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Can we call $y=1$ as tangent to $y=\sin x$.

Definition of tangent is a straight line which touches the curve at exactly one point. But $y=1$ is tangent to $y=\sin x $ at infinite points. Can we call it as a tangent?

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In addition to what eyeballfrog just mentionned, by your definition the red line

$y=sin(1)+cos(1)(x-1)$

would not be a tangent to the blue sinusoid because it has another intersection point with it (near $x=-2$, on the figure below).

tangent at sin(1)

So your definition of tangent is not true globally, but only locally [*]. Plus having only $1$ intersection point is not sufficient, the slopes of the straight line and the curve should also be the same.

[*] And even locally, the line $y=1$ is a tangent to well... the constant function $y(x)=1$, with infinitely many common points.

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    Ok so red line is indeed a tangent to sinusoid in the neighbourhood of the point where it touches. Am i Right? – 2017-01-09
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    We simply call the red line, THE tangent to the sinusoid at the point $(1,sin(1))$. – 2017-01-09
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    ok now i clearly understood – 2017-01-09
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Your definition of tangent is a little off. A tangent line has the same slope as a curve at their intersection. $y = 1$ is tangent to $y = \sin x$ at every intersection point.

As an example of a single intersection line that is not tangent, the line $x = 0$ intersects $y = x^2$ only once, but is not tangent to it.

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    @UmeshShankar Similarly the curve $y=(x-1)^2(x+1)^2$ shares a comment tangent of $y=0$ at $=\pm1$. – 2017-01-09
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    @eyeballfrog Consider however that the vertical $y$ axis (line equation: $x=0$) is tangent to the curve $y=\sqrt[3]{x}$ at the origin, but the slope is undefined. – 2017-01-09
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    @dxiv There shouldn't be any problem with defining the slope to be $\infty$, using the one-point compactification of R. – 2017-01-09
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    @eyeballfrog It's not entirely straightforward. For example, in most contexts the vertical axis is *not* considered to be tangent to the curve $y=\sqrt[3]{x^2}$ which has a "cusp" at the origin. – 2017-01-09
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    @dxiv which is fine because the curve isn't differentiable there. – 2017-01-09
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    @eyeballfrog But the one-sided derivatives exist, and they are both equal to $\infty$ if you simply use `the one-point compactification` to define vertical tangents. I understand where you are going with that line of thought, all I am saying is that it requires careful definitions alongside i.e. *it's not entirely straightforward*. – 2017-01-09