Trigonometry Review Find sin(  /4) = cos(  /4) = tan(  /4) = Find sin(  /4) = cos(  /4) = tan(  /4) = csc(  /4) = sec(  /4) = cot(  /4) = csc(  /4) = sec(  /4) = cot(  /4) =

Evaluate tan(  /4) A. Root 2 B. 2 C. Root 2 /2 D. 2 / Root 2 E. 1

Trigonometry Review sin(2  /3) = cos(2  /3) = tan(2  /3) = sin(2  /3) = cos(2  /3) = tan(2  /3) = csc(2  /3) = sec(2  /3) = cot(2  /3) = csc(2  /3) = sec(2  /3) = cot(2  /3) =

Evaluate sec(2  /3) A. -1 B. -2 C. -3 D. Root(3) E. 2 / Root(3)

Evaluate cos(  /2) A. -1 B C. 1 D. 0.0

Evaluate sin(  /3) A B. 0.5 C D

If y = sec(  ), find y if  =

Squeeze Theorem If f(x) g(x) h(x) on an open interval containing a, and then

Rule 4 is a Theorem

Theorem -> Sector Area = x/2 Theorem -> Sector Area = x/2

Proof.. ½ sin(x)cos(x) ½ x ½ sin(x)/cos(x) ½ sin(x)cos(x) ½ x ½ sin(x)/cos(x) cos(x) x/sin(x) 1/cos(x) cos(x) x/sin(x) 1/cos(x) 1 1 therefore 1 therefore 1

sin(.1)= sin(.01)= sin(.001= sin(.0001)= sin( )=

Rule 5 is a Theorem = 0 Proof cos(A+B)=cos(A)cos(B)-sin(A)sin(B) If A = B = x/2 cos(x)= 1 -

2sin 2 (x/2)= 1-cos(x)

= 0

Equation of Lines Write the equation of a line that passes through (-3, 1) with a slope of – ½.

Equation of Lines Write the equation of a line that passes through (-3, 1) with a slope of – ½.

Equation of Lines Write the equation of a line that passes through (-3, 1) with a slope of – ½.

Equation of Lines Write the equation of a line that passes through (-3, 1) with a slope of – ½.

Passes through (0,1) with a slope of -3. What is the missing blue number?

Write the equation of the line tangent to y = x + sin(x) when x = 0 given the slope there is 2. A. y = 2x + 1 B. y = 2x C. y = 2x

Write the equation of the line tangent to y = x + sin(x) when x = 0 given the slope there is 2. A. y = 2x + 1 B. y = 2x C. y = 2x