Nonagon Property!

Geometry Level 2

sec ⁡ 2 ( π 9 ) + sec ⁡ 2 ( 2 π 9 ) + sec ⁡ 2 ( 4 π 9 ) = ? \large \sec^{2} \left ( \dfrac{\pi}{9} \right ) + \sec^{2} \left (\dfrac{2 \pi}{9}\right) + \sec^{2} \left (\dfrac{4 \pi}{9}\right) = \, ?


The answer is 36.

View wiki
Danish Ahmed by Danish Ahmed , 24, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Louis W
Mar 12, 2015

Someone with an easier solution feel free to write it and I won't be at you, because this one is a doozey the way I did it. But I got it.

WARNING! WARNING! Trigonometry Overload Commencing!

Trig Identity: sec ⁡ θ = 1 cos ⁡ θ \sec\theta=\frac{1}{\cos\theta}

sec ⁡ 2 π 9 + sec ⁡ 2 2 π 9 + sec ⁡ 2 4 π 9 = 1 cos ⁡ 2 π 9 + 1 cos ⁡ 2 2 π 9 + 1 cos ⁡ 2 4 π 9 \sec^{2}\frac{\pi}{9}+\sec^{2}\frac{2\pi}{9}+\sec^{2}\frac{4\pi}{9}=\frac{1}{\cos^{2}\frac{\pi}{9}}+\frac{1}{\cos^{2}\frac{2\pi}{9}}+\frac{1}{\cos^{2}\frac{4\pi}{9}} = cos ⁡ 2 2 π 9 cos ⁡ 2 4 π 9 + cos ⁡ 2 π 9 cos ⁡ 2 4 π 9 + cos ⁡ 2 π 9 cos ⁡ 2 2 π 9 cos ⁡ 2 π 9 cos ⁡ 2 2 π 9 cos ⁡ 2 4 π 9 =\frac{\cos^{2}\frac{2\pi}{9}\cos^{2}\frac{4\pi}{9}+\cos^{2}\frac{\pi}{9}\cos^{2}\frac{4\pi}{9}+\cos^{2}\frac{\pi}{9}\cos^{2}\frac{2\pi}{9}}{\cos^{2}\frac{\pi}{9}\cos^{2}\frac{2\pi}{9}\cos^{2}\frac{4\pi}{9}}

First, the numerator.

Trig Identities: cos ⁡ θ cos ⁡ ϕ = 1 2 [ cos ⁡ ( θ − ϕ ) + cos ⁡ ( θ + ϕ ) ] \cos\theta\cos\phi=\frac{1}{2}[\cos(\theta-\phi)+\cos(\theta+\phi)]

cos ⁡ ( π − θ ) = − cos ⁡ θ \cos(\pi-\theta)=-\cos\theta

cos ⁡ π 3 = 1 2 , cos ⁡ 2 π 3 = − 1 2 \cos\frac{\pi}{3}=\frac{1}{2}, \cos\frac{2\pi}{3}=-\frac{1}{2}

N u m = Num= [ 1 2 ( cos ⁡ 2 π 9 + cos ⁡ 2 π 3 ) ] 2 + [ 1 2 ( cos ⁡ π 3 + cos ⁡ 5 π 9 ) ] 2 + [ 1 2 ( cos ⁡ π 9 + cos ⁡ π 3 ) ] 2 [\frac{1}{2}(\cos\frac{2\pi}{9}+\cos\frac{2\pi}{3})]^{2}+[\frac{1}{2}(\cos\frac{\pi}{3}+\cos\frac{5\pi}{9})]^{2}+[\frac{1}{2}(\cos\frac{\pi}{9}+\cos\frac{\pi}{3})]^{2} = 1 4 ( cos ⁡ 2 π 9 − 1 2 ) 2 + 1 4 ( 1 2 − cos ⁡ 4 π 9 ) 2 + 1 4 ( cos ⁡ π 9 + 1 2 ) 2 =\frac{1}{4}(\cos\frac{2\pi}{9}-\frac{1}{2})^{2}+\frac{1}{4}(\frac{1}{2}-\cos\frac{4\pi}{9})^{2}+\frac{1}{4}(\cos\frac{\pi}{9}+\frac{1}{2})^{2} = 1 4 [ cos ⁡ 2 π 9 + cos ⁡ 2 2 π 9 + cos ⁡ 2 4 π 9 + cos ⁡ π 9 − cos ⁡ 2 π 9 − cos ⁡ 4 π 9 + 3 4 ] =\frac{1}{4}[\cos^{2}\frac{\pi}{9}+\cos^{2}\frac{2\pi}{9}+\cos^{2}\frac{4\pi}{9}+\cos\frac{\pi}{9}-\cos\frac{2\pi}{9}-\cos\frac{4\pi}{9}+\frac{3}{4}]

Look at the 3 cosines that aren't squared:

Trig Identity: cos ⁡ θ + cos ⁡ ϕ = 2 cos ⁡ ( θ + ϕ 2 ) cos ⁡ ( θ − ϕ 2 ) \cos\theta+\cos\phi=2\cos(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2}) .

cos ⁡ π 9 − cos ⁡ 2 π 9 − cos ⁡ 4 π 9 = cos ⁡ π 9 − ( cos ⁡ 2 π 9 + cos ⁡ 4 π 9 ) \cos\frac{\pi}{9}-\cos\frac{2\pi}{9}-\cos\frac{4\pi}{9}=\cos\frac{\pi}{9}-(\cos\frac{2\pi}{9}+\cos\frac{4\pi}{9}) = cos ⁡ π 9 − 2 cos ⁡ π 3 cos ⁡ π 9 = cos ⁡ π 9 − cos ⁡ π 9 = 0 =\cos\frac{\pi}{9}-2\cos\frac{\pi}{3}\cos\frac{\pi}{9}=\cos\frac{\pi}{9}-\cos\frac{\pi}{9}=0

Well that was nice. Now look at the 3 squared cosines:

Trig Identity: cos ⁡ 2 θ = 1 2 ( 1 + cos ⁡ 2 θ ) \cos^{2}\theta=\frac{1}{2}(1+\cos2\theta) and one from earlier.

cos ⁡ 2 π 9 + cos ⁡ 2 2 π 9 + cos ⁡ 2 4 π 9 \cos^{2}\frac{\pi}{9}+\cos^{2}\frac{2\pi}{9}+\cos^{2}\frac{4\pi}{9} = 1 2 ( 1 + cos ⁡ 2 π 9 ) + 1 2 ( 1 + cos ⁡ 4 π 9 ) + 1 2 ( 1 + cos ⁡ 8 π 9 ) =\frac{1}{2}(1+\cos\frac{2\pi}{9})+\frac{1}{2}(1+\cos\frac{4\pi}{9})+\frac{1}{2}(1+\cos\frac{8\pi}{9}) = 1 2 ( 1 + cos ⁡ 2 π 9 ) + 1 2 ( 1 + cos ⁡ 4 π 9 ) + 1 2 ( 1 − cos ⁡ π 9 ) =\frac{1}{2}(1+\cos\frac{2\pi}{9})+\frac{1}{2}(1+\cos\frac{4\pi}{9})+\frac{1}{2}(1-\cos\frac{\pi}{9}) = 3 2 − 1 2 ( cos ⁡ π 9 − cos ⁡ 2 π 9 − cos ⁡ 4 π 9 ) = 3 2 − 1 2 ( 0 ) = 3 2 =\frac{3}{2}-\frac{1}{2}(\cos\frac{\pi}{9}-\cos\frac{2\pi}{9}-\cos\frac{4\pi}{9})=\frac{3}{2}-\frac{1}{2}(0)=\frac{3}{2}

Put the numerator back together: 1 4 ( 3 2 + 3 4 ) = 9 16 \frac{1}{4}(\frac{3}{2}+\frac{3}{4})=\frac{9}{16}

Earlier trig identities can be used on the denominator as well: * cos ⁡ π 2 = 0 \cos\frac{\pi}{2}=0

D e n = ( cos ⁡ π 9 cos ⁡ 2 π 9 cos ⁡ 4 π 9 ) 2 = [ 1 2 ( cos ⁡ π 9 + cos ⁡ π 3 ) cos ⁡ 4 π 9 ] 2 Den=(\cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9})^{2}=[\frac{1}{2}(\cos\frac{\pi}{9}+\cos\frac{\pi}{3})\cos\frac{4\pi}{9}]^{2} = [ 1 2 ( cos ⁡ π 9 + 1 2 ) cos ⁡ 4 π 9 ] 2 = [ 1 2 ( cos ⁡ 4 π 9 cos ⁡ π 9 + 1 2 cos ⁡ 4 π 9 ) ] 2 =[\frac{1}{2}(\cos\frac{\pi}{9}+\frac{1}{2})\cos\frac{4\pi}{9}]^{2}=[\frac{1}{2}(\cos\frac{4\pi}{9}\cos\frac{\pi}{9}+\frac{1}{2}\cos\frac{4\pi}{9})]^{2} = [ 1 2 ( 1 2 ( cos ⁡ π 3 + cos ⁡ 5 π 9 ) + 1 2 cos ⁡ 4 π 9 ) ] 2 = [ 1 4 ( 1 2 + cos ⁡ 5 π 9 + cos ⁡ 4 π 9 ) ] 2 =[\frac{1}{2}(\frac{1}{2}(\cos\frac{\pi}{3}+\cos\frac{5\pi}{9})+\frac{1}{2}\cos\frac{4\pi}{9})]^{2}=[\frac{1}{4}(\frac{1}{2}+\cos\frac{5\pi}{9}+\cos\frac{4\pi}{9})]^{2} = [ 1 4 ( 1 2 + 2 cos ⁡ π 2 cos ⁡ π 18 ) ] 2 = [ 1 4 ( 1 2 + 0 ) ] 2 = 1 64 =[\frac{1}{4}(\frac{1}{2}+2\cos\frac{\pi}{2}\cos\frac{\pi}{18})]^{2}=[\frac{1}{4}(\frac{1}{2}+0)]^{2}=\frac{1}{64}

Now put it all back together:

9 16 1 64 = 36 □ \frac{\frac{9}{16}}{\frac{1}{64}}=36\space\space\space\Box

23 Helpful 4 Interesting 5 Brilliant 3 Confused

Here is the outline of an alternative solution: To simplify the exposition, let me replace sec ⁡ ( π 9 ) \sec(\frac{\pi}{9}) by sec ⁡ ( 8 π 9 ) \sec(\frac{8\pi}{9}) ; since we are squaring, the answer will be the same. Applying the triple angle formula for cos to t = 2 π 9 t =\frac{2\pi}{9} , t = 4 π 9 t =\frac{4\pi}{9} and t = 8 π 9 t =\frac{8\pi}{9} we find that 4 cos ⁡ 3 t − 3 cos ⁡ t = cos ⁡ ( 3 t ) = − 1 2 4\cos^3t - 3\cos t = \cos(3t) = -\frac{1}{2} in all three cases. Thus a = cos ⁡ ( 2 π 9 ) a=\cos(\frac{2\pi}{9}) , b = cos ⁡ ( 4 π 9 ) b=\cos(\frac{4\pi}{9}) , and c = cos ⁡ ( 8 π 9 ) c=\cos(\frac{8\pi}{9}) are the roots of the polynomial 4 x 3 − 3 x + 1 2 4x^{3}-3x+\frac{1}{2} . By Vieta's Formula we have a b c = − 1 8 abc = -\frac{1}{8} , a + b + c = 0 a+b+c=0 and a b + b c + a c = − 3 4 ab+bc+ac = - \frac{3}{4} . Now we find the answer 1 a 2 + 1 b 2 + 1 c 2 = ( b c ) 2 + ( a c ) 2 + ( a b ) 2 ( a b c ) 2 = ( a b + b c + a c ) 2 − 2 a b c ( a + b + c ) ( a b c ) 2 = ( a b + b c + a c ) 2 ( a b c ) 2 = 36 \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} = \frac{(bc)^2+(ac)^2+(ab)^2}{(abc)^2}=\frac{(ab+bc+ac)^2-2abc(a+b+c)}{(abc)^2} = \frac{(ab+bc+ac)^2}{(abc)^2}=36 .

Otto Bretscher - 6 years, 3 months ago

Log in to reply

good explanation sir

Akash Einstein - 5 years, 2 months ago

Saved a soul sir!!

Sanmit Mandal - 2 years, 7 months ago

hats off .

gundu cat - 6 years, 2 months ago
Siddharth Singh
Mar 14, 2015

I just found the number of lines joining the points in the nonagon. 8*9/2=36.

8 Helpful 1 Interesting 0 Brilliant 0 Confused

Why plz ? :$

Rohan Nuckchady - 6 years, 2 months ago

can u pls explain how did tht corelate to the question

Johnson B. Vadakkethala - 3 years, 8 months ago

I have to also find the correlation of formula with figure

Aman Pandey - 3 years, 6 months ago
Lab Bhattacharjee
Mar 30, 2016

http://math.stackexchange.com/questions/175736/evaluate-tan-220-circ-tan-240-circ-tan-280-circ

3 Helpful 0 Interesting 1 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...