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Problem of the week #65 - June 25th, 2013

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Jameson

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Jan 26, 2012
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Thank you to anemone for this problem!

Prove \(\displaystyle \tan 55^{\circ}\tan 65^{\circ}\tan 75^{\circ}=\tan85^{\circ}\).
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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Jameson

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Jan 26, 2012
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Congratulations to the following members for their correct solutions:

1) MarkFL
2) Opalg
3) Sudharaka

Thank you again to anemone for this great problem!!

Solution (from MarkFL):
Let's begin with the left side:

\(\displaystyle \tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)\)

Using the product to sum identity for the tangent function:

\(\displaystyle \tan(\alpha)\tan(\beta)=\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{\cos(\alpha-\beta)-\cos(\alpha+\beta)}\)

we may write:

\(\displaystyle \tan\left(65^{\circ} \right)\tan\left(55^{\circ} \right)=\frac{\cos\left(10^{\circ} \right)-\cos\left(120^{\circ} \right)}{\cos\left(10^{\circ} \right)+\cos\left(120^{\circ} \right)}\)

Given that \(\displaystyle \cos\left(120^{\circ} \right)=-\frac{1}{2}\) we now have:

\(\displaystyle \tan\left(65^{\circ} \right)\tan\left(55^{\circ} \right)=\frac{\cos\left(10^{\circ} \right)+\frac{1}{2}}{\cos\left(10^{\circ} \right)-\frac{1}{2}}=\frac{2\cos\left(10^{\circ} \right)+1}{2\cos\left(10^{\circ} \right)-1}\)

Hence, we may write:

\(\displaystyle \tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{2\cos\left(10^{\circ} \right)+1}{2\cos\left(10^{\circ} \right)-1}\cdot\tan\left(75^{\circ} \right)\)

Using the identity:

\(\displaystyle \tan(\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)}\)

we have:

\(\displaystyle \tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\left(2\cos\left(10^{\circ} \right)+1 \right)\sin\left(75^{\circ} \right)}{\left(2\cos\left(10^{\circ} \right)-1 \right)\cos\left(75^{\circ} \right)}\)

Distributing, there results:

\(\displaystyle \tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{2\sin\left(75^{\circ} \right)\cos\left(10^{\circ} \right)+\sin\left(75^{\circ} \right)}{2\cos\left(75^{\circ} \right)\cos\left(10^{\circ} \right)-\cos\left(75^{\circ} \right)}\)

Using the product-to-sum identity:

\(\displaystyle 2\sin(\alpha)\cos(\beta)=\sin(\alpha+\beta)+\sin( \alpha-\beta)\)

we have:

\(\displaystyle 2\sin\left(75^{\circ} \right)\cos\left(10^{\circ} \right)=\sin\left(85^{\circ} \right)+\sin\left(65^{\circ} \right)\)

Using the product-to-sum identity:

\(\displaystyle 2\cos(\alpha)\cos(\beta)=\cos(\alpha+\beta)+\cos( \alpha-\beta)\)

we have:

\(\displaystyle 2\cos\left(75^{\circ} \right)\cos\left(10^{\circ} \right)=\cos\left(85^{\circ} \right)+\cos\left(65^{\circ} \right)\)

And now we may state:

\(\displaystyle \tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\sin\left(85^{\circ} \right)+\sin\left(65^{\circ} \right)+\sin\left(75^{\circ} \right)}{\cos\left(85^{\circ} \right)+\cos\left(65^{\circ} \right)-\cos\left(75^{\circ} \right)}\)

Rearranging, we have:

\(\displaystyle \tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\sin\left(85^{\circ} \right)+\sin\left(75^{\circ} \right)+\sin\left(65^{\circ} \right)}{\cos\left(85^{\circ} \right)-\left(\cos\left(75^{\circ} \right)-\cos\left(65^{\circ} \right) \right)}\)

Using the sum-to-product identity:

\(\displaystyle \sin(\alpha)+\sin(\beta)=2\sin\left(\frac{\alpha+ \beta}{2} \right)\cos\left(\frac{\alpha-\beta}{2} \right)\)

we have:

\(\displaystyle \sin\left(75^{\circ} \right)+\sin\left(65^{\circ} \right)=2\sin\left(70^{\circ} \right)\cos\left(5^{\circ} \right)\)

Using the sum-to-product identity:

\(\displaystyle \cos(\alpha)-\cos(\beta)=-2\sin\left(\frac{\alpha+ \beta}{2} \right)\sin\left(\frac{\alpha-\beta}{2} \right)\)

we have:

\(\displaystyle -\left(\cos\left(75^{\circ} \right)-\cos\left(65^{\circ} \right) \right)=2\sin\left(70^{\circ} \right)\sin\left(5^{\circ} \right)\)

Thus, we now have:

\(\displaystyle \tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\sin\left(85^{\circ} \right)+2\sin\left(70^{\circ} \right)\cos\left(5^{\circ} \right)}{\cos\left(85^{\circ} \right)+2\sin\left(70^{\circ} \right)\sin\left(5^{\circ} \right)}\)

Using the co-function identities:

\(\displaystyle \cos(\alpha)=\sin\left(90^{\circ}-\alpha \right)\)

\(\displaystyle \sin(\alpha)=\cos\left(90^{\circ}-\alpha \right)\)

we have:

\(\displaystyle \cos\left(5^{\circ} \right)=\sin\left(85^{\circ} \right)\)

\(\displaystyle \sin\left(5^{\circ} \right)=\cos\left(85^{\circ} \right)\)

and we now may write:

\(\displaystyle \tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\sin\left(85^{\circ} \right)+2\sin\left(70^{\circ} \right)\sin\left(85^{\circ} \right)}{\cos\left(85^{\circ} \right)+2\sin\left(70^{\circ} \right)\cos\left(85^{\circ} \right)}\)

Factoring, we get:

\(\displaystyle \tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\sin\left(85^{\circ} \right)\left(1+2\sin\left(70^{\circ} \right) \right)}{\cos\left(85^{\circ} \right)\left(1+2\sin\left(70^{\circ} \right) \right)}\)

Dividing out common factors, we now have:

\(\displaystyle \tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\sin\left(85^{\circ} \right)}{\cos\left(85^{\circ} \right)}\)

Using the identity:

\(\displaystyle \tan(\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)}\)

we have:

\(\displaystyle \tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\tan\left(85^{\circ} \right)\)

Shown as desired.


Alternate solution (from Opalg):
The equation $\tan (3x) = \tan 75^\circ$ has solutions $x=25^\circ,\ 85^\circ,\ 145^\circ$. (The first solution is obvious. For the others, notice that if you add $60^\circ$ to $x$, then $3x$ increases by $180^\circ$ so its tan is unaltered.)

Next, $\tan(3x) = \dfrac{3t-t^3}{1-3t^2}$, where $t = \tan x$. So the equation $\tan (3x) = \tan 75^\circ$ becomes $3t-t^3 = (1-3t^2)\tan 75^\circ$, or $t^3 - 3(\tan 75^\circ)t^2 - 3t +\tan 75^\circ = 0$. The product of the roots of this cubic equation is equal to the negative of the constant term, $-\tan 75^\circ$. But the roots are $\tan25^\circ,\ \tan85^\circ,\ \tan145^\circ$. Therefore $$\tan25^\circ \tan85^\circ \tan145^\circ = -\tan 75^\circ.$$
Now use the relations $\tan(90^\circ - x) = \dfrac1{\tan x}$ and $\tan(90^\circ + x) = -\dfrac1{\tan x}$ to write that as $$\Bigl(\frac1{\tan65^\circ}\Bigr) \tan85^\circ \Bigl(-\frac1{\tan 55^\circ}\Bigr) = -\tan 75^\circ,$$ so that $\tan55^\circ \tan65^\circ \tan75^\circ = \tan 85^\circ$.
 
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