# If sin cos tan are just a ratio of two sides of right angled triangle then why do we use theta with these functions?

Like if I say what I am learning is we have a right-angled triangle. It has three sides. The ratio of the altitude and hypotenuse is sine. Simply sine represents specific sides of the triangle. Then why we use theta(angle) in this case. I am having depression because I cant understand why is it there?

• math.stackexchange.com/questions/2485739/… – Peter Oct 23 '17 at 11:35
• Since $\sin$, etc, are unchanged when you scale the triangle (as the scale factor cancels in the ratio) they are in fact functions of the angles of the triangle. – lulu Oct 23 '17 at 11:37

As you defined it, in a right-angled triangle $\triangle XYZ$ (with right angle Y) the function "sine in the triangle $\triangle XYZ$ on the vertex $X$" should depend on $|YZ|$ (the magnitude of opposite side) and $|XZ|$ (the magnitude of the hypothenuse). Let us denote this function as $\sin(\triangle XYZ;X)$ for a moment.

The following is a theorem: if $\triangle ABC$ and $\triangle ADE$ are both right-angled triangles (or more generally, if they are aligned similar triangle with a common vertex $A$), then $$\sin(\triangle ABC,A):=\frac{|BC|}{AC}=\frac{|DE|}{AE}:=\sin(\triangle ADE;D)$$

Hence, the value $\sin(\triangle ABC;A)$ does not depend on the actual magnitudes of the sizes, but rather on the magnitude of the angle $A$. If we call $\theta$ this magnitude, we have our more usual notation $\sin(\theta)$. Note that writing $\sin(\triangle ABC;A)$ all the time will be very messy, and since it only depends on the magnitude of an angle, the short-cut is preferable.

• @DavidK Oh, yeah, it should be vertex. – Darío G Oct 23 '17 at 12:58

Because the ratio is depends on the angle. Think about 2 similar triangles with ratio of 1:2. Say for the small one there is $A,B,C$ and the $3$ angles are $\theta,\alpha,90°(\text{or }\frac{\pi}2 radians)$($\theta$ is between $A,B$, $\alpha$ is the angle between $B,C$) so the large triangle is $2A,2B,2C$ which the same angles. Now for the small triangle $\sin\theta=\frac CB$ and for the large you have $\sin\theta=\frac{2C}{2B}=\frac CB$ so you can see that for similar triangles the functions are the same. There are more fundamentals reason but you will probably study them in the future, this is just a little insight for this decision

$sin( \theta )$ = Altitude/Hypotenuse.
A triangle has three sides and three angles.
$\theta$ is one of the angle, $sin( \theta )$ specify which angle we are referring to and which of the ratio.

Because if we didn't specify and the triangle was turned 180° and then flipped, we would not be able to distinguish between $sin( \theta )$ and $sin( 90°- \theta)$, this $(90°-\theta)$ is the angle above $\theta$ in a right angled triangle.

$\theta, (90° - \theta)$ are complimentary angles, also because $sin$ is a function, and every function must contain some variable mapped in it