Trigonometry (advanced) : Angles in 2nd Quadrant

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Angles in 2nd Quadrant

what you'll learn...

From 2nd to 1st Quadrant

2nd quadrant » $90 + θ$ $90+theta$
    → $θ$ $theta$ measured with y axis
    → complementary functions » $180 - θ$ $180-theta$
    → $θ$ $theta$ measured with x axis
    → equivalent functions

From 3rd to 1st Quadrant

third quadrant » $180 + θ$ $180+theta$
    → $θ$ $theta$ measured with x axis
    → equivalent functions » $270 - θ$ $270-theta$
    → $θ$ $theta$ measured with y axis
    → complementary functions

From 4th to 1st Quadrant

4th quadrant » $270 + θ$ $270+theta$
    → $θ$ $theta$ measured with y axis
    → complementary functions » $- θ$ $-theta$
    → $θ$ $theta$ measured with x axis
    → equivalent functions

Complementary Angles 1st Quadrant

1st quadrant » $90 - θ$ $90-theta$
→ $θ$ $theta$ measured with y axis
→ complementary functions

second quadrant

The angle $ω = 90 + θ$ $omega = 90+theta$ is shown in the figure as point P. The similar triangle with angle theta is given by $Q (x, y)$ $Q(x,y)$ in 1st quadrant. Note: Given $Q (x, y)$ $Q(x,y)$ , the coordinate of $P$ $P$ is $(P_{x}, P_{y})$ $(color(coral)(P_x),color(deepskyblue)(P_y))$ :
$P_{x} = - y$ $color(coral)(P_x=-y)$ and $P_{y} = x$ $color(deepskyblue)(P_y=x)$

$\tan (90 + θ) = \frac{- x}{y} = - \cot θ$ $tan (90+theta) = (-x)/y = - cot theta$

$\tan (90 + θ)$ $tan (90+theta)$ (tan of the given angle)
$= \frac{P_{y}}{P_{x}}$ $quad = (P_y)/(P_x)$ (by definition of $\tan$ $tan$ )
$= \frac{x}{- y}$ $quad = x/(-y)$ (substituting the values)
$= - \cot θ$ $quad = - cot theta$ (equivalently in 1st quadrant.)

Note: learners can work out this for $\sin$ $sin$ and $\cos$ $cos$ .

For angles given as $90 + θ$ $90+theta$ ,

• the angle is in 2nd quadrant, so x projection is negative and y projection is positive.

• the $∠ θ$ $/_theta$ is measured with y-axis, and so the similar triangle in 1st quadrant will have $∠ θ$ $/_theta$ with x-axis, thereby swapping the x and y projections.

the trigonometric ratios are:

• $\sin (90 + θ) = \cos θ$ $sin (90+theta) = cos theta$

• $\cos (90 + θ) = - \sin θ$ $cos (90+theta) = - sin theta$

• $\tan (90 + θ) = - \cot θ$ $tan (90+theta) = - cot theta$

The angle $ω = 180 - θ$ $omega = 180-theta$ is shown in the figure as point P. The similar triangle with $∠ θ$ $/_theta$ is given by $Q (x, y)$ $Q(x,y)$ in 1st quadrant.

Note: Given $Q (x, y)$ $Q(x,y)$ , the coordinates of $P$ $P$ is $(P_{x}, P_{y})$ $(color(coral)(P_x),color(deepskyblue)(P_y))$ :
$P_{x} = - x$ $color(coral)(P_x=-x)$ and $P_{y} = y$ $color(deepskyblue)(P_y=y)$

$\tan (180 - θ) = \frac{y}{- x} = - \tan θ$ $tan (180-theta) = y/(-x) = - tan theta$

$\tan (180 - θ)$ $tan (180-theta)$ (tan of the given angle)
$= \frac{P_{y}}{P_{x}}$ $quad = (P_y)/(P_x)$ (by definition of $\tan$ $tan$ )
$= \frac{y}{- x}$ $quad = y/(-x)$ (substituting the values)
$= - \tan θ$ $quad = - tan theta$ (equivalently in 1st quadrant.)

Note: learners can work out this for $\sin$ $sin$ and $\cos$ $cos$ .

summary

For angles given as $180 - θ$ $180-theta$ , the trigonometric ratios are:

• $\sin (180 - θ) = \sin θ$ $sin (180-theta) = sin theta$

• $\cos (180 - θ) = - \cos θ$ $cos (180-theta) = - cos theta$

• $\tan (180 - θ) = - \tan θ$ $tan (180-theta) = - tan theta$

third quadrant

The angle $ω = 180 + θ$ $omega = 180+theta$ is shown in the figure as point P. The similar triangle with angle theta is given by $Q (x, y)$ $Q(x,y)$ in 1st quadrant.

Note: Given $Q (x, y)$ $Q(x,y)$ , the coordinates of $P$ $P$ is $(P_{x}, P_{y})$ $(color(coral)(P_x),color(deepskyblue)(P_y))$ :
$P_{x} = - x$ $color(coral)(P_x=-x)$ and $P_{y} = - y$ $color(deepskyblue)(P_y=-y)$

$\tan (180 + θ) = - \frac{y}{- x} = \tan θ$ $tan (180+theta) = -y/-x = tan theta$

$\tan (180 + θ)$ $tan (180+theta)$ (tan of the given angle)
$= \frac{P_{y}}{P_{x}}$ $quad = (P_y)/(P_x)$ (by definition of $\tan$ $tan$ )
$= \frac{- y}{- x}$ $quad = (-y)/(-x)$ (substituting the values)
$= \tan θ$ $quad = tan theta$ (equivalently in 1st quadrant.)

Note: learners can work out this for $\sin$ $sin$ and $\cos$ $cos$ .

For angles given as $180 + θ$ $180+theta$ , the trigonometric ratios are:

• $\sin (180 + θ) = - \sin θ$ $sin (180+theta) = - sin theta$

• $\cos (180 + θ) = - \cos θ$ $cos (180+theta) = - cos theta$

• $\tan (180 + θ) = \tan θ$ $tan (180+theta) = tan theta$

The angle $ω = 270 - θ$ $omega = 270-theta$ is shown in the figure as point P. The similar triangle with angle theta is given by $Q (x, y)$ $Q(x,y)$ in 1st quadrant.

Note: Given $Q (x, y)$ $Q(x,y)$ , the coordinates of $P$ $P$ is $(P_{x}, P_{y})$ $(color(coral)(P_x),color(deepskyblue)(P_y))$ :
$P_{x} = - y$ $color(coral)(P_x=-y)$ and $P_{y} = - x$ $color(deepskyblue)(P_y=-x)$

What is $\tan (270 - θ) = \frac{- x}{- y} = \cot θ$ $tan (270-theta)= (-x)/(-y)= cot theta$

$\tan (270 - θ)$ $tan (270-theta)$ (tan of the given angle)
$= \frac{P_{y}}{P_{x}}$ $quad = (P_y)/(P_x)$ (by definition of $\tan$ $tan$ )
$= \frac{- x}{- y}$ $quad = (-x)/(-y)$ (substituting the values)
$= \cot θ$ $quad = cot theta$ (equivalently in 1st quadrant.)

Note: learners can work out this for $\sin$ $sin$ and $\cos$ $cos$ .

summary

For angles given as $270 - θ$ $270-theta$ , the trigonometric ratios are:

• $\sin (270 - θ) = - \cos θ$ $sin (270-theta) = - cos theta$

• $\cos (270 - θ) = - \sin θ$ $cos (270-theta) = - sin theta$

• $\tan (270 - θ) = \cot θ$ $tan (270-theta) = cot theta$

fourth quadrant

The angle $ω = 270 + θ$ $omega = 270+theta$ is shown in the figure as point P. The similar triangle with angle theta is given by $Q (x, y)$ $Q(x,y)$ in 1st quadrant.

Note: Given $Q (x, y)$ $Q(x,y)$ , the coordinates of $P$ $P$ is $(P_{x}, P_{y})$ $(color(coral)(P_x),color(deepskyblue)(P_y))$ :
$P_{x} = y$ $color(coral)(P_x=y)$ and $P_{y} = - x$ $color(deepskyblue)(P_y=-x)$

$\tan (270 + θ) = \frac{- x}{y} = - \cot θ$ $tan (270+theta) = (-x)/y = - cot theta$

$\tan (270 + θ)$ $tan (270+theta)$ (tan of the given angle)
$= \frac{P_{y}}{P_{x}}$ $quad = (P_y)/(P_x)$ (by definition of $\tan$ $tan$ )
$= \frac{- x}{y}$ $quad = (-x)/y$ (substituting the values)
$= - \cot θ$ $quad = - cot theta$ (equivalently in 1st quadrant.)

Note: learners can work out this for $\sin$ $sin$ and $\cos$ $cos$ .

For angles given as $270 + θ$ $270+theta$ , the trigonometric ratios are:

• $\sin (270 + θ) = - \cos θ$ $sin (270+theta) = - cos theta$

• $\cos (270 + θ) = \sin θ$ $cos (270+theta) = sin theta$

• $\tan (270 + θ) = - \cot θ$ $tan (270+theta) = - cot theta$

The angle $ω = - θ$ $omega = -theta$ is shown in the figure as point P. The similar triangle with angle theta is given by $Q (x, y)$ $Q(x,y)$ in 1st quadrant.

Note: Given $Q (x, y)$ $Q(x,y)$ , the coordinates of $P$ $P$ is $(P_{x}, P_{y})$ $(color(coral)(P_x),color(deepskyblue)(P_y))$ :
$P_{x} = x$ $color(coral)(P_x=x)$ and $P_{y} = - y$ $color(deepskyblue)(P_y=-y)$

$\tan (- θ) = \frac{- y}{x} = - \tan θ$ $tan (-theta) = (-y)/x = - tan theta$

$\tan (- θ)$ $tan (-theta)$ (tan of the given angle)
$= \frac{P_{y}}{P_{x}}$ $quad = (P_y)/(P_x)$ (by definition of $\tan$ $tan$ )
$= \frac{- y}{x}$ $quad = (-y)/x$ (substituting the values)
$= - \tan θ$ $quad = - tan theta$ (equivalently in 1st quadrant.)

Note: learners can work out this for $\sin$ $sin$ and $\cos$ $cos$ .

summary

For angles given as $- θ$ $-theta$ , the trigonometric ratios are:

• $\sin (- θ) = - \sin θ$ $sin (-theta) = - sin theta$

• $\cos (- θ) = \cos θ$ $cos (-theta) = cos theta$

• $\tan (- θ) = - \tan θ$ $tan (-theta) = - tan theta$

negative angle

The angle $ω = 90 - θ$ $omega = 90-theta$ is shown in the figure as point P. The similar triangle with angle theta is given by $Q (x, y)$ $Q(x,y)$ in 1st quadrant.

For angles given as $90 - θ$ $90-theta$ , the trigonometric ratios are:

• $\sin (90 - θ) = \cos θ$ $sin (90-theta) = cos theta$

• $\cos (90 - θ) = \sin θ$ $cos (90-theta) = sin theta$

• $\tan (90 - θ) = \cot θ$ $tan (90-theta) = cot theta$

For angles measured in reference to x axis, the trigonometric ratios remain the same when equivalently represented in the 1st quadrant.

For angles measured in reference to y axis, the trigonometric ratios are equivalently complementary ratios in the 1st quadrant.

summary

Trigonometric Ratios in 1st Quadrant:
For Angles in reference to x-axis:
$\sin (180 + θ) = - \sin θ$ $sin(180+theta) = -sin theta$
$\sin (180 - θ) = \sin θ$ $sin(180-theta) = sin theta$
$\cos (180 \pm θ) = - \cos θ$ $cos(180+-theta) = -cos theta$
$\tan (180 + θ) = \tan θ$ $tan(180+theta) = tan theta$

$\tan (180 - θ) = - \tan θ$ $tan(180-theta) = -tan theta$

$\sin (- θ) = - \sin θ$ $sin(-theta) = -sin theta$
$\cos (- θ) = \cos θ$ $cos(-theta) = cos theta$
$\tan (- θ) = - \tan θ$ $tan(-theta) = -tan theta$

For Angles in reference to y-axis:
$\sin (90 \pm θ) = \cos θ$ $sin(90+-theta) = cos theta$
$\cos (90 + θ) = - \sin θ$ $cos(90+theta) = -sin theta$
$\cos (90 - θ) = \sin θ$ $cos(90-theta) = sin theta$
$\tan (90 + θ) = - \cot θ$ $tan(90+theta) = -cot theta$

$\tan (90 - θ) = + \cot θ$ $tan(90-theta) = +cot theta$

$\sin (270 \pm θ) = - \cos θ$ $sin(270+-theta) = -cos theta$
$\cos (270 \pm θ) = \sin θ$ $cos(270+-theta) = sin theta$
$\tan (270 \pm θ) = - \tan θ$ $tan(270+-theta) = -tan theta$

Outline

It is advised to do the firmfunda version of "basics of Trigonometry" course before doing this.

The outline of material to learn "Advanced Trigonometry" is as follows.
Note: go to detailed outline of Advanced Trigonometry

→ Unit Circle form of Trigonmetric Values

→ Trigonometric Values in all Quadrants

→ Trigonometric Values or any Angles : First Principles

→ Understanding Trigonometric Values in First Quadrant

→ Trigonometric Values in First Quadrant

→ Trigonometric Values of Compound Angles: Geometrical Proof

→ Trigonometric Values of Compound Angles: Algebraic Proof

→ Trigonometric Values of Compound Angles: tan cot

→ Trigonometric Values of Compound Angles: more results