# How can I represent the root of $\sin x=\cos x+\tan x$ as “length”?

Quesiton:
Represent the root(s) of $\sin x=\cos x+\tan x$ as length on rectangular coordinate.

For example, if $x=2$, you represent it as "the length between $(0,0)$ and $(2,0)$".

How can I solve this?

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Why would "the length between $(0,0)$ and $(u,0)$, where u is a solution to $\sin\,u=\cos\,u+\tan\,u$ not suffice?" –  Guess who it is. Jul 23 '11 at 18:16
That's right. But I want to know the exact value of root, and I want to represent it as "length". –  Sqrt4 Jul 23 '11 at 18:25
Generally there is no reason to expect a simple expression for a transcendental equation such as yours... but FWIW: express everything in terms of either sine or cosine, solve the resulting algebraic equation for sine or cosine, and then take the arcsine/arccosine of that... –  Guess who it is. Jul 23 '11 at 18:26
Thanks. I asked this just for fun. I saw this problem on internet. –  Sqrt4 Jul 23 '11 at 18:32

The equation $\sin x=\cos x+\tan x$ is equivalent to

$$\frac{2\tan \frac{x}{2}}{1+\tan ^{2}\frac{x}{2}}=\frac{1-\tan ^{2}\frac{x}{2% }}{1+\tan ^{2}\frac{x}{2}}+\frac{2\tan \frac{x}{2}}{1-\tan ^{2}\frac{x}{2}}.$$

Set $\tan \frac{x}{2}=y$. Then

$$2y=1-y^{2}+\left( 2y\right) \frac{1+y^{2}}{1-y^{2}},$$

or, equivalently

$$y^{4}+4y^{3}-2y^{2}+1=0.$$

Then $x=2\arctan y$, where $y$ are the solutions of this quartic (see this computation in Wolfram Alpha).

The direct computation in Wolfram Alpha gives solution(s) in terms of $\arccos(R(x))$ where $R(x)$ is a function with too many radicals.

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Of course, pick the appropriate solutions of the quartic, and then remember that you get an infinite number of solutions due to periodicity... –  Guess who it is. Jul 23 '11 at 19:22

I think I found another method to solve this problem. Let me know any errors of this.

There's an unit circle which center is (0, 0). And A(x', y') is on this circle.
Then we know $\sin x=y'$, $\cos x=x'$, and $\tan x= \frac{y'}{x'}$.

$$y' = x' + \frac{x'}{y'}$$
$$x'y' = (x')^2 + y'$$
$$(x'-1)y' = (x')^2$$
So, $y' = x'+1+\frac{1}{x'-1}$. And $(x')^2+(y')^2=1$.

We can get the intersection points of these.
Start at (1, 0), follow the circle's circumference, and end at that intersection.
The length of that curve is same with the roots of $\sin x = \cos x + \tan x$.

(I checked with calculator and these links: (1) , (2) )

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In the first displayed equation, the fraction is upside-down (but by the next equation, everything is the way it should be). –  Gerry Myerson Jul 24 '11 at 4:41