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So sine of angle $A$ is just a ratio. It is the ratio of the length of the opposite or perpendicular of angle $A$ and the hypotenuse.
Cosine of angle $A$ is also just a ratio. It is the ratio of the length of base of $A$ and the hypotenuse.
Tangent of angle $A$ is the ratio of the length of the perpendicular to $A$ and length of the base of $A$.
So far so good. These are just ratios.
Sine of $A$ makes sense since it determines how big angle $A$ is.
But the cosine of angle $A$ or tangent of $A$ is not intuitive to me.
For example in what way the length of base of $A$ and the hypotenuse's length affect the angle of $A$? Their length doesn't seem relevant to how many degrees it has. Same for tangent.
Can someone please help me on this?

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    $\begingroup$ If you have a fixed hypotenuse and you increase the base, the angle has to decrease. For example, consider a ladder (hypotenuse) resting against a wall (opposite). If you pull the bottom of the ladder away from the wall (increasing the base), the ladder will get lower (decreasing the angle with the floor). $\endgroup$ – Rahul May 18 '16 at 21:15
  • $\begingroup$ "For example in what way the length of base of A and the hypotenuse's length affect the angle of A? Their length doesn't seem relevant to how many degrees it has. Same for tangent." ... Um... why do you think it has no relevence? Given a hypotentuse there is only one possible right triangle with a given b so the the angle most certainly is determined. Like wise given two legs of a right triangle there is only one right triangle with the sides so the angle is determined. I don't understand your confusion. $\endgroup$ – fleablood May 18 '16 at 21:19
  • $\begingroup$ @Rahul: For a fixed hypotenuse if I increase the base then the right angle is gone right? Take the triangle in my OP. Also in the ladder example the ladder gets lower but hypotenuse increases right? $\endgroup$ – Jim May 18 '16 at 21:19
  • $\begingroup$ @fleablood: I didn't understand this Given a hypotentuse there is only one possible right triangle with a given b $\endgroup$ – Jim May 18 '16 at 21:21
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    $\begingroup$ By "fixed hypotenuse" I meant a hypotenuse of fixed length (but it can slide, like a ladder sliding down a wall: learner.org/courses/learningmath/measurement/images/session5/…). $\endgroup$ – Rahul May 18 '16 at 21:36
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Here’s a somewhat different way to define these functions that might make the relationships to sides of triangle clearer.

Consider the unit circle with center $O$ and two rays from $O$ that intersect the circle at points $A$ and $B$, with an acute angle $\theta=\angle AOB$ between them, as shown below.

trig function diagram

Instead of defining the trigonometric functions as ratios of sides of triangles, define them as the lengths of certain line segments as follows:

  • $\sin\theta = \overline{BC}$, the perpendicular to $\overrightarrow{OA}$ that passes through $B$.
  • $\tan\theta=\overline{AD}$, the length of the segment tangent to the circle at $A$ that intersects $\overrightarrow{OB}$ at $D$.
  • $\sec\theta=\overline{OD}$, the distance to the intersection of $\overrightarrow{OB}$ with the tangent to the circle at $A$.

The “co” functions are the lengths of the corresponding line segments drawn on the other side of $\overrightarrow{OB}$, using the ray $\overrightarrow{OE}$, which is perpendicular to $\overrightarrow{OA}$.

$\triangle{OCB}$ is congruent to $\triangle{BFO}$, so $\overline{BF}=\overline{OC}$. Moreover, $\overline{OB}=1$, so we have the familiar ratios of sides of a right triangle for the sine and cosine.

Now, $\triangle{OCB}$ and $\triangle{OAD}$ are similar, so $\overline{AD}:\overline{BC}::\overline{OA}:\overline{OC}$ and $\overline{OA}=1$, and we have $$\tan\theta=\overline{AD}={\overline{BC}\over\overline{OC}}={\sin\theta\over\cos\theta}.$$ From these same triangles and the previous equation we find that $$\sec\theta = {\tan\theta\over\sin\theta} = \frac1{\cos\theta}.$$

The remaining two functions can be related to $\triangle{OCB}$ and the other functions via similar considerations.

If we change the radius of the circle, it’s clear that the length of all of these line segments change proportionally, i.e., the ratios of their lengths to the radius are constant across similar triangles. Since we’re taking a radius as the hypotenuse of the triangle, this also means that their ratios to the hypotenuse are constant for a given angle, which leads to the more familiar definitions of these functions as ratios of sides of a right triangle.

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  • $\begingroup$ How did you demonstrate how the ratios determine the angles. I am not sure what do you try to show me when you define sin8 as just a length $\endgroup$ – Jim May 20 '16 at 19:52
  • $\begingroup$ @Jim Once these functions are defined in this way for a right triangle with unit hypotenuse, they can be extended to any right triangle via similarity, which gives you the ratios of sides with which you’re already familiar. As far as determining the angles, since you’re working with right triangles, the lengths of any two sides uniquely determine the length of the third via the Pythagorean theorem, and so also determine all of the angles of the triangle. Starting with lengths instead of ratios should make this fairly obvious. $\endgroup$ – amd May 20 '16 at 21:40
  • $\begingroup$ The idea behind this answer is actually exactly what you're looking for. It defines the $\cos$ function in a way that is not dependent on a ratio. Since angles are best understood in terms of circles, this approach allows one to see the conn between the angle and the trig functions more directly. $\endgroup$ – rnrstopstraffic May 21 '16 at 2:39
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Every right angle with have distinct sides a, b, and hypotenuse h, always with the condition $a^2 + b^2 = h^2$. Each triangle with an (a,b,h) sides will have a distinct angle with with a one-to-one coorespondence between angle $\theta$ and the set of $(a/h, b/h, 1)$ where $(a/h, b/h, 1)$ represents a class of similar right triangles-- similar up to a scaling factor.

$a/h$ is unique to the angle. We can that $sin$. $b/h$ is unique to the angle we call that $cos$. $a/b = (a/h)/(b/h)$ is also unique. That is $tan$.

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You say you understand that $A/H$ determines how big the angle is. But $B/H = \sqrt{1 - A/H}$ by the pythagorean theorem. If one determines how big the angle is, the other has to also determine how big it is.

Also if you just flip the triangle sideways, the height becomes the base, and the base becomes the height. So whatever was true about the height of one must be true about the base of the other.

Now if $A/H$ determines how big the angle is, so must $\frac{A/H}{\sqrt{1 - A/H}} = \frac{A/H}{B/H} = A/B = \tan$.

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  • $\begingroup$ Thanks for explaining this. Could you please clarify a bit the part will have a distinct angle with with a one-to-one coorespondence between angle θ and the set of (a/h,b/h,1)(a/h,b/h,1) $\endgroup$ – Jim May 18 '16 at 22:01
  • $\begingroup$ Let's consider for simplicity sake only triangles with fixed hypotenuse h. Then for a triangle with angle $\theta$ there will be one and only one triangle with sides a,b,h. And for every triangle with sides a',b', h there will be one and only one angle $\theta'$. For any given a we can find b= $\sqrt{h^2 -a^2}$ and therefore we can find $\theta$. For any given b we can find a = $\sqrt{h^2-b^2}$ and therefore find $\theta$. For any given $a/b = c$ we can solve for $a/\sqrt{h^2 - a^2} = c$ to find out what a and b are and therefore find $\theta$. $\endgroup$ – fleablood May 18 '16 at 22:16
  • $\begingroup$ About your point: "B/H = sqrt(1 - A/H)....Don't you mean sqrt(1 - (A/H)^2)? I see your point that using the pythagorean theory we end up with a formula that expresses B/H in terms of A/H but it is not too intuitive, is it? $\endgroup$ – Jim May 20 '16 at 19:51
  • $\begingroup$ Yes, I meant the square. That was a typo. As for whether I think that expressing A in terms of B was intuitive. Yes, I do think that is utterly basic and the absolutely fundimental to trigonometry. So yes, I do expect it to be intuitive. $sin^2 + cos^2 = 1$ and the trig functions uniquely determined by the angle is the idea of trig. It should become the very first thing to consider. $\endgroup$ – fleablood May 21 '16 at 3:58
  • $\begingroup$ Even if using the pythagorean theorem isn't intuitive (which it should be) turning a right triangle on its side so that the leg becomes a base and the base becomes a leg should be. $\endgroup$ – fleablood May 21 '16 at 4:03

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