Radians
Graphics
Transformations of
Trigonometric Functions
Equivalent Trigonometric Functions
Compound & Double Angle Formulas
100
Convert angle from degrees to radians, in exact form. 200° =
200° = 180°x, x = 10/9, 200° = 10π/9
100
What are the graphs of secx, cscx and cotx? (0<x<2π)
secx-Graph b, cscx-Graph c, cotx-Graph d
100
What is the parent function y=sinx point. (0≤x≤2π)
(0,0) (π/2,-1) (2π,0) (3π/2,-1) (2π,0)
100
Write an expression that is equivalent to each of the following expressions, using the related acute angle. tan5π/3 =
tan(π+θ) = tanθ, tan(5π/3) = tan(π+2π/3)
100
2×sinπ/12×cosπ/12
=sin2(π/12) =sin(π/6) =sin30° =1/2
200
How many degree is 1 radian?
2π radians =360°, π radians ＝180°, π radians/π = 180°/π, 1 radian= 180°/π =57.3°
200
Stretch graph for 0
a=-2=max-min/2, k=2=2π/period, period=π d=-π/4, c=2=max+min/2 2max=0, max=0, min=4
200
State the period, amplitude, horizontal translation, and equation of the axis for this trigonometric function. y=1/3sin(2(x-π/4))-2
a=1/3 k=2 d=π/4 c=-2 Period= π = 180, Amplitude= a=1/3, Horizontal translation: Horizontal shift right π/4 units Equation of the axis= -2
200
Use the cofunction identities to write an expression that is equivalent to the following expression. cos(5π/16)=
cosθ=sin(π/2-θ), cos(5π/16)= sin(π/2-3π/16)
200
cos⁡(15°)＝
cos⁡(15°) = (cos⁡45°-30°) = cos⁡(45°-30°) = cos⁡(45°) (cos30°)+sin⁡(45°) sin⁡(30°) = (1/√2)(√3/2)+(1/√2)(1/2) = (√3+1)/(2√2)
300
State an equivalent expression in terms of the related acute angle. 1.csc⁡(-π/3), 2.sec 7π/4
1.csc⁡(-π/3)=csc⁡(4π/3), 2.sec 7π/4=sec π/4
300
Use equation (x/k+d, ay+c) to stretch graph for y=3cos(2x+π). Graph e
y=3cos(2x+π) change to y=3cos(2(x+π/2)), a=3,k=2,c=-π/2,d=0, in the parent function, x=0, y=1, x=π/2, y=0, x=π, y=-1, x=3π/2, y=0, x=2π, y=1. in y=3cos(2x+π), (x/2, 3y-π/2) x=0, y=(6-π)/2 x=π/4, y=-π/2 x=π/2, y=(-6-π)/2 x=3π/4, y=-π/2 x=π, y=5π/2
300
How to change the the sin graph to the cos graph.
sin(π/2-a)=sin(π/2)cosa-cos(π/2)sina=cosa
300
State whether the following are true or false. For those that is false, justify your decision. cot(π/2+θ) = tanθ
if θ=π/3, Ls=cot(π/2+θ) =cot(π/2+π/3) =cot(5π/6) =1/-tan(5π/6), Rs=tan(π/3)≠1/-tan(5π/6)≠Ls
300
Simplify each expression， sin⁡(π+x)+sin⁡(π-x)＝
sin⁡(π+x)+sin⁡(π-x) ＝sin(π)cosx+cos(π)sinx + sin(π)cosx-cos(π)sinx =2sin(π)cosx =sin2x
400
Show that(2sin)^2 θ-1=(sin)^2 θ-(cos)^2 θ of 11π/6
2(sin^2 (11π/6))-1=2(〖-1/2)〗^2-1, (sin^2 11π/6)-(cos^2 11π/6) =(〖-1/2)〗^2-(〖√3/2)〗^2 =1/4-3/4 =1/2, 2(sin^2 (11π/6))-1 =(sin^2 11π/6)-(cos^2 11π/6)
400
Accounting to the graph and write the function
y=7 sin (5(x-π/4)-3
400
The graph is trigonometric function( y=sinx) has a period of π and an amplitude of 4. And then, this function is reflection in x-axis and horizontal translation is shift right π/6 units.
a=-4 k=2 d=π/6 c= 7 The equation is y=-4sin(2(x-π/6))+7
400
Write an equation that is equivalent to y=-5sin(x-π/2)-8, using the cosine function.
y=-5sin(x-π/2)-8 =-5sin(-(π/2-x))-8 =-5(-cosx)-8 =5cosx-8
400
Which of the following identities could you use to help you prove that 2tanx/(1+tan^2(x))=sin2x? a) 1+tan^2(x)=sec^2(x), b) sin2x=2sincosx, c) tanx=sinx/cosx, d) all of these.
Ls=2tanx/(1+tan^2(x)) =(2sinx/cosx)/(1+sin^2(x)/cos^2(x)) =(2sinx/cosx)/((sin^2(x)+cos^2(x))/cos^2/(x)) =(2sinx/cosx)/(1/cos^2(x)) =(2sinx/cosx)*cos^2(x) =2sinxcosx=sin2x=Rs, b)sin2x=2sinxcosx
500
(2√3-3)/4
The terminal arm of θ is in the fourth quadrant. If cotθ=-√3, the calculate sinθcotθ-cos^2 θ.
500
Describe all transformation (Very important).....
a is reflection and vertical stretch or compression k is reflection and vertical stretch or compression D is the horizontal translation C is the vertical translation
500
Describe this function and sketch the graph. y=-1/2sin(3(x-π/3))+4
a=-1/2 reflection in X-axis and horizontal stretch 1/2 k=3 vertical compression 1/3 d=π/3 vertical shift right π/3 units c=4 horizontal shift up 4 unit
500
y=-6cos(x+ π/2)+4 =-6(cosxcos(π/2)-sinxsin(π/2))+4 =-6(-sinx*1)+4 =6sin+4
Use the sine function to write an equation that is equivalent to y=-6cos(x+ π/2)+4.
500
Prove cosC-cosD=-sin((C+D)/2)sin((C-D)/2).
Let C=x+y and let D=x-y, cosC-cosD =cos(x+y)-cos(x-y) =cosxcosy-sinxsiny-(cosxcosy-sinxsiny) =-2sinxsiny (C+D)/2=(x+y+x-y)/2=x, (C-D)/2=(x+y-x+y)/2=y. So cosC-cosD=-2sin((C+D)/2)sin((C-D)/2)
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