# In my textbook,the coordinate (x,y) by sine and cosine addition formula seems to form a circle,is that a coincidence?

In my textbook,the coordinate (x,y) by sine and cosine addition formula seems to form a circle,is that a coincidence ?Since the addition formula once was defined by acute case and to prove by geometry with acute case,why does the obtuse and reflex case also work to form a circle?

@J.M-my high school trig book
@joriki-it is when we add a angle to a acute angle we get a obtuse angle or reflex angle...

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Which textbook? – J. M. Jul 22 '11 at 18:14
Could you explain what you mean by "the coordinate (x,y) by sine and cosine addition formula seems to form a circle"? – joriki Jul 22 '11 at 18:14
The expression $(\cos,\sin)$ traces out a circle, and one can see why by drawing a line from the center to any point on the unit circle and then making a right triangle with the line as hypotenuse. At any rate, I don't see what the trigonometric addition formulas for sine and cosine have to do with anything, unless you want to involve 2D rotation matrices into the picture. – anon Jul 23 '11 at 5:51
@abcde: You might be shocked to know that (cosh, sinh) form a hyperbola. – Mehrdad Jul 23 '11 at 7:18

No, it's not a coincidence. The points $(x,y)$ of the form $$(x,y) = (\cos t,\sin t)$$ are precisely the points on the unit circle: the circle of radius $1$ centered at the origin. (In fact, this is one way in which the functions sine and cosine may be defined).

To see that every point of this form is in the unit circle, remember that the unit circle corresponds precisely to all the points $(x,y)$ whose distance to the origin is $1$. The distance from $(x,y)$ to $(0,0)$ is $\sqrt{x^2+y^2}$; so $(x,y)$ is on the unit circle if and only if $\sqrt{x^2+y^2}=1$, which occurs if and only if $x^2+y^2=1$. (This is "the equation of the circle").

If you have a point of the form $(\cos t,\sin t)$, then evaluating $x^2+y^2$ gives $$x^2+y^2 = \cos^2t + \sin^2 t = 1,$$ so the point is on the unit circle.

Conversely, if you have a point $(x,y)$ on the unit circle, consider the right triangle that vertices at $(0,0)$, $(x,0)$, and $(x,y)$. The angle $t$made by the sides from $(0,0)$ to $(x,0)$ and the hypotenuse $(0,0)$ to $(x,y)$ has cosine $\frac{x}{1} = x$ (remember the hypotenuse has length $1$, since $(x,y)$ is in the unit circle, and the adjacent side has length $x$, wince it goes from $(0,0)$ to $(x,0)$). And the sine of the angle is $\frac{y}{1}=y$ (since the opposite side goes from $(x,0)$ to $(x,y)$, so it has length $y$). So $x=\cos t$ and $y=\sin t$; that is, every point on the unit circle is of the form $(\cos t,\sin t)$ for some $t$.

Put these two together and it tells you that every point on the circle is of the form $(\cos t,\sin t)$ for some $t$, and that every point of the form $(\cos t,\sin t)$ is on the unit circle.

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So,my question is:does the a coordinate on the unit circle rely on the addition formula? – abcde Jul 22 '11 at 20:14
@abcde: The proofs of properties of the sine and cosine depend entirely on how you define the functions sine and cosine. Absent information about how you defined them, questions about whether something relies on something else are meaningless; one can define sine and cosine in terms of the unit circle, in which case everything about sine and cosine would rely on the unit circle, and the unit circle would not "depend" or "rely" on anything about sines and cosines. Or one can define sines and consines differently, and then deduce things about the unit circle. – Arturo Magidin Jul 22 '11 at 20:18
@abcde (cont): But since you're not saying what the definitions you are using are, your question cannot be answered (assuming that I have interpeted it correctly, which I'm afraid is a very big assumption to make...) – Arturo Magidin Jul 22 '11 at 20:18
Fine,thank you.i geuss that is the way it had to be – abcde Jul 22 '11 at 20:21
@abcde: No, it's not. You could try to give enough information to make answering your question possible. Of course, if you don't want to do that, then certainly that is the way it had to be. – Arturo Magidin Jul 22 '11 at 20:23

I assume that you observed something like the following. We have the addition formulas $$\sin(x+y)=\sin x \cos y+\cos x\sin y.$$ $$\cos(x+y)=\cos x\cos y -\sin x\sin y.$$

Square both right-hand sides, and add. After a while, using $\sin^2 x+\cos^2x=1$ and $\sin^2 y+\cos^2 y=1$, we obtain $$\sin^2(x+y)+\cos^2(x+y)=1.$$

The fact that $\sin^2 u+\cos^2 u=1$ was explained to you in class or in your book in terms of right-angled triangles. That explanation makes sense, but only when the angle $u$ is less than $90^\circ$, or, in radians, if $u<\pi/2$.

However, even if $x$ and $y$ are less than $90^\circ$, it is possible for $x+y$ to be greater than $90^\circ$. I think you are asking why the sum of the squares is still $1$, even if we go beyond $90^\circ$.

Here is a partial answer. A "right" way to define the trigonometric functions is in terms of the unit circle, as described in one of the answers. That way, we can define the trigonometric functions uniformly for all numbers $x$.

There is also a high school way to do it. I cannot draw a picture, so you will have to draw while reading. Imagine a point $P$ in the second quadrant, say the point with coordinates $(-3,4)$. Look at the angle that we have to turn the positive $x$-axis, counterclockwise, to get to $P$. This is an obtuse angle $\theta$. We want to define what we mean by $\sin\theta$ and $\cos\theta$.

Drop a perpendicular from $P$ to the $x$-axis, meeting the $x$ axis at $Q$. Then $\triangle PQO$ is right-angled, where $O$ is the origin. Let $\phi=\angle QOP$, in the ordinary sense. Then $\phi=180^\circ -\theta$, if we are working in degrees. Define $\sin\theta$ to be $\sin\phi$, and $\cos\theta$ to be $-\cos\phi$. This makes sense in terms of right triangles, since $\phi$ is an acute angle. With this definition, it is built in that $\sin^2 u +\cos^2 u=1$ even for angles between $90^\circ$ and $180^\circ$. This is because $\sin^2\phi+\cos^2\phi=1$, since $\phi$ is an acute angle. But we have only changed a sign; squaring takes care of that.

In a roughly similar way, we can define $\sin$ and $\cos$ for angles between $180^\circ$ and $270^\circ$, and also between $270^\circ$ and $360\circ$ (in each case, drop perpendiculars to the $x$-axis, and make the sensible choice of signs.)

So four different definitions, depending on which quadrant we are in! This is a very big nuisance, since when we try to extend the usual trigonometric identities, almost every little proof breaks up into a number of cases. So just for the sake of not being driven crazy by petty detail, we are almost forced to make a more sensible definition of $\sin$ and $\cos$, such as the one based on the unit circle.

It would now be a good idea to read part of the Wikipedia article on the Trigonometric Functions.

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