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I am confused about the Unit Circle explanation for the trigonometric ratios for angles greater than 90 degrees.

It seems that for the first (top right) quadrant, $\sin(\theta)$ is equivalent to the y-coordinate, because

  1. $\sin(\theta)$ = opposite / hypotenuse
  2. $\sin(\theta)$ = opposite [hypotenuse is 1]
  3. $\sin(\theta)$ = y-coordinate [length of opposite side is the y-coordinate]

then, as theta extends in counter-clockwise manner to the top left quadrant, it is assumed that $\sin(\theta)$ is the still the value of the y-coordinate.

From what I understand, the basis for this is that because $\sin(\theta)$ is equivalent to the y-coordinate in the first quadrant, this extends to all quadrants. But this does not make sense to me because the $\sin(\theta)$ = o/h equation was applicable in the first quadrant but not in the others.

It seems to me that there are two definitions for the sine function:

  • The relationship between the opposite side and the hypotenuse for an acute angle in a right-angled triangle

  • The y-coordinate of a point along the unit circle, with angle theta (counter-clockwise from the x-axis)

The co-existence of these two definitions is making it confusing for me as it is not clear to me how we can get from the first to the second. How can we have two right angles in a triangle? It is just not possible. How would an "obtuse $sin(\theta)$" look like on the unit circle.

Confused to the core.

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    $\begingroup$ There are several near-duplicates of your question in the handy list of related questions to the right. $\endgroup$ – amd Nov 8 '17 at 6:14
  • $\begingroup$ Checked them all. Some of them aren't even answered. $\endgroup$ – Saksham Sharma Nov 8 '17 at 6:18
  • $\begingroup$ @amd Could you please answer it sir? $\endgroup$ – Saksham Sharma Nov 8 '17 at 8:46
  • $\begingroup$ Your "two" definitions are actually one and the same. The hypotenuse is actually the radius of the unit circle. What is said to be $∅$, is the angle between the hypotenuse and the x-axis of the quadrant you are in, which is the -ve x-axis for the 2nd and 3rd quadrants and +ve x-axis for the 1st and 4th. Note that this angle is always acute. $\endgroup$ – Mainak Roy Nov 8 '17 at 9:39
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    $\begingroup$ Historically, the trigonometric functions were first defined with respect to the circle. The special case of trigonometric ratios came later. $\endgroup$ – N. F. Taussig Nov 8 '17 at 10:02
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I think your specific confusion with non-first-quadrant angles and trigonometric functions is coming from the way negative "side lengths" are considered. For example, when $\theta=\frac{5\pi}4$. This looks like this on the unit circle:

enter image description here

Notice that $\sin\left(\frac{5\pi}4\right) = -\frac{\sqrt{2}}2.$ This is because the hypotenuse of the triangle is (of course) $1$, but the "opposite" is $-\frac{\sqrt{2}}2$.

In considering the sine function as a geometric tool, it is often useful to consider it as $\frac OH$. This idea, however, is generally less comprehensible when considering sine as a function or with respect to the unit circle. In these cases, the geometric interpretation slightly breaks down (due to negative numbers).

Additionally, it helps to visualize the angle to understand why negative numbers get involved. The angle $\frac{5\pi}4$ looks like:

enter image description here

not:

enter image description here

For a more intuitive understanding of how this works, check out this desmos demo.

If you have any questions for clarification, feel free to ask.

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