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Recently, I had a look at this question on Quora about how the values of integer tangents from $\tan(1^{\circ})$ to $\tan(90^{\circ})$ sum to $1$. I tried to solve the question on my own here, and got stuck here:

$$ \tan(θ) = \frac{1}{\tan(90 - \theta)} $$

$$ \tan(θ)\tan(90 - \theta) = 1 $$

Please help and, if possible, find more than one solution for this identity (since I know it is commonly used).

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    $\begingroup$ Draw a right triangle. Label one of the acute angles $\theta$. Then the other angle has measure $\frac{\pi}{2}-\theta$. Suppose the leg opposite of $\theta$ has length $a$, and the leg adjacent to $\theta$ has length $b$. Then $\tan\theta=\frac{a}{b}$. Use the same picture to see that the tangent of the other acute angle, $\frac{\pi}{2}-\theta$, is $\frac{b}{a}$. $\endgroup$ – symplectomorphic Apr 11 '17 at 2:36
  • $\begingroup$ @TobyMak: Two answers have been given. You should choose and "accept" an answer that satisfies you. (To accept move your mouse to the left margin near the top of the answer and then click there getting a green tick symbol). You can always wait for a few hours before accepting an answer to wait for other better answers. $\endgroup$ – P Vanchinathan Apr 12 '17 at 4:10
  • $\begingroup$ Oh! I didn't know you could do that, thanks! $\endgroup$ – Toby Mak Apr 12 '17 at 6:33
  • $\begingroup$ I noticed that someone has downvoted on my post. Please explain how I can improve and how I should do this. $\endgroup$ – Toby Mak May 29 '17 at 0:00
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Worth remembering:

The cosine of an angle is the sine of its complement.

I am pretty sure that is why this function is called cosine. I guess in the days when these values had to be looked up on tables, it was easier to have another column in the table than to add a step to the arithmetic.

The complement of an angle $\theta$ is $90-\theta$. So \begin{align*} \cos(\theta) &= \sin(90-\theta) \\ \sin(\theta) &= \cos(90-\theta) \end{align*} Since tangent is sine divided by cosine: $$ \tan(90-\theta) = \frac{\sin(90-\theta)}{\cos(90-\theta)} = \frac{\cos(\theta)}{\sin(\theta)} = \cot(\theta) $$ Again, the cotangent of an angle is the tangent of its complement. Finally, since $\cot(\theta) = \frac{1}{\tan(\theta)}$, your identity is proved.

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  • $\begingroup$ I think it should be sin(90-theta)/cos(90-theta) $\endgroup$ – Toby Mak Apr 11 '17 at 2:42
  • $\begingroup$ @TobyMak right you are. I fixed it. $\endgroup$ – Matthew Leingang Apr 11 '17 at 2:43
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    $\begingroup$ "I am pretty sure that is why this function is called cosine." Yes. The word "cosine" means "complementary sine". Likewise for the other co-functions. $\endgroup$ – Blue Apr 11 '17 at 3:10
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Draw a right angled traingle. The two acute angles can be called $\theta$ and $90-\theta$. Now the definition says tan of angle is the ratio of opposite side by adjacent side. Do this for these two angles in the triangle you have just drawn. Lo and behold you will get th eproof you want.

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  • $\begingroup$ Thanks! I never realised the solution was this simple! $\endgroup$ – Toby Mak Apr 11 '17 at 2:36
  • $\begingroup$ @P Vanachinathan Is there another solution for this? $\endgroup$ – Toby Mak Apr 11 '17 at 2:41
  • $\begingroup$ This is a simple solution which you have agreed. If you want more complicated one take the infinite Taylor series expansion for tan function substitute $\theta$ and $\pi/2-\theta$ multiply these two series and check if you get 1. $\endgroup$ – P Vanchinathan Apr 11 '17 at 2:48

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