A brief note on Trigonometric Functions and Their Graphs . Read it before a test. Try each question at the end before you open its answer.
The unit circle Turn the point A ( 1 , 0 ) A(1, 0) A ( 1 , 0 ) through an angle x x x on the circle of radius 1 1 1 . The new point is P = ( cos x , sin x ) P = (\cos x, \sin x) P = ( cos x , sin x ) . This defines sine and cosine for every real x x x . Both lie between − 1 -1 − 1 and 1 1 1 . sin 2 x + cos 2 x = 1 \sin^2 x + \cos^2 x = 1 sin 2 x + cos 2 x = 1 , 1 + tan 2 x = sec 2 x 1 + \tan^2 x = \sec^2 x 1 + tan 2 x = sec 2 x and 1 + cot 2 x = cosec 2 x 1 + \cot^2 x = \operatorname{cosec}^2 x 1 + cot 2 x = cosec 2 x . tan x = sin x cos x \tan x = \dfrac{\sin x}{\cos x} tan x = cos x sin x and cot x = cos x sin x \cot x = \dfrac{\cos x}{\sin x} cot x = sin x cos x . sec x = 1 cos x \sec x = \dfrac{1}{\cos x} sec x = cos x 1 and cosec x = 1 sin x \operatorname{cosec} x = \dfrac{1}{\sin x} cosec x = sin x 1 . Signs and values Positive in quadrants I, II, III and IV: all; sin and cosec; tan and cot; cos and sec. To find a value, take away whole turns of 2 π 2\pi 2 π , find the acute angle with the x-axis, and give the sign of the quadrant. cos ( − x ) = cos x \cos(-x) = \cos x cos ( − x ) = cos x and sin ( − x ) = − sin x \sin(-x) = -\sin x sin ( − x ) = − sin x . x x x sin x \sin x sin x cos x \cos x cos x tan x \tan x tan x 0 0 0 0 0 0 1 1 1 0 0 0 π 6 \tfrac{\pi}{6} 6 π 1 2 \tfrac{1}{2} 2 1 3 2 \tfrac{\sqrt{3}}{2} 2 3 1 3 \tfrac{1}{\sqrt{3}} 3 1 π 4 \tfrac{\pi}{4} 4 π 1 2 \tfrac{1}{\sqrt{2}} 2 1 1 2 \tfrac{1}{\sqrt{2}} 2 1 1 1 1 π 3 \tfrac{\pi}{3} 3 π 3 2 \tfrac{\sqrt{3}}{2} 2 3 1 2 \tfrac{1}{2} 2 1 3 \sqrt{3} 3 π 2 \tfrac{\pi}{2} 2 π 1 1 1 0 0 0 not defined
Periods, domains and ranges Sin, cos, sec and cosec have period 2 π 2\pi 2 π . Tan and cot have period π \pi π . Tan and sec are not defined at the odd multiples of π 2 \dfrac{\pi}{2} 2 π . Cot and cosec are not defined at the multiples of π \pi π . Sin and cos take the values in [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] . Tan and cot take every real value. Sec and cosec take the values in ( − ∞ , − 1 ] ∪ [ 1 , ∞ ) (-\infty, -1] \cup [1, \infty) ( − ∞ , − 1 ] ∪ [ 1 , ∞ ) . Graphs y = sin x y = \sin x y = sin x is a wave through the origin. y = cos x y = \cos x y = cos x is the same wave moved π 2 \dfrac{\pi}{2} 2 π to the left. y = tan x y = \tan x y = tan x rises between asymptotes at the odd multiples of π 2 \dfrac{\pi}{2} 2 π . For y = a sin ( b x + c ) + d y = a\sin(bx + c) + d y = a sin ( b x + c ) + d with b > 0 b > 0 b > 0 : the amplitude is ∣ a ∣ |a| ∣ a ∣ ; the period is 2 π b \dfrac{2\pi}{b} b 2 π ; the shift is − c b -\dfrac{c}{b} − b c sideways; the range is [ d − ∣ a ∣ , d + ∣ a ∣ ] [d - |a|, d + |a|] [ d − ∣ a ∣ , d + ∣ a ∣ ] . A worked example Find sec 23 π 6 \sec \dfrac{23\pi}{6} sec 6 23 π .
Step 1. Whole turns. 23 π 6 = 2 π + 11 π 6 \dfrac{23\pi}{6} = 2\pi + \dfrac{11\pi}{6} 6 23 π = 2 π + 6 11 π .
Step 2. The quadrant. 11 π 6 \dfrac{11\pi}{6} 6 11 π is in quadrant IV, where cos and sec are positive. The acute angle is π 6 \dfrac{\pi}{6} 6 π .
Step 3. The value. cos 11 π 6 = 3 2 \cos \dfrac{11\pi}{6} = \dfrac{\sqrt{3}}{2} cos 6 11 π = 2 3 , so sec 23 π 6 = 2 3 \sec \dfrac{23\pi}{6} = \dfrac{2}{\sqrt{3}} sec 6 23 π = 3 2 .
Practice with new numbers
A program to try The program reads three waves from their numbers and prints the amplitude, the period and the range of each.
Run it. Check each line against the rules above.
1. Add a wave of your own, such as "y = 3 sin(2x - 1) + 2": (3, 2, -1, 2). What are its amplitude and its range?
Mistakes to avoid Reading sin 2 x \sin^2 x sin 2 x as sin ( x 2 ) \sin(x^2) sin ( x 2 ) . It is ( sin x ) 2 (\sin x)^2 ( sin x ) 2 . Skipping the sign. The size comes from the acute angle, and the sign from the quadrant. The period of sin b x \sin bx sin b x as 2 π b 2\pi b 2 π b . It is 2 π b \dfrac{2\pi}{b} b 2 π . Dividing by zero. tan π 2 \tan \dfrac{\pi}{2} tan 2 π and cosec π \operatorname{cosec} \pi cosec π are not defined. Degrees in Python. math.sin(30) is the sine of 30 radians. Write math.sin(math.radians(30)) for 30 ∘ 30^\circ 3 0 ∘ . Check yourself In which quadrant is x x x when cos x < 0 \cos x < 0 cos x < 0 and cot x < 0 \cot x < 0 cot x < 0 ? Stuck? Show a hint Show the answerFind sin ( − 7 π 3 ) \sin\left(-\dfrac{7\pi}{3}\right) sin ( − 3 7 π ) . Stuck? Show a hint Show the answerWhat are the amplitude and the period of y = − 3 cos x 4 y = -3\cos \dfrac{x}{4} y = − 3 cos 4 x ? Stuck? Show a hint Show the answerWhy does sec x \sec x sec x never take a value between − 1 -1 − 1 and 1 1 1 ? Stuck? Show a hint Show the answerWhich is larger, sin 2 \sin 2 sin 2 or sin 3 \sin 3 sin 3 ? The angles are in radians. Stuck? Show a hint Show the answer