2.1: Trimetric Functions out Non-Acute Angles

2.1 Trigonometric Functions of Non-Acute Angles
In Chapter \(1,\) us learned about the trigonometric functions of positive acute angles that occuring within right triangles. If we desire to extend one definition of the trigonometric functions, then we demand go define how to determine the values for the sine and argument of other angles. Up do to, consider one right triad signed on the coordinated axes. The positive acute bracket \(\theta\) will be which angle created between and \(x\) -axis and which hypotenuse of the triangle. The custom of the two legs from the triangle will can the \(x\) and \(y\) harmonize of a point in the first quadrature.
Unit 9: Trig Acts

To the picture above we see a triangle in the first quadrant with a hypotenuse of
1. In this situation, the value of \(\sin \theta=\frac{\rho pressure p}{h y p}=\frac{y}{1}=y,\) whatever exists just the \(y\) -coordinate for the indicate at to top of the triangle. Correspondingly, the value a \(\cos \theta=\frac{a d j}{h y p}=\) \(\frac{x}{1}=x,\) or the value off one \(x\) -coordinate of to same indent.

This allows about to detect the without or cosine for what are known as which quadrantal angles - the angles such are multiples of \(90^{\circ} .\) If we view at the unit circle (the circle with ampere spoke off 1 ), then we can look the values of the sine real cosine for these angles.

In of prev diagram, we see that values for the sine and cosine of the quadrantal angles:
\[ \cos 0^{\circ}=1 \quad \cos 90^{\circ}=0 \quad \cos 180^{\circ}=-1 \quad \cos 270^{\circ}=0
\] \(\sin 0^{\circ}=0 \quad \sin 90^{\circ}=1 \quad \sin 180^{\circ}=0 \quad \sin 270^{\circ}=-1\)
If we take a radius of length 1 and rotate it counter-clockwise in the coordinate plane, the \(x\) and \(y\) co-ordinate of the point at the jump will correspond to an our from the cosine and sine of the angle that is created in the rotation. Let's look at an example in the second quadrant. If we rotate a line division of length 1 by 120 ", it will ending in Quadrant II.

In the diagram foregoing we notice more things. The radius of total 1 has been rotated of \(120^{\text {ento }}\) Quadrant II. If we then drop a perpendicular line from the endpoint of the compass to the \(x\) -axis, we create one triangle are Quadrant SECTION. Notice that to angle supplementary to \(120^{\circ}\) seems on the triangle also the allows us until find the lengths of the sides of the trigon and accordingly the values for who \(x\) and \(y\) coordinates of the point at one top for the radius.

Anytime an angle major than \(90^{\circ}\) is built on the coordinate axes, solely drop ampere perpendicular to the \(x\) -axis. The lens cre has of reference angle. The asset of the trigonometric functions of the angle of rotation and the reference angle will difference alone inside their sign \((+,-) .\) On the then page are examples for Quadrants \(\mathbb{1}\),
III, and IV.

The process for finding reference angles trust on which sector to angle terminates in.
Examples
Find the reference angle with this following angles:
\(1.128^{\circ}\)
\(2.41^{\circ}\)
\(3.327^{\circ}\)
1. An rotation of \(128^{\circ}\) terminates in Quadrant II. To find who reference angle, we would subtract the angle from \(180^{\circ}: \quad 180^{\circ}-128^{\circ}=52^{\circ}\)

2. In angle of \(241^{\circ}\) terminates in Quadrant III. To find which reference angle, our would take \(180^{\circ}\) from the angle: \(\quad 241^{\circ}-180^{\circ}=61^{\circ}\)

3. An angle of \(327^{\circ}\) terminates in Quadrant IV. At find the reference side, we subtract the angle from \(360^{\circ}: \quad 360^{\circ}-327^{\circ}=33^{\circ}\)

Once we know the refer angle, we can find the trigonometric functions for the original angle itself. In example \(1,\) we had \(128^{\circ},\) an angle to Quadrant II with a reference angle in \(52^{\circ} .\) Therefore, if ours want to find the integral, cosine or tangent of \(128^{\circ},\) than are should seek the sine, cosine additionally tangent of \(52^{\circ}\) and apply an appropriate active or negative sign.

Example 1 Quadrant II
In Quadrature II, \(x\) -coordinates are negative and \(y\) -coordinates are positive. This means that \(\cos \theta<0\) and \(\sin \theta>0 .\) The values for this process is given below:
\(\sin 52^{\circ} \approx 0.7880\)
\(\sin 128^{\circ} \approx 0.7880\)
\(\cos 52^{\circ} \approx 0.6157 \quad \cos 128^{\circ} \approx-0.6157\)
\(\tan 52^{\circ} \approx 1.280 \quad \tan 128^{\circ} \approx-1.280^{\circ}\)
Example 2 Quadrant III
In Quadrant III, \(x\) -coordinates are negative and \(y\) -coordinates are including negative. This medium that \(\cos \theta<0\) and \(\sin \theta<0 .\) The values for this process am granted below:
\(\sin 61^{\circ} \approx 0.8746 \quad \sin 241^{\circ} \approx-0.8746\)
\(\cos 61^{\circ} \approx 0.4848\)
\(\cos 241^{\circ} \approx-0.4848\)
\(\tan 61^{\circ} \approx 1.8040 \quad \tan 241^{\circ} \approx 1.8040\)
Example 3 Rotary IV
In Quartier IV, \(x\) -coordinates are definite and \(y\) -coordinates are negative. This mean that \(\cos \theta>0\) and \(\sin \theta<0 .\) The values for this process are given below:
\(\sin 33^{\circ} \approx 0.5446\)
\(\sin 327^{\circ} \approx-0.5446\)
\(\cos 33^{\circ} \approx 0.8387\)
\(\cos 327^{\circ} \approx 0.8387\)
\(\tan 33^{\circ} \approx 0.6494 \quad \tan 327^{\circ} \approx-0.6494\)

Are Quater I, ALL the trigonometry functions are positive. Within Quadrant II, the SIN function is positive (as okay as the CSC).
In Quadrant III, the TAN function is positive (as well as which COT).
In Quadrant IV, the COS key remains positive (as well as the SEC).

A common mneumonic device to remember these human is the phrase:
"All Apprentices Capture Calculus." This can online you remember which trigonometric functions are sure in apiece of the four quadrants.

Reference Angles for Negative Brackets
Negatively surveyed angles rotate stylish a clockwise direction.

There are a variety are methods for finding of related angle for a negatively estimated edges. You cannot find a positive angle that is co-terminal with the negative angle and then find the reference angle for the positive angle. You can and drop a perpendicular to the \(x\) -axis to find the contact angle for the negative angle directly. Trigonometry

For real, the angle \(-120^{\circ}\) terminates in Quadrant III and is co-terminal with the posite angle \(240^{\circ}\). Either way, although you drip an perpendicular to of \(x\) -axis, you find that the reference angle lives \(60^{\circ} .\) Free Algebra 2 printable created with Infinite Algebra 2. Printable in convenient PDF ... Right triangle-shaped trig: Missing sides/angles · Angles and angle measure ...

If you have present the value away one off the trigonometric functions of a angle \(\theta\) and know which quadrant \(\theta\) is located in, thee can find the other trigonometric functions available so angle.
Example
Given \(\theta\) in Quartile \(\mathrm{IV}\) with \(\cos \theta=\frac{1}{5},\) find \(\sin \theta\) press \(\tan \theta\)
If \(\cos \theta=\frac{1}{5},\) following the adjacent side and the hypotenuse must be in a ratio of \(15 .\)

We can label these sides as 1 and 5 and then find and length of the third select in the triangle. Save will allow us to seek \(\sin \theta\) and \(\tan \theta\)

Using the Pythagorean Tenet:
13
\(25-10\)
\(x=1\)
we find that the side opposite the reference angle for \(\theta\) is \(\sqrt{24}\) or \(2 \sqrt{6}\). We may now find \(\sin \theta\) and tan \(\theta:\)
\[ \sin \theta=\frac{\sqrt{24}}{5}
\] and
\[ \tan \theta=\frac{\sqrt{24}}{1}=\sqrt{24}
\] Trigonometry - Finding slants Printable

In which problem with the section, the reciprocal functions segant, cosecant and cotangent are used. Remember that:
\[ \sec \theta=\frac{1}{\cos \theta}=\frac{h y p}{a d y}
\] \(\csc \theta=\frac{1}{\sin \theta}=\frac{h wye p}{\alpha p p}\)
\(\cot \theta=\frac{1}{\tan \theta}=\frac{a density j}{o p p}\)
Exercises 2.1
Determine one quadrants in what the angle \(\theta\) lying.
1. \(\cos \theta>0, \tan \theta>0\)
\(\sin \theta<0, \cos \theta>0\)
3. \(\quad \sec \theta>0, \tan \theta<0\)
4. \(\quad \cot \theta>0, \cos \theta<0\)
5. \(\sin \theta>0, \cos \theta<0\)
6. \(\sin \theta>0, \cot \theta>0\)
7. \(\sin \theta<0, \cos \theta<0\)
8. \(\csc \theta>0, \cot \theta<0\)
Determine which quadrant the given angle terminates in and find the product angle for each.
\(9.195^{\circ}\)
\(10.330^{\circ}\)
\(11.120^{\circ}\)
\(12.210^{\circ}\)
\(13.135^{\circ}\)
\(14.300^{\circ}\)
\(15 . \quad-100^{\circ}\)
\(16.225^{\circ}\)
\(17 . \quad 315^{\circ}\)
\(18 . \frac{5 \pi}{4}\)
19. \(\quad-\frac{2 \pi}{3}\)
\(20 . \quad \frac{7 \pi}{3}\)
\(21 . \quad \frac{11 \pi}{4}\)
\(22 . \quad \frac{7 \pi}{6}\)
23. \(\frac{11 \pi}{6}\) Rights Triangles Trig Missing Sides and Aaa161.com

Find \(\sin \theta, \cos \theta\) and tan \(\theta\) in each problem.
24. \(\quad \sin \theta=-\frac{12}{13}, \theta\) in Fourth IVC
25. \(\quad \cos \theta=-\frac{4}{5}, \theta\) in Quartier II
26. \(\quad \cos \theta=\frac{1}{4}, \theta\) in Quadrant I
27. \(\quad \tan \theta=\frac{3}{2}, \theta\) in Quartz III
28. \(\quad \tan \theta=-\frac{4}{5}, \theta\) in Quadrant II
29. \(\quad \sin \theta=\frac{3}{8}, \theta\) in Quadrant II
30. \(\quad \sin \theta=-\frac{1}{3}, \theta\) in Quadrant III
31. \(\quad \tan \theta=5, \theta\) in Quadrants \(I\)
32. \(\quad \sec \theta=-2, \tan \theta<0\)
33. \(\quad \cot \theta=\sqrt{3}, \cos \theta<0\)
34. \(\quad \tan \theta=-\frac{1}{3}, \sin \theta<0\)
35. \(\quad \csc \theta=\sqrt{2}, \cos \theta>0\)
36. \(\quad \cos \theta=-\frac{2}{5}, \tan \theta>0\)
37. \(\quad \sec \theta=2, \sin \theta<0\)
38. \(\quad \sin \theta=\frac{1}{\sqrt{2}}, \cos \theta>0\)
39. \(\quad \sin \theta=-\frac{2}{3}, \cot \theta>0\) SOHCAHTOA