Expert: Sherry Wallin - 11/29/2009

Question
QUESTION: hey i have a couple of questions. frist is probability.1) spares of particular component are produced by two firms,bestbits and lesserprod tests show tht,on average,1 in 200 components produced by bestbits fail within one year of fitting, and 1 in 50 components produced by lesserprod fail within one year of fitting. given tht 20percent of the components sold and fitted are made by bestbits and 80 percent by lesserprod, wht is the probability tht a component chosen at random from those sold fitted will fail within year of fitting?
find the proportion of components sold and fitted tht would need to be made by bestbits for this probability to be 0.01?

now the trignometry questions,1) given tht sin(A+B)=2sin(A+B),show tht tanA=3tanB.Hence or otherwise find the solution of the equations; sin(A+30)=2sin(A-30) in the range 0<A<2pie.

2)show that tanA+cotA=2/sin2A

3) prove that sinA/1+cosA = 1-cosA/sinA
thank you soooooo much if u help me then ur a life savor!!:D

ANSWER: Sara~  lots of questions!

1/200 = .5% and 1/50 = 2%, these are the failure probabilities
Since bestbits supplies only 20% of the product that means they will have .005(.2)= .001 or .1% of theirs that will fail. Now the other company will have .02(.8)=.016 or 1.6% of their product fail. So the overall probability of failure is .001 + .016 = .017 or 1.7 out of every 100 parts fail within 1 year.So the probability of a part failing chosen at random is .017.

Let x = # produced out of 100, i.e. x/100 and write .02x +(.005)(100-x) = .01 solve for x and you will get 66.6 parts per 100 need to be produced by bestbits for the failure rate to be .01
check: .02(33.3%) +.005[(100-33.3)%] = .02(.333) + .005(.667) = .00666 + .003335 = .009995 ~= .01



#1 is written wrong, it isn't true, perhaps you meant sin(A+B)=2sin(A-B)?
#2 start with tan A + cot A = sinA/cosA + cosA/sinA find a common denominator
= [sin^2A + cos^2 A]/sinA*cosA
= 1/sinA*cosA multiply top and bottom by 2
=2/2sinA*cosA
=2/sin2A       2sinAcosA = sin2A

#3  is a two stepper: multiply the top and bottom of the left hand side by the conjugate of 1+cosA
which is (of course) 1 - cosA
[(1-cosA)*sinA]/(1+cosA)(1-cosA)= [(1-cosA)*sinA]/[1-cos^2 A]
= (1-cosA)*sinA]/sin^2A    since 1-cos^2 A = sin^2 A
=(1-cosA)/sinA  cancel sin A with sin A

Math Prof


---------- FOLLOW-UP ----------

QUESTION: yes for the first trigonometry questions its sin(A+B)=2sin(A-B).

Answer
sin(A+B) = 2sin(A-B)  use rules for sin of a sum and sin of a difference
sinAcosB+cosAsinB = 2(sinAcosB-cosAsinB)=2sinAcosB - 2cosAsinB
0 = sinAcosB - 3cosAsinB   subtract sinAcosB+cosAsinB from both sides
3cosAsinB = sinAcosB   add 3cosAsinB to both sides
3sinB = tanAcosB  divide both sides by cosA
3sinB/cosB = tan A  divide both sides by cosB
3tanB = tan A  sinB/cosB = tanB

let B = 30 so that tanA = 3tan30 = sqrt3 -> tan^-1(sqrt3) = A -> A = 60deg
sin(A+B) = Sin(A + 30) = 2sin(A-30)-> sin(60+30) = 2sin(60-30) -> sin 90 = 2sin30 -> 1 = 2(1.2) = 1 which is true. This just proves that A is 60 degrees but you need to think about where else in (0,2pi) that this will happen? Hint the tangent has the same sign in QI and QIII. So it must be when the angle measure 240 degrees. Check it out: Is sin(240+30) ?= 2sin(240-30)
Is sin(270) = 2sin(210)? is -1 = 2(-1/2) -1 yes

Math Prof