For a point P(x, y) on the terminal side of an angle q:
Define r: r = \text{distance from origin to P}
Sine Function:
\sin q = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{r}
\csc q = \frac{r}{y} = \frac{1}{\sin q}
Cosine Function:
\cos q = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{r}
\sec q = \frac{r}{x} = \frac{1}{\cos q}
Tangent Function:
\tan q = \frac{\text{opposite}}{\text{adjacent}} = \frac{y}{x} = \frac{\sin q}{\cos q}
\cot q = \frac{x}{y} = \frac{1}{\tan q} = \frac{\cos q}{\sin q}
Circle Measures:
A circle contains 360^{\circ} or 2\pi radians.
Conversion Formula:
180^{\circ} = \pi \text{ radians}
Convert 36^{\circ} to radians:
Apply Conversion:
36^{\circ} = 36^{\circ} \cdot \frac{\pi}{180^{\circ}} = \frac{2\pi}{10} = \frac{\pi}{5} \text{ radians}
Definition:
The reference angle is formed by the terminal side of the angle and the x-axis.
Quadrant Calculations:
Quadrant II: R = 180^{\circ} - q
Quadrant III: R = q - 180^{\circ}
Quadrant IV: R = 360^{\circ} - q
Quadrant I:
All trigonometric functions are positive.
Quadrant II:
Sine is positive.
Quadrant III:
Tangent is positive.
Quadrant IV:
Cosine is positive.
Mnemonic:
"All Students Take Calculus" (
Reciprocal Signs:
Sine and cosecant have the same sign.
Cosine and secant have the same sign.
Tangent and cotangent have the same sign.
Common Angles:
Using the ratios of common triangles with angles 30^{\circ}, 45^{\circ}, and 60^{\circ} allows us to determine the trigonometric functions of these special angles.
Values:
\sin 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}
\sin 30^{\circ} = \frac{1}{2}
\cos 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}
\cos 30^{\circ} = \frac{\sqrt{3}}{2}
\tan 45^{\circ} = \frac{1}{1} = 1
\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}
Sine and Cosine:
If (x, y) is a point on the unit circle, then \cos q = x and \sin q = y.
Values for Multiples of 90°:
\cos 0^{\circ} = 1
\cos 90^{\circ} = 0
\cos 180^{\circ} = -1
\cos 270^{\circ} = 0
\sin 0^{\circ} = 0
\sin 90^{\circ} = 1
\sin 180^{\circ} = 0
\sin 270^{\circ} = -1
Tangent Values:
\tan 0^{\circ} = 0
\tan 90^{\circ} = \text{undefined}
Cotangent Values:
\cot 0^{\circ} = \text{undefined}
\cot 90^{\circ} = 0
Secant Values:
\sec 0^{\circ} = 1
\sec 90^{\circ} = \text{undefined}
Cosecant Values:
\csc 0^{\circ} = \text{undefined}
\csc 90^{\circ} = 1
Calculation Ability:
At this point, you should be able to calculate the trig functions of all angles with reference angles of 30^{\circ}, 45^{\circ}, 60^{\circ}, as well as all multiples of 90^{\circ}.
\sin^2 q + \cos^2 q = 1
1 + \cot^2 q = \csc^2 q
\tan^2 q + 1 = \sec^2 q
Derivation:
The last two can be derived from the first by dividing by either \sin^2 q or \cos^2 q
\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B
\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B
Let A = B in the addition formulas:
\sin(2A) = 2 \sin A \cos A
\cos(2A) = \cos^2 A - \sin^2 A
a^2 = b^2 + c^2 - 2bc \cos A
\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}
From the Unit Circle:
\cos q = x and \sin q = y.
Sine Function:
As we move around the circle, y = \sin(x) measures the y-value of the point, starting at 0 and increasing to 1 at x = \frac{\pi}{2}.
Cosine Function:
y = \cos(x) measures the x-value of the point, starting at 1 and decreasing to 0 at x = \frac{\pi}{2}.
x | y = sin(x) | y = cos(x) |
---|---|---|
0 | 0 | 1 |
\pi/2 | 1 | 0 |
\pi | 0 | -1 |
3\pi/2 | -1 | 0 |
2\pi | 0 | 1 |
Definition:
The amplitude of y = A \sin(x) is |A|. A vertical stretch.
Definition:
The period of y = \sin(Bx) is \frac{2\pi}{B}. A horizontal shrink.
Example:
For y = \sin(x) + 1, the graph is shifted vertically by 1 unit.
Example:
For y = \cos(x - \frac{\pi}{2}), set x - \frac{\pi}{2} = 0 to find the shift.
Here, x = \frac{\pi}{2}, so the shift is to the right by \frac{\pi}{2}.
Example: Solve \sin^2 x = 1 for all 0 \le x < 2\pi
Rearrange:
\sin^2 x - 1 = 0
Factor:
(\sin x + 1)(\sin x - 1) = 0
Solve:
\sin x = 1 or \sin x = -1
Solutions:
x = \frac{\pi}{2}, \frac{3\pi}{2}
Example: Solve \cos 2x = 1 for 0 \le x < 2\pi
\cos 2x = \cos^2 x - \sin^2 x = 1
(1 - \sin^2 x) - \sin^2 x = 1
1 - 2\sin^2 x = 1
\sin^2 x = 0 \implies \sin x = 0
Solutions:
x = 0, \pi
\cos 2x = 1 \implies 2x = 0
Add One Period:
2x = 0 + 2\pi
Solve for x:
x = 0, \pi
Example: Solve \sin^2 3x = 1 for 0 \le x < 2\pi
\sin 3x = \pm 1
\sin 3x = 1 at 3x = \frac{\pi}{2}
\sin 3x = -1 at 3x = \frac{3\pi}{2}
Solve for x:
x = \frac{\pi}{6}, \frac{\pi}{2}
Period of \sin 3x:
The period of \sin 3x is \frac{2\pi}{3}, so add this to each solution until the values exceed 2\pi
Complete Set of Solutions:
x \in { \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} }
Importance:
Being able to identify the major ones, especially the double angle formulas and the famous \sin^2x + \cos^2x = 1 is very important.
Example: Prove \cos 2x + 2\sin^2 x = 1
Start with the more difficult side:
\cos 2x + 2\sin^2 x = (\cos^2 x - \sin^2 x) + 2\sin^2 x
= \cos^2 x + (-\sin^2 x + 2\sin^2 x)
= \cos^2 x + \sin^2 x = 1
Example: Prove (\sec x)(\cot x)(\csc x) = \csc^2 x
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