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The question is "Find the tangent to y = 2 - sin(x); Where x = Pi"

I've solved this equation but looking at my work is not helping.

First I get the derivative of the question which nets me " - cos X" by using the power rule/trig derivatives. Than I substitute x for Pi which gets me a positive 1.

From here it gets weird because now I have y - 2 = cos (x) * (x - Pi). I'm not sure why the 2 is set up like that.

Next I have y = cos x*x-cos(x) Pi + 2 = y = x - Pi + x. Which is the answer.

I did this with a tutor but I can't understand why after getting a 1 I have to loop around and do all of this extra work. Can someone please explain why I am doing this and for what purpose? What does tangent in this scenario do for helping me in graphing or finding derivatives?

EDIT: Better asked, What am I suppose to do after I find the Gradient as @Patrick Stevens has pointed out?

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You're right in that you found the gradient of the tangent to be $1$. Now to define the line, you just need to find a single point on it; and we already know that $y=2-\sin(x)$ takes the value $2$ at $x=\pi$, so:

The tangent is the line through $(\pi, 2)$ whose gradient is $1$.

To answer the question "Why am I finding a tangent?": probably mainly for practice. As far as I'm aware, explicitly finding tangents is not a particularly useful skill. However, it's a very good exercise for developing your general algebraic fluency. Finding derivatives is a lot more useful (and is necessary to find the tangent).

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    I kinda follow what your saying but why is it that I did what I did in this problem? I'm referring to after finding 1 as my gradient why I get " y - 2 = cos(x) (x-Pi)" as a new step? Why does the two go outside of the equation? Why is the cos now positive (if it should be)? And why is it that i'm subtracting x and Pi? I think A MUCH BETTER way of phrasing my question is HOW do I find the Tangent after I find the Gradient? – 2017-02-19
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    The tangent is precisely specified as "the line through $(\pi, 2)$ whose gradient is $1$". You can turn this into $y=mx+c$ form by noting that the gradient, $m$, is $1$; then $y=x+c$ for some $c$. We have $2 = \pi + c$ (since $(\pi, 2)$ is on the line) so $c = 2-\pi$; hence $y=x+2-\pi$. – 2017-02-19
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    I don't really understand what you did, I'm afraid. – 2017-02-19