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I know that I should be using the property $$\tan A\tan B+\tan B\tan C+\tan C\tan A=1+\sec A\sec B\sec C.$$

Now I don't know how to calculate $\sec A\sec B\sec C$ without using a calculator as the angles aren't the regular ones. So my question is how will I be able to find the required value without using the calculator?

Juniven Acapulco
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nikola
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3 Answers3

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HINT

Use the formula $cos(\frac {\pi} 7)cos(\frac {2\pi} 7)cos(\frac {3\pi} 7)=\frac 1 8$ and the fact that $cos(\frac {4\pi} 7)=-cos(\frac {3\pi} 7)$.

For a proof of the formula see this

Also you can take a look at this

  • Any chance, you can explain the calculations because the link you gave contains mathematics that isn't in my course. I also have to find the value of sec^2(A)+sec^2(B)+sec^2(C). – nikola Dec 26 '16 at 13:08
  • @nikola See this: https://socratic.org/questions/how-do-you-prove-cos-pi-7-cos-2pi-7-cos-3pi-7-1-8 –  Dec 26 '16 at 13:11
  • any chance, you also know how to find the value of sec^2(A)+sec^2(B)+sec^2(C), please. – nikola Dec 26 '16 at 13:25
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    @nikola You have to post another question –  Dec 26 '16 at 13:26
  • Well, thanks I think I can pick up from here. – nikola Dec 26 '16 at 13:27
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After knowing the angles $A=\frac{\pi}{7}$, $B=\frac{2\pi}{7}$ and $C=\frac{4\pi}{7}$

$$\tan A\tan B+\tan B\tan C+\tan C\tan A=1+\sec A\sec B\sec C.$$

Substituting angles you should get $$\tan A\tan B+\tan B\tan C+\tan C\tan A=-7$$

  • Could the angles be calculated using values of trigonometric functions of 30, 45 and 60 degrees, like we find the value of 36 degree – nikola Dec 26 '16 at 13:00
  • That requires trigonometric transformations,and i will try to add it –  Dec 26 '16 at 13:02
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You have to calculate:

$$\cos \left(\frac{\pi}{7}\right) \cdot \cos \left(\frac{2\pi}{7}\right) \cdot \cos \left(\frac{4\pi}{7}\right)=P \quad (1)$$

We will use that

$$\sin(2x)=2\sin x \cos x \quad (2)$$

Multiply both sides of the equation $(1)$ by $\sin \left(\frac{\pi}{7}\right)$:

$$\sin \left(\frac{\pi}{7}\right) \cdot\cos \left(\frac{\pi}{7}\right) \cdot \cos \left(\frac{2\pi}{7}\right) \cdot \cos \left(\frac{4\pi}{7}\right)=P\cdot \sin \left(\frac{\pi}{7}\right)$$

Now using $(2)$ for $x=\pi/7$ we get:

$$\frac{1}{2}\cdot \sin \left(\frac{2\pi}{7}\right) \cdot \cos \left(\frac{2\pi}{7}\right) \cdot \cos \left(\frac{4\pi}{7}\right)=P\cdot \sin \left(\frac{\pi}{7}\right)$$

Now using $(2)$ for $x=2\pi/7$ we get:

$$\frac{1}{4}\cdot \sin \left(\frac{4\pi}{7}\right) \cdot \cos \left(\frac{4\pi}{7}\right)=P\cdot \sin \left(\frac{\pi}{7}\right)$$

Now using $(2)$ for $x=4\pi/7$ we get:

$$\frac{1}{8}\cdot \sin \left(\frac{8\pi}{7}\right) =P\cdot \sin \left(\frac{\pi}{7}\right)$$

Now remember that

$$\sin \left(\frac{8\pi}{7}\right)=\sin \left(\pi+\frac{\pi}{7}\right)=-\sin \left(\frac{\pi}{7}\right)$$

then:

$$P=-\frac{1}{8}$$

and:

$$1+\sec A\cdot \sec B \cdot \sec C= 1-8=-7$$

Arnaldo
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