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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.

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    $\begingroup$ Historically, the trigonometric functions were first defined in terms of the circle. The right triangle definitions came later. $\endgroup$ – N. F. Taussig Jul 10 '16 at 10:18
  • $\begingroup$ This answer may help you merge the two notions in your mind. $\endgroup$ – Blue Jul 10 '16 at 14:02
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    $\begingroup$ @N.F.Taussig this made alot of sense to me. So the trig functions are defined by the unit circle, and the right triangle definitions are just a side property of trig $\endgroup$ – Llama.new Jul 11 '16 at 5:32
  • $\begingroup$ Hi @Blue , that was a great explanation for the angle sum and difference equations for all angles between 0 and 90, but I still cant wrap around the jump from acute to obtuse angles. From the post, I understand that the idea of trig functions have extended beyond the first quadrant and acute angles, but the link in between is not fully clear (kind of like an exploration). $\endgroup$ – Llama.new Jul 11 '16 at 8:27
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Extending to other quadrants

If trigonometric identity formulas hold true so long as we are within the first quadrant, shouldn't we be able to extend our trig functions to all quadrants in this manner?

Mainly, you'd want the following identities.

$$\cos(\alpha+\theta)=\cos(\alpha)\cos(\theta)-\sin(\alpha)\sin(\theta)$$

$$\sin(\alpha+\theta)=\sin(\alpha)\cos(\theta)+\cos(\alpha)\sin(\theta)$$

Since we know $\sin\left(\frac\pi4\right)=\cos\left(\frac\pi4\right)=\frac{\sqrt2}2$, we can derive $\sin\left(\frac\pi2\right)=1$, $\cos\left(\frac\pi2\right)=0$, and so forth.

Similarly, we can derive what $\sin(-\theta)$ is by using $\sin(\theta-\theta),$ $\cos(\theta-\theta)$, and $\cos^2+\sin^2=1$, the Pythagorean identity. Two equations, and you can solve for $\cos(-\theta),\sin(-\theta)$ by substitution. Use Pythagorean identity for simplifying the answer.

It just happens to be that on the unit circle, the hypotenuse is by definition $1$, so...

$$\sin=\frac{\text{opp}}{\text{hyp}}=\text{opp}=y$$

$$\cos=\frac{\text{adj}}{\text{hyp}}=\text{adj}=x$$

And since both this definition and the one above derived by trig identities come out the same for $\theta>90\deg$ or $\pi$, then both are equally correct.

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  • $\begingroup$ This is one good reason why we extend sine and cosine the way we do. We could cite other reasons, such as their power series, their derivatives, and other properties that would become a lot less useful if we chose any other definitions. But all of it is just more of the same case you just made: why make things ugly and incomplete when you can make them completely beautiful? $\endgroup$ – David K Jul 18 '18 at 19:28
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For all angles, on the unit circle $\cos\theta=x,\sin\theta=y$. So

$$\begin{matrix}&0°&&90°&&180°&&270°&&360°\\\cos&&+&&-&&-&&+\\\sin&&+&&+&&-&&-\end{matrix}$$

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