Download MAT 155P Chapter 4 Formulas

MAT 155P - PRECALCULUS MATHEMATICS PLUS
Chapter 4 Formulas
Radian Measure
 
Arc Length
s  r
s
r
Even Identities
Odd Identities
sec   t   sec t 
csc   t    csc t 
sin   t    sin t 
cos   t   cos t 
tan   t    tan t 
cot   t    cot t 
Pythagorean Identities
tan2  t   1  sec2  t 
1  cot2  t   csc2  t 
Cofunction Identities

tan    cot  90
sec    csc  90



sin    cos 90o  
o
o

cot    tan  90
csc    sec  90



cos    sin 90o  
o
*If  is in radians, replace 90o with
o

2
How to Find the Coterminal Angles of a Given Angle
1. If   360 (2 ) , subtract appropriate multiples of 360 (2 ) to obtain  .
2. If   0 , add appropriate multiples of 360 (2 ) to obtain  .
How to Find Reference Angles for Angles Less Than 360 o (2π)
Quadrant
Angle 
Reference angle α
I
0˚ <  < 90 
α= 
II
90˚ <  < 180 ˚
α = 180 ˚ – 
III
180˚ <  < 270 ˚
α =  – 180˚
IV
270˚ <  < 360 ˚
α = 360 ˚ – 
If  is measure in radians and 0    2 , replace 180 with π in these calculations.
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How to Find Reference Angles for Angles Greater Than 360 o ( 2 )
1. Subtract appropriate multiples of 360 , until you obtain an angle between 0 and 360 , from .
2. Find the reference angle for the result from step 1.
How to Find Reference Angles for Angles Less Than -360o (- 2 )
1. Add appropriate multiples of 360o until you obtain an angle between 0 and 360 , to  .
2. Find the reference angle for the result from step 1.
Steps for Evaluating Trigonometric Functions for Any Angle
1. If   360 or   0 , find a coterminal angle between 0 and 360 . Otherwise go to Step 2.
2. Find reference angle.
3. Find trigonometric function value for the reference angle.
4. Determine the sign of the trigonometric function based on the quadrant in which  lies to prefix the
appropriate sign of the function value in Step 3.
Finding Exact Values
   
,
1. For   sin 1 x find the value of  in 
that satisfies sin   x .
 2 2 
2. For   cos 1 x find the value of  in  0,   that satisfies cos  x .
  
,  that satisfies tan   x .
 2 2
3. For   tan 1 x find the value of  in  
Cancellation Equations for Inverse Sine, Cosine, and Tangent Functions
f 1 ( f ( x))  sin 1 (sin x)  x
where


x
2
2
f ( f 1 ( x))  sin(sin 1 x)  x where -1  x  1
f 1 ( f ( x))  cos 1 (cos x)  x
where 0  x  
f ( f 1 ( x))  cos(cos 1 x)  x
where -1  x  1
f 1 ( f ( x))  tan 1 (tan x)  x
where
f ( f 1 ( x))  tan(tan 1 x)  x
where   x  


x
2
2
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