Test 3: Sections 7.3-7.6 and 4.1
Simplifying Expressions using Double-Angle Identities
1. (cos^2)7x-(sin^2)7x = identity: cos2A=(cos^2)A-(sin^2)A substitute "7x" = (cos^2)7x - (sin^2)7x = cos2(7x) = cos (14x)
How to Solve a Trigonometric Equation
1. Decide whether equation is linear or quadratic in form so you can determine the solution method. 2. If only one trigonometric function is present, solve the equation for that function. 3. If more than one is present, rearrange equation so that one side equals 0. Then try to factor and set each factor equal to 0. 4. If equation is quadratic in form, but not factorable, use the quadratic formula. Check that solutions are in the desired interval. 5. Try using identities to change form of equation.
Multiple Angle Identities
1. Write sin3x in terms of sinx. sin3x = sin(2x+x) =sin2x(cosx)+cos2x(sinx) (2sinxcosx)cosx+(cos^2x-sin^2x)sinx =2sinxcos^x+cos^2xsinx-sin^3x =2sinx(1-sin^2x)+(1-sin^2x)sinx-sin^3x =2sinx-2sin^3x+sinx-sin^3x-sin^3x =3sinx-4sin^3x
Verifying Identities
1. sin(π/6+θ) + cos(π/3+θ) = cosθ = start on left side. [sin(π/6)cosθ+cos(π/6)sinθ)] + [cos(π/3)cosθ -sin(π/3)sinθ)] = 1/2cosθ + 1/2cosθ = cosθ
How to find cos87ºcos93º-sin87ºsin93º
=cos(87º+93º) =cos(180º) = -1
One-to-One Function
Each range element is only used once. Ex. f(x)=x^3 is one-to-one. f(2) is not equal to f(-2). g(x)=x^2 is not one-to one. g(2) is equal to g(-2).
Inverse Function
F^-1. Change x and y. Solve for y. Then replace y with f^-1. >Domain of f is the range of f^-1. >Range of f is the domain of f^-1. >If point (a,b,) is on graph of f, point (b,a) is on the graph of f^-1.
How to find cos15º
cos(45º-30º) =cos45cos30+sin45sin30 =square root of 2/2 x square root of 3/2 + square root of 2/2 x 1/2 =square root of 6 plus square root of 2 over 4.
Cofunction Identities
cos(90-θ) = sinθ sin(90-θ) = cosθ tan(90-θ) = cotθ cot(90-θ) = tanθ sec(90-θ) = cscθ csc(90-θ) = secθ
Cosine sum and difference Identities
cos(a+b)= cosAcosB-sinAsinB cos(a-b)= cosAcosB+sinAsinB
How to find cos5π/12
cos(π/6+π/4) which is also (2π/12+3π/12) =cos(π/6)cos(π/4)-sin(π/6)sin(π/4) =square root of 3 over two x square root of 2 over 2 minus 1/2 x square root of 2 over 2. =square root of 6 minus square root of 2 over 4.
Double-Angle Identities
cos2A=cos^2A-sin^2A cos2A=2(cos^2)A-1 cos2A=1-2(sin^2)A sin2A=2sinAcosA tan2A=(2tanA)/(1-tan^2A)
Product-to-Sum Identities
cosAcosB = 1/2[cos(A+B) + cos(A-B)] sinAsinB = 1/2[cos(A-B) - cos(A+B)] sinAcosB = 1/2[sin(A+B) + sin(A-B)] cosAsinB = 1/2[sin(A+B) - sin (A-B)]
Using cofunction Identities
ex. Find value of θ or x that satisfies each of the following: 1. cotθ=tan25º tan(90-θ) = tan25º 90-θ = 25º θ = 65º 2. sinθ=cos(-30º) cos(90º-θ) = -30º θ=120º 3. csc(3π/4) = sec x csc(3π/4) = csc(π/2 - x) **replace 90º with π/2 3π/4 = π/2 - x x= - π/4
Half-Angle Identities
sin x/2=±sqrt[(1-cos x)/2] cos x/2=±sqrt[(1+cos x)/2] tan x/2=±sqrt[(1-cos x)/(1+cos x)] tan x/2= [(sinA)/(1+cosA)] tan x/2= [(1-cosA)/(sinA)]
Sin of a Sum or Difference
sin(A+B) = sinAcosB + cosAsinB sin(A-B) = sinAcosB - cosAsinB
Sum-to-Product Identities
sinA + sinB = 2sin(A+B/2)cos(A-B/2) sinA-sinB = 2cos(A+B/2)sin(A-B/2) cosA+cosB = 2cos(A+B/2)cos(A-B/2) cosA-cosB = -2sin(A+B/2)sin(A-B/2)
Tangent of a Sum or Difference
tan(A+B) = tanA+tanB/1-tanAtanB tan(A-B) = tanA-tanB/1+tanAtanB
Inverse Sin Function
y=Sin^-1x or y=arcsinx or x=sin y "Y is the number/angle in the interval [-pi/2, pi/2] whose sin is x." Quadrants I and IV.
Inverse Cosine Function
y=cos^-1x or y=arccos x or x=cos y "Y is the number/angle in the interval [0, pi] whose cosine is x." Quadrants I and II.
Inverse Cotangent Function
y=cot^-1 or y=arccotx or x=cot y _____________ 0<y<pi. Quadrants I and II.
Inverse Cosecant Function
y=csc^-1x or y=arccscx or x=csc y. ______________ -pi/2</=y</=pi/2. Not equal to 0. Quadrants I and IV.
Inverse Secant Function
y=sec^-1x or y=arcsecx or x=sec y _____________ 0</=y</=pi. Not equal to pi/2. Quadrants I and II.
Inverse Tangent Function
y=tan^-1x or y=arctanx or x=tan y "Y is the number/angle in the interval [-pi/2, pi/2] whose tangent is x." Quadrants I and IV.