Math 130 Final Review

Find all solutions of the equation: (a) sqrt(3) tan x-1=0 (b) csc^2x-2=0 (c) 4sin^2x-4sinx+1=0 (d) cos3x=-sin3x

(a) Add 1 to both sides Divide sqrt 3 from both sides tanx=1/sqrt 3 Multiply sqrt 3 to top & bottom to rationalize x=npi+pi/6 (b) Add 2 to both sides sqrt both sides cscx=+-sqrt(2) >>> sinx=+- sqrt(2/2) sinx=sqrt(2/2) x=2npi+pi/4 & x=2npi+3pi/4 sinx=-sqrt(2/2) x=2npi+5pi/4 & x=2npi+7pi/4 (c) (2sinx-1)^2=0 2sinx-1=0 Add 1 to both sides Divide both sides by 2 sinx= 1/2 x=2npi+pi/6 & 2npi+5pi/6 (d) Subtract -cos3x from both sides -1=tan3x 3x=npi-pi/4 x=1/3npi-pi/12

Find all solutions of the equation in the interval [0, 2pi): (a) cotx-3tanx=0 (b) 2cos^2x-sinx=1

(a) (1/tanx)-3tanx=0 1-3tan^2x=0 -3tan^2x=-1 tan^2x=1/3 tanx=+-sqrt(3)/3 tanx=sqrt(3)/3 x=pi/6, 7pi/6 tanx=-sqrt(3)/3 x=5pi/6, 11pi/6 (b) 2(1-sin^2x)-sinx-1=0 2-2sin^2x-sinx-1=0 -2sin^2x-sinx+1=0 2sin^2x+sinx-1=0 (2sinx-1)(sinx+1)=0 2sinx-1=0 sinx=1/2 x=pi/6, 5pi/6 sinx+1=0 sinx=-1 x=3pi/2

Write the product as a sum: sin5xcosx

1/2(sin(5x+x)+sin(5x-x)) 1/2(sin6x+sin4x)

Write the sum as a product: sin7x+sin3x

2sin((7x+3x)/2)cos((7x-3x)/2) 2sin(10x/2)cos(4x/2) 2sin5xcos2x

Find the exact value of the expression, if it is defined. (a) sin^-1 (-sqrt 2/2) (b) cos^-1 (sqrt 3/2) (c) tan^-1 (-sqrt 3) (d) tan (cos^-1(sqrt3/2)) (e) cos (sin^-1(-3/5))

Memorization! (a) -pi/4 (b) pi/6 (c) -pi/3 (d) tan(pi/6)= sqrt 3/3 (e) cos t=sqrt(1-sin^2 t) =sqrt(1-(-3/5)^2) =sqrt(1-(9/25)) =sqrt(16/25) =4/5

Verify the identity: (a) (1-cosb)(1+cosb)=1/csc^2b (b) 2cos^2x-1=1-2sin^2x (c) tan(x-pi/4)=tanx-1/tanx+1 (d) cos(x+y)+cos(x-y)=2cosxcosy

1+cosb-cosb-cos^2b 1-cos^2b=sin^2b (a) sin^2b=1/csc^2b (b) 2cos^2x-1=1-2sin^2x (It's an identity) (tanx-tanpi/4)/(1+tanxtanpi/4) tanx-1/1+tanx (c) tanx-1/tanx+1 (d) cosxcosy-sinxsiny+cosxcosy+sinxsiny 2cosxcosy

Find: (i) the reference number of t; (ii) the terminal point P(x,y) on the unit circle determined by t; (iii) sin t & cos t (a) t= 7pi/4 (b) t=-2pi/3

Draw the unit circle, just the 1st Quadrant! (a) 2pi+pi/4 Q4 (i) t=pi/4 (ii) P(sqrt 2/2, -sqrt 2/2) (iii) sin t=-sqrt 2/2 cos t= sqrt 2/2 (b) pi-pi/3 Q3 (i) t=pi/3 (ii) P(-1/2, -sqrt 3/2) (iii) sin t=-sqrt 3/2 cos t=-1/2

The graph of one complete period of a sine or cosine curve is given. (i) Find the amplitude, period, & the horizontal shift. (ii) Write the equation that represents the curve in the form y=asink(x-b) or y=acosk(x-b).

Equation: -2sinpi(x-1) Amplitude: 2 Period: 2pi/pi=2 Shift: Right 1 unit

Use the laws of Logarithms to find the exact value of the expression: log100 - log18-log50 all w/ base 3

Factor out the (-) between 18 & 50: log100-(log18+log50) Multiply 18 & 50: log900 Condense: log(100/900) Reduce: log(1/9) Convert to exponent: log3^-2 w/ base 3 Use loga^x w/ base a= x -2

The turkey population in a certain region has a relative growth rate of 5% per year. It is estimated that the population in 2000 was 1,200. (a) Find a function n(t)= n0e^rt that models the population t years after 2000. (b) Use the function from part (a) to estimate the turkey population in the year 2015. (c) After how many years will the turkey population reach 1,900?

Plug in 1,200 for n & 5% for r: (a) n(t)=1,200e^0.05t Plug in 15 for t: n(t)=1,200e^(0.05)(15) (b) 2,540 Set equation (a) equal to 1,900 Divide by 1,200 Apply ln of both sides Divide by 0.05 t=9.2 years

Find the exponential function f(x)=a^x whose graph is given. Point: (-2,16)

Rewrite as y=a^2 Plug values in: 16=a^-2 Also written as 16=1/a^2 Multiply both sides by a^2: 16a^2=1 Divide both sides by 16: 1/16=a^2 Take square root from both sides: a=1/4 Rewrite as f(x)=(1/4)^x

Find exact solutions of the exponential or logarithmic equation: (a) 3^2x-3^x-6=0 (b) log(x+3)+logx=1

Substitute the middle term for a letter: Let 3^x =t Rewrite w/ that letter: t^2-t-6=0 Factor: (t-3)(t+2)=0 t=3 or t=-2 Substitute middle term back in: 3^x=3 or 3^x=-2 Solve both to get x Plug both in to see which satisfies the equation 3^x=3 (a) x=1 Take away the 2nd log to make it simpler: log(x+3)x=1 The single x has log base of 10, so 10 replaces the 1 Distribute and subtract 10 from both sides: x^2+3x-10=0 Factor: (x+5)(x-2)=0 x=-5 or x=2 Plug in solutions to find which works (b) x=2

Express the equation in logarithmic form. (a) 3^2x=10 (b) 5^3=y (c) e^x+1=3

Use formula: a^y=x >>> logx w/ base a (a) log10 w/ base 3 = 2x (b) logy w/base 5 = 3 (c) ln3 = x+1

Find the values of all the trigonometric functions of t from the given information. sin t = -3/5, terminal point of t is in Q4

1) sin t= -3/5 2) Find csc t: 1/(-3/5)= (-5/3) 3) Find cos t: x^2+y^2=1 x^2+(-3/5)^2=1 x^2+9/25=1 Subtract 9/25 from both sides sqrt both sides x=4/5 4) sec t= 1/(4/5)= 5/4 5) tan t=(-3/5)/(4/5)=-3/4 6) cot t= 1/(-3/4)= -4/3

Prove the identity: (a) 2(tanx-cotx)/tan^2x-cot^2x=sin2x (b)1/1-sinx-1/1+sinx=2sectanx (c)sin3x+sin7x/cos3x-cos7x=cot2x (d)cos87+cos33=sin63 (e)(cosx+cosy)^2+(sinx-siny)^2=2+2cos(x+y) (f)sin^2(x/2)cos^2(x/2)=1/4sin^2x

2tanx-cotx/(tanx-cotx)(tanx+cotx) 2/tanxcotx=2/(sinx/cosx)+(cosx/sinx) 2sinxcosx/sin^2x+cos^2x 2sinxcosx (a) sin2x ((1+sinx)/(1-sinx)(1+sinx))-((1-sinx/(1-sinx)(1+sinx)) ((1+sinx)-(1-sinx))/((1-sinx)(1+sinx)) 2sinx/1-sin^2x=2sinx/cos^2x 2(1/cosx)*(sinx/cosx) (b) 2secxtanx (2sin(3x+7x/2)cos(3x-7x/2))/-2sin(3x+7x/2)sin(3x-7x/2) (cos(-2x)/-sin(-2x)) cos2x/sin2x (c) cot2x 2cos(87+33/2)cos(87-33/2) 2cos60cos27 2*1/2cos27 cos27=cos(90-63) (d) sin63 (cosx+cosy)(cosx+cosy)+(sinx-siny)(sinx-siny) cos^2x+2cosxcosy+cos^2y+sin^2x-2sinxsiny-sin^2y 2+2cosxcosy-2sinxsiny 2+2(cosxcosy-sinxsiny) (e) 2+2cos(x+y) (1-cosx/2)(1+cosx/2) (1-cos^2x/4) (f) 1/4sin^2x

Find the amplitude, period, & horizontal shift of the function, & graph one complete period: y=-2cos2(x-pi/4)

Amplitude: -2(absolute value)=2 Period: 2pi/k= 2pi/2= pi Shift: pi/4 to the right(subtraction) Graph: Start (pi/4, -2) Middle (3pi/4, 2) End (5pi/4, -2)

A 50-gal barrel is filled completely with pure water. Salt water with a concentration of 0.3 lb/gal is then pumped into the barrel, and the resulting mixture overflows at the same rate. The amount of salt in the barrel at time t is given by Q(t)=15(1-e^-0.04t) where t is measured in minutes and Q(t) is measured in pounds. (a) How much salt is in the barrel after 5 minutes? (b) How many minutes does it take when 10 pounds of salt remains in the barrel?

Plug 5 minutes into equation given (a) 2.72lbs Set original equation given equal to 10 minutes Divide both sides by 15 Subtract 1 from both sides Divide both sides by -1 to get rid of (-) on e Apply ln to both sides Divide both sides by -0.04 (b) t=27.5

Use law of logarithms to expand expression: log(x^2 sqrt(y(z+1)^5/z^3) all have base 3

Put log in front of everything, put 1/2 in front instead of sqrt, add/subtract: logx^2+1/2(logy+log(z+1)^5-logz^3 Put exponents in front of log: 2logx+1/2(logy+5logz+1-3logz)

Find sin x/2, cos x/2, & tan x/2 from the given information: cscx=-3/2, 270 degrees<x<360 degrees

Q4 sinx/2= +-sqrt(1-cosx/2) cscx=-3/2 sinx= 1/(3/2)=-2/3 sin x/2= sqrt(1-cosx/2) sqrt(1-sqrt(5/3)/2) "sqrt(3-sqrt(5)/6)" sin^2+cos^2=1 Subtract sin^2x from both sides Sqrt both sides cosx=sqrt(1-sin^2x sqrt(1-(-2/3)^2) sqrt(1-(4/9)) sqrt(5)/3 cos x/2=-sqrt(1+cosx/2) -sqrt(1+sqrt(5/3)/2) "-sqrt(3+sqrt(5)/6)" tan x/2=sinx/1+cosx (-2/3)/1+(sqrt(5/3)) "-2/(3+sqrt(5))"

Find the domain of the function g(x)=log(x^2-2x-8) w/ base 2

Set up as you would a normal factoring problem: x^2-2x-8=0 Factor: (x-4)(x+2)=0 x=4 or x=-2 Draw out a number line w/ these 2 solutions on it Pick a number from each section & plug into factoring equation: (x-4)(x+2)=0 If solution is greater than 0, it's included in domain If solution is less than 0, it's not included in domain D: (-infinity, -2) in union w/ (4, +infinity)

Use law of logarithms to combine the expression to a single quantity of logarithm: 3logx+1/3log(x-1)-2log(x+3)-log(2x-1)

Turn coefficients into exponents: logx^3+log(x-1)^1/3-log(x+3)^2-log(2x-1) Put one log out in front to get rid of individuals & multiply/divide: log(x^3(x-1)^1/3/(x+3)^2(2x-1))

Describe how to evaluate log7 w/ base 9 with your calculator and evaluate it.

Use change of base formula & plug into calculator logx w/ base b= logx w/ base a/logb w/ base a (log7/log9)= .885622

If $18,000 is invested at an interest rate of 5.25% per year, find the value of the investment after the 10 years if compounded (a) quarterly; (b) weekly; (c) daily; (d) continuously.

Use equations: A=P(1+(r/n))^nt A=Pe^rt Plug in values (a) $30,324.51 (b) $30,420.20 (c) $30,427.09 (d) $30,428.26

Express the equation in exponential form. (a) log8 w/ base x = 3 (b) log(1/125) w/ base 5 = -3 (c) ln(x+1) = 2

Use formula: logx w/ base a = y >>> a^y=x (a)x^3=8 (b)5^-3=1/125 (c)e^2=x+1

Use the definition or properties of the logarithmic function to find x. (a) log7 w/ base 49 = x (b) log16 w/ base x = 4 (c) x=3^log10 w/base 3 (d) x=e^ln12

Use formula: logx w/ base a = y >>> a^y=x 49^x=7 (a) x=1/2 x^4=16 (b) x=4 Use formula: a^logx w/ base a = x (c) x=10 (d) x=12

Find the exact value of the expression: (a) cos 15 degrees (b) sin pi/8 (c) cos 55 degrees cos 10 degrees+sin 50 degrees sin 10 degrees (d) cos 67.5 degrees+cos 22.5 degrees (e) sin(cos^-1(-4/5)) (f) cos^-1(cos(-7pi/6))

cos(45-30)=cos45cos30+sin45sin30 sqrt(2/2)sqrt(3/2)+sqrt(2/2)(1/2) (a) (sqrt(6)sqrt(2))/4 sin (pi/4)/2=sqrt((1-cos pi/4/)2) sqrt(1-sqrt(2/2)/2) (sqrt(2-sqrt2)/4) (b) (sqrt(2-sqrt2)/4) cos(55-10) cos45 (c) sqrt(2/2) cos 135/2+cos45/2 (sqrt(1+cos135)/2)+(sqrt(1+cos45)/2) (sqrt(1-sqrt(2/2)/2)+(sqrt(1+sqrt(2/2)/2) (d) sqrt(2-sqrt(2))/2+sqrt(2+sqrt(2))/2 cos^-1(-4/5)=t cos t=-4/5 t in Q2 sin t >0 sin t = sqrt(1-cos^2 t) sqrt(1-(-4/3)^2) (e) 3/5 cos(-7pi/6)=cos(7pi/6) 7pi/6 in Q3 t=pi/6 cos^-1(cos 7pi/6) cos^-1(-cos pi/6) cos^-1(-sqrt 3/2) (f) 5pi/6

Rewrite the expression as an algebraic expression in x: (a) tan(sin^-1x) (b) cos(cos^-1x-sin^-1x)

t=sin^-1x sin t=x tan t=sin t/cos t cos t=sqrt(1-sin^2 t) sqrt(1-x^2) (a) tan t=x/sqrt(1-x^2) cos^-1x=t cos t=x sin t=sqrt(1-cos^2 t) sqrt(1-x^2) sin^-1x=b sin b=x cos b=sqrt(1-sin^2 b) sqrt(1-x^2) cos(t-b)=cos t cos b+sin t sin b xsqrt(1-x^2)+sqrt(1-x^2)x 2xsqrt(1-x^2)