ}+15^{\circ})\Big) \\ && \mathbf{ \cos(85^{\circ})\cdot\cos(15^{\circ}) = \frac{1}{2}\Big(\cos (70^{\circ}) + \cos (100^{\circ})\Big) } \\\\ && \cos(35^{\circ})\cdot\cos(25^{\circ}) = \frac{1}{2}\Big(\cos (35^{\circ}-25^{\circ}) + \cos (35^{\circ}+25^{\circ})\Big) \\ && \mathbf{ \cos(35^{\circ})\cdot\cos(25^{\circ}) = \frac{1}{2}\Big(\cos (10^{\circ}) + \cos (60^{\circ})\Big) } \\\\ &=& \frac{ \frac{1}{2}\Big(\cos (70^{\circ}) - \cos (100^{\circ})\Big) \cdot \frac{1}{2}\Big(\cos (10^{\circ}) - \cos (60^{\circ})\Big) } { \frac{1}{2}\Big(\cos (70^{\circ}) + \cos (100^{\circ})\Big) \cdot \frac{1}{2}\Big(\cos (10^{\circ}) + \cos (60^{\circ})\Big) } \\\\ &=& \frac{ \Big(\cos (70^{\circ}) - \cos (100^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) - \cos (60^{\circ})\Big) } { \Big(\cos (70^{\circ}) + \cos (100^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) + \cos (60^{\circ})\Big) } \quad | \quad \cos(100^{\circ}) = \cos(90^{\circ}+10^{\circ})=-\sin(10^{\circ}) \\\\ &=& \frac{ \Big(\cos (70^{\circ}) + \sin(10^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) - \cos (60^{\circ})\Big) } { \Big(\cos (70^{\circ}) - \sin(10^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) + \cos (60^{\circ})\Big) } \quad | \quad \cos(70^{\circ}) = \cos(90^{\circ}-20^{\circ})=\sin(20^{\circ}) \\\\ &=& \frac{ \Big(\sin(20^{\circ}) + \sin(10^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) - \cos (60^{\circ})\Big) } { \Big(\sin(20^{\circ}) - \sin(10^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) + \cos (60^{\circ})\Big) } \quad | \quad \cos(60^{\circ}) = \frac{1}{2} \\\\ &=& \frac{ \Big(\sin(20^{\circ}) + \sin(10^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) - \frac{1}{2}\Big) } { \Big(\sin(20^{\circ}) - \sin(10^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) + \frac{1}{2}\Big) } \\\\ &=& \frac{ \sin(20^{\circ})\cdot \cos (10^{\circ}) - \frac{1}{2}\cdot \sin(20^{\circ})+\sin(10^{\circ})\cdot \cos (10^{\circ}) - \frac{1}{2} \cdot \sin(10^{\circ}) } { \sin(20^{\circ})\cdot \cos (10^{\circ}) + \frac{1}{2}\cdot \sin(20^{\circ})-\sin(10^{\circ})\cdot \cos (10^{\circ}) - \frac{1}{2} \cdot \sin(10^{\circ}) } \quad | \quad \sin(10^{\circ})\cdot \cos (10^{\circ}) = \frac{1}{2}\cdot \sin(20^{\circ}) \\\\ &=& \frac{ \sin(20^{\circ})\cdot \cos (10^{\circ}) - \frac{1}{2}\cdot \sin(20^{\circ})+\frac{1}{2}\cdot \sin(20^{\circ}) - \frac{1}{2} \cdot \sin(10^{\circ}) } { \sin(20^{\circ})\cdot \cos (10^{\circ}) + \frac{1}{2}\cdot \sin(20^{\circ})-\frac{1}{2}\cdot \sin(20^{\circ}) - \frac{1}{2} \cdot \sin(10^{\circ}) } \quad | \quad \frac{1}{2}\cdot \sin(20^{\circ})-\frac{1}{2}\cdot \sin(20^{\circ}) = 0 \\\\ &=& \frac{ \sin(20^{\circ})\cdot \cos (10^{\circ}) + 0 - \frac{1}{2} \cdot \sin(10^{\circ}) } { \sin(20^{\circ})\cdot \cos (10^{\circ}) + 0 - \frac{1}{2} \cdot \sin(10^{\circ}) } \\\\ &=& \frac{ \sin(20^{\circ})\cdot \cos (10^{\circ}) - \frac{1}{2} \cdot \sin(10^{\circ}) } { \sin(20^{\circ})\cdot \cos (10^{\circ}) - \frac{1}{2} \cdot \sin(10^{\circ}) } \\\\ &=& 1 \\ \hline \end{array}$$

heureka  Mar 29, 2017
#2
+91256
+1

Thanks Heureka :)

I have only just started working through this.

I am sure it is very clear and I will have no problem following it.

However, I would not have been able to put that string of logic together myself.

Maybe as I work through it some light may come on for me,

I hope so  :))

Melody  Mar 29, 2017

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