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

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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

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)

Trig. Derivatives sin’(x) = cos(x) cos’(x) = - sin(x)

Trig. Derivatives sin’(x) = cos(x) sin’(x) =

sin’(x) =. sin’(x) =

Rule 4 says. A. 0 B. 0.5 C. 1 D. 1.5

Rule 5 says. A. 0 B. 0.5 C. 1 D. 1.5

sin’(x) =. sin’(x) =

Trig. Derivatives sin’(x) = cos(x) cos’(x) = - sin(x)

If y = sin(x) + 2x 2, find dy/dx A. - cos(x) + 4x B. cos(x) + 4 C. cos(x) + 4x

cos’(x) = - sin(x) cos’(x) = =

cos’(x) = - sin(x) cos’(x) = =

Trig. Derivatives sin’(x) = cos(x) cos’(x) = - sin(x) sin’(x) = cos(x) cos’(x) = - sin(x) Slope of the sin graph when x = 0 Slope of the sin graph when x = 0 A) sin’(0) = cos(0) = A) sin’(0) = cos(0) =

Trig. Derivatives sin’(x) = cos(x) cos’(x) = - sin(x) sin’(x) = cos(x) cos’(x) = - sin(x) Slope of the sin graph when x = 0 Slope of the sin graph when x = 0 A) sin’(0) = cos(0) = 1 A) sin’(0) = cos(0) = 1

Trig. Derivatives sin’(x) = cos(x) cos’(x) = - sin(x) sin’(x) = cos(x) cos’(x) = - sin(x) A) sin’(0) = cos(0) = 1 A) sin’(0) = cos(0) = 1 B) sin’(  /4) = cos(  /4) = B) sin’(  /4) = cos(  /4) =

Trig. Derivatives sin’(x) = cos(x) cos’(x) = - sin(x) sin’(x) = cos(x) cos’(x) = - sin(x) A) sin’(0) = cos(0) = 1 A) sin’(0) = cos(0) = 1 B) sin’(  /4) = cos(  /4) = B) sin’(  /4) = cos(  /4) = 0.707

Trig. Derivatives sin’(x) = cos(x) cos’(x) = - sin(x) sin’(x) = cos(x) cos’(x) = - sin(x) A) sin’(0) = cos(0) = 1 A) sin’(0) = cos(0) = 1 B) sin’(  /4) = cos(  /4) = B) sin’(  /4) = cos(  /4) = C) sin’(-  /3) = cos(-  /3) = C) sin’(-  /3) = cos(-  /3) =

Trig. Derivatives sin’(x) = cos(x) cos’(x) = - sin(x) sin’(x) = cos(x) cos’(x) = - sin(x) A) sin’(0) = cos(0) = 1 A) sin’(0) = cos(0) = 1 B) sin’(  /4) = cos(  /4) = B) sin’(  /4) = cos(  /4) = C) sin’(-  /3) = cos(-  /3) = 0.5 C) sin’(-  /3) = cos(-  /3) = 0.5

x= 0, 2  /3, - 3  /4 cos’(x) = - sin(x) cos’(x) = - sin(x) A) cos’(0) = - sin (0) = 0 A) cos’(0) = - sin (0) = 0 B) cos’(-3  /4) = - sin(5  /4) = B) cos’(-3  /4) = - sin(5  /4) = C) cos’(2  /3) = - sin(2  /3) = C) cos’(2  /3) = - sin(2  /3) =

Evaluate cos’(  /2) A. -1 B C. 1 D

Evaluate sin’(  /3) A B. 0.5 C D

Trig. Derivatives sin’(x) = cos(x) cos’(x) = - sin(x) sin’(x) = cos(x) cos’(x) = - sin(x) tan’(x) = sec 2 (x) cot’(x) = - csc 2 (x) tan’(x) = sec 2 (x) cot’(x) = - csc 2 (x) sec’(x) = sec(x)tan(x) csc’(x) = -csc(x)cot(x) sec’(x) = sec(x)tan(x) csc’(x) = -csc(x)cot(x)

Trig. Derivatives Theorem tan’(x) = sec 2 (x) Theorem tan’(x) = sec 2 (x) Proof : tan’(x) = [sin(x)/cos(x)]’ Proof : tan’(x) = [sin(x)/cos(x)]’

Trig. Derivatives Theorem tan’(x) = sec 2 (x) Theorem tan’(x) = sec 2 (x) tan’(  /4) = tan’(  /4) =

Trig. Derivatives Theorem tan’(x) = sec 2 (x) Theorem tan’(x) = sec 2 (x) tan’(  /4) = sec 2 (  /4) = 2 while tan(  /4) = tan’(  /4) = sec 2 (  /4) = 2 while tan(  /4) = 1

Trig. Derivatives Theorem cot’(x) = - csc 2 (x) Theorem cot’(x) = - csc 2 (x) Proof : cot’(x) = [cos(x)/sin(x)]’ Proof : cot’(x) = [cos(x)/sin(x)]’

Trig. Derivatives Theorem sec’(x) = sec(x)tan(x) Theorem sec’(x) = sec(x)tan(x) Proof : sec’(x) = [1/cos(x)]’ Proof : sec’(x) = [1/cos(x)]’

Trig. Derivatives Theorem csc’(x) = - csc(x)cot(x) Theorem csc’(x) = - csc(x)cot(x) Proof : csc’(x) = [1/sin(x)]’ Proof : csc’(x) = [1/sin(x)]’

Trig. Derivatives sin’(x) = cos(x) cos’(x) = - sin(x) sin’(x) = cos(x) cos’(x) = - sin(x) tan’(x) = sec 2 (x) cot’(x) = - csc 2 (x) tan’(x) = sec 2 (x) cot’(x) = - csc 2 (x) sec’(x) = sec(x)tan(x) csc’(x) = - csc(x)cot(x) sec’(x) = sec(x)tan(x) csc’(x) = - csc(x)cot(x)

If y = tan(x) sec(x) find the velocity and y’(  /3) sec’(x) = sec(x)tan(x) tan’(x) = sec 2 (x) sec’(x) = sec(x)tan(x) tan’(x) = sec 2 (x) y ’ = tan(x)sec(x)tan(x) + sec(x)sec 2 (x) y ’ = tan(x)sec(x)tan(x) + sec(x)sec 2 (x) y’=sec(x)[sec 2 (x)-1] + sec 3 (x)=2sec 3 (x)-sec(x) y’=sec(x)[sec 2 (x)-1] + sec 3 (x)=2sec 3 (x)-sec(x) y’(  /3) = 2sec 3 (  /3)-sec(  /3) = y’(  /3) = 2sec 3 (  /3)-sec(  /3) = sin 2 x+cos 2 x=1 dividing by cos 2( x) sin 2 x+cos 2 x=1 dividing by cos 2( x) tan 2 (x)+1=sec 2 (x) tan 2 (x)+1=sec 2 (x)

If y = tan(x) cos(x) find the acceleration and y’’(  /3) y’ = cos(x) y’ = cos(x) y’’ = -sin(x) y’’(  /3)= y’’ = -sin(x) y’’(  /3)=

If y = tan(x) + cos(x) find the initial acceleration, y’’(0) tan’(x) = sec 2 (x) sec’(x) = sec(x)tan(x) tan’(x) = sec 2 (x) sec’(x) = sec(x)tan(x) y’ = sec(x)sec(x) - sin(x) y’’ = sec(x) sec(x)tan(x) + sec(x) sec(x)tan(x) - cos(x) = 2 sec 2 (x) tan(x) – cos(x) y’’(0) = 2 * 1 *

y” = 2 sec 2 (x) tan(x) – cos(x) y”(0) = -1

If y = sec(x), find the acceleration, y’’(0) using the product rule on sec’(x).

Find the slope of the tangent line to y = x + sin(x) when x = 0

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

Displacement x(t) = 2 cos(t) find x(0). A. x(0) = 2 cos(t) B. x(0) = -2 C. x(0) = 2

Displacement x(t) = 2 cos(t) find x(0). A. x(0) = 2 cos(t) B. x(0) = -2 C. x(0) = 2

Displacement x(t) = 2 cos(t) find x(  /2). A. x(  /2) = 1 B. x(  /2) = 0 C. x(  /2) =

Displacement x(t) = 2 cos(t) find x(  /2). A. x(  /2) = 1 B. x(  /2) = 0 C. x(  /2) =

Displacement x(t) = 2 cos(t) find x’’(t) = a(t) A. a(t) = 2 sin(t) B. a(t) = 2 cos(t) C. a(t) = -2 cos(t)

If f(x) = x n then f ' (x) = n x (n-1) Proof, when n is a natural number. n = 1 Proof : Lim [f(x+h)-f(x)]/h = Lim (x + h - x)/h = Lim h/h = 1 What is the derivative of x grandson? One grandpa, one.

If f(x) = x n then f ' (x) = n x (n-1) Assume true when n = k Assume that k is fixed and k >= 1 and f(x) = x k and that f '(x) = k x (k-1) f(x) = x k and that f '(x) = k x (k-1)

If f(x) = x n then f ' (x) = n x (n-1) Prove true when n = k + 1 Proof : Suppose f(x) = x k+1 = x x k f ’(x) = x [x k ]’ + [x k ] 1 = x [kx k-1 ] + x k = x [kx k-1 ] + x k = k x k + x k = (k+1) x k = k x k + x k = (k+1) x k