|
Tutors Answer Your Questions about Trigonometry-basics (FREE)
Question 185262: I'm very confused!
Please help!
The Equations I know are:
Theta = arch length/radius
Angular velocity = theta/time
1. A wheel is rotating at an angular velocity of 1.2 pi radians/s while a point on the circumference of the wheel travels at 9.6 pi m in 10 seconds.
a) How many revolutions does the wheel make in 1 min?
Click here to see answer by edjones(3298)   |
Question 185484This question is from textbook
: Find the exact values of the six trigonometric functions of theta if the terminal side of theta in standard position contains the point (-5,-4). Please help! This question is from textbook
Click here to see answer by stanbon(26291)  |
Question 185583: a fairgrounds most popular attraction is a roller-coaster ride known as the terror run. one of the stretches of track is called the missile path and is in the form of a parabolic curve,starting at point A on the curve and ending at B. B is 180m horizontally from A and the highest point of the curve is 100m above A and B
a safety engineer examined the structure and observed that points A and B were likely to be damaged due to the steepness of the missile path near these points. the owner is puzzled. he can see no way to make the missile path less steep near A and B and to keep the height of the ride the same.
(a)how can this be done?
(b)is it possable to comply with the engineers instructions and also raise the height of the missile path?
Click here to see answer by stanbon(26291) |
Question 187426: #1) If tan of theta = (6/5) and cos of theta < 0, use the fundamental identities to evaluate the other five trigonometric functions of theta.
#2) use the fundamental identities to simplify csc squared if beta (1-cos squared of beta)
#3) factor and simplify. (sec "to the power of 4" x- tan "to the power of 4" x/ sec "squared" x + tan "squared" x)
Click here to see answer by stanbon(26291) |
Question 187494: I had a test today, and could not crack this problem I really would appreciate the help so I can understand it later on.
Prove:
cotx*cosx=cotx-cosx
first i found a common denominator, but after that i was stuck.
Click here to see answer by Alan3354(6092)  |
Question 188285: Verify that the equation is an identity.
I tried solving on the right side
= [(1 + cos x) / 2] - [(1 - cos x) / 2]
= [1 + cosx - 1 + cos x]/2
= 2cosx/2
= cosx
I have no clue how to get sin2x over 2sinx from this and I don't know what I did wrong.
Click here to see answer by Alan3354(6092) |
Question 188283: Verify that this equation is an identity...
sin(4y) = 4sin(y)cos(y) - 8sin³(y)
I tried working on the right side and this is as far as I got.
First I factored...
4sinycosy(1 - 2sin^2y)
then I substituted 1-cos^2y for sin^y
4sinycosy(1-2(1-cos^2y)
4sinycisy(1-2+2cos^2y)
4sinycosy(2cos^2y-1)
and I don't know what to do next to get 4siny
Click here to see answer by Edwin McCravy(2922) |
Question 188286: If a person bends at the waist with a straight back making an abgle of theta degrees with the horizontal, the the force F exerted on the back muscles can be modeled by the equation...
F = .6Wsin(theta + 90degrees)/ sin 12
where W is the weight of the person.
Use an identity to show that F is approximately equal to 2.9W cos theta.
I had no clue how to do this one.
Click here to see answer by jim_thompson5910(13794) |
Question 188997: A Ferris wheel of diameter 40m has its centre 21m above the ground. The wheel rotates once every 30s. (Draw a graph to show a rider's height above the ground during a 1 min ride if possible please) Or else could you write the equation for the graph.
Please help! Thanks!
Click here to see answer by solver91311(5072)  |
Question 189614: Hi-- I am really struggling with this problem. I can't seem to get the right answer. If you could help me, that would be greatly appreciated. Thanks! :)
Find sin theta/2, given that sin theta=1/4, and theta terminates in quadrant I.
Answer Options:
A. sqrt (8+2sqrt15 /4)
B. sqrt (8-2sqrt15 /4)
C. sqrt (6) / 4
D. sqrt (10) / 4
I have tried this problem, but this is what I got:
sin theta = 1/4, theta lies in Quadrant I
sin theta/2 = √[(1-cos theta)/2]
since sin theta = 1/4, then cos theta = √15/ 4.
sin theta/2 = √([1 - √15/4]/ 2) = √[(4 - √15)/8] = [√2(4-√15)]/4 : This is not one of my answer choices.
Click here to see answer by Alan3354(6092) |
|
Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790
|
| |
