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April 16, 2014

Homework Help: Math: Trigonometry

Recent Homework Questions About Trigonometry

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trig
POOP
Thursday, November 21, 2013 at 6:03pm

Trig - Law of sines and cosines
Thanks a lot
Thursday, November 21, 2013 at 4:04pm

Trig-Medians and law of cosines and sines
thanks a lot man!
Wednesday, November 20, 2013 at 3:59pm

Trig-Medians and law of cosines and sines
median length x = BC length median hits BC at D triangle ABC 4^2 = 3^2 + x^2 - 6 x cos B triangle ABD x^2 = 3^2 + (x/2)^2 - 3 x cos B ================================= 16 = 9 + x^2 - 6x cos B x^2 = 9 +x^2/4 -3 x cos B x^2 -6 x cos B -7 = 0 3x^2/4 +3 x cos B - 9 = 0 x^2 - 6 x ...
Wednesday, November 20, 2013 at 12:30pm

Trig-Medians and law of cosines and sines
In triangle ABC, we have AB=3 and AC=4. Side BC and the median from A to BC have the same length. What is BC? Not making sense to me, I think the answer must be simple, but I don't know how to solve I applied the law of sines but to no avail. Help is appreciated, thanks.
Wednesday, November 20, 2013 at 11:53am

Trig
since sin(34) = cos(56) sin(48) = cos(42) the expression evaluates to 2
Wednesday, November 20, 2013 at 12:11am

Trig
sin^2(34)+sin^2(48)+sin^2(56)+sin^2(42)
Tuesday, November 19, 2013 at 10:18pm

Trig - Law of sines and cosines
I started with AB, the vertical diagonal of the square. I then drew square ALMN At M, I drew the base BC of the equilateral triangle perpendicular to AM, letting BM = MC = 2 Joi AB and AC for the triangle. Let the intersection of ML and AB be P, and the intersection of MN and ...
Tuesday, November 19, 2013 at 9:37pm

Trig
√(100+100cot^2 x) √(100(1+cot^2 x)) √(100csc^2 x) 10cscx
Tuesday, November 19, 2013 at 9:24pm

Trig - Law of sines and cosines
ABC is an equilateral triangle with side length 4. M is the midpoint of BC, and AM is a diagonal of square ALMN. Find the area of the region common to both ABC and ALMN. I drew the diagram but I don't know how to find the answer? I think it has something to do with the law...
Tuesday, November 19, 2013 at 8:09pm

Trig
sqrt(100+(10cotx)^2)
Tuesday, November 19, 2013 at 8:02pm

Pre-Cal
these are all done the same way. You have to solve a quadratic in some trig function. #1. 2sec^2 4x - 3sec 4x - 5 = 0 (2sec 4x - 5)(sec 4x + 1) = 0 So, sec 4x = 5/4 or sec 4x = -1/4 There are two angles for each solution in [0,360°) where this is true. But that means there...
Tuesday, November 19, 2013 at 12:19am

Calculus
1) The period of a trig. function y=sin kx is 2pi/k. Then period of y=sin^2(pi.x/a) should be 2pi/(pi/a)=2a, but somewhere it is given as a. Which is correct? 2) The period of r=sin^3(theta/3) is given as 3pi. How is it worked out? Is it because after theta=0, the function ...
Monday, November 18, 2013 at 12:18am

Trig
V1 + V2 = 1200[70o] + 900[205o] X = 1200*cos70 + 900*cos205 = -405.3 Y = 1200*sin70 + 900*sin205 = 747.3 tanAr = Y/X = 747,3/-405.3 = -1.84376 Ar = -61.53o = Reference angle. A = -61.53 + 180 = 118.5o Vr = Y/sinA = 747.3/sin118.5 = 850.3[118.5o] = Resultant vector.
Saturday, November 16, 2013 at 6:16pm

Trig
2 vectors act on a bolt. vector 1(1200/70 degrees) vector 2(900/205 degrees) then E = ? must show work!
Friday, November 15, 2013 at 8:57am

Trig
Dunno what E is, but sounds like you just want to add two vectors. Bad news is you gotta convert to rectangular coordinates to do that.
Friday, November 15, 2013 at 6:27am

Trig
the resultant velocity is 320.0 mph @ 103° so, now you know how far it goes in 4 hours.
Friday, November 15, 2013 at 6:22am

Trig
Plane leaves airport with heading of 110 degrees at 300 mph, the wing out of the southwest is 42 mph. locate the plane after 4 hours? find the distance back to airport and angle from due east?
Friday, November 15, 2013 at 1:14am

Trig
2 vectors act on a bolt. vector 1(1200/70 degrees) vector 2(900/205 degrees) then E = ? must show work!
Friday, November 15, 2013 at 12:37am

Algebra 2 Trig - Probability Question
help solving a probability problem: probability that a person taking a survey is male "given" that he preferred a European automobile if 112 males prefer european autos out of 188, and 216 of a total of 452 individuals were surveyed were male. Additional information...
Thursday, November 14, 2013 at 9:13am

Trig
The magnitude of the resultant R is R^2 = 650^2 + 300^2 R = appr 715.89 if Ø is the angle it makes with the longer vector, then tanØ = 300/650 = ... Ø = appr 24.8°
Wednesday, November 13, 2013 at 8:56pm

Trig
2 vectors at a Right angle to each other are pulling on a bolt. vector 1 = 650; vector 2 = 300 then R = ?
Wednesday, November 13, 2013 at 7:35pm

Trigonometry - Identities and proof
Two of the standard conversion formulas are: sinA - sinB = 2sin((A-B)/2) cos( (A+B)/2) and cosA - cosB = - 2sin( (A-B)/2) sin( (A+B)/2) see: near bottom of page under: Sum-to-Product Formulas http://www.sosmath.com/trig/Trig5/trig5/​trig5.html RS = -(sinx - siny)/(cosx...
Wednesday, November 13, 2013 at 11:59am

Trigonometry - Identities and proof
Show that cot((x+y)/2) = - (sin x - sin y)/(cos x - cos y) for all values of x and y for which both sides are defined. I tried manipulating both sides in terms of trig identities but I don't really have a solution....help would be appreciated, thanks.
Wednesday, November 13, 2013 at 11:23am

Math
absolute means global. It's possible to have several relative extrema, and some, all, or none may be absolute extrema. A parabola has one relative extremum, which is absolute. A cubic such as y=(x-1)(x-2)(x-3) has two relative extrema, but no absolutes at all, since the ...
Sunday, November 10, 2013 at 4:38pm

trig
¤©¤·¤&f​rac34;¤·¤°&curr​en;·¤©
Friday, November 8, 2013 at 1:53am

math iv
it doesn't infer anything, though you may infer that it's a problem involving the first quadrant. Or a problem on acute triangles Or principal values of trig functions . . .
Thursday, November 7, 2013 at 12:00pm

Math-Trigonometry
Show that if A, B, and C are the angles of an acute triangle, then tan A + tan B + tan C = tan A tan B tan C. I tried drawing perpendiculars and stuff but it doesn't seem to work? For me, the trig identities don't seem to plug in as well. Help is appreciated, thanks.
Wednesday, November 6, 2013 at 5:21pm

Math - Trigonometry
If sin theta +cos theta =1.2, then what is sin^3 theta + cos^3 theta? Hmm...I don't understand how to proceed. I know I must apply a trig Identity, but which one? Thanks in advance
Tuesday, November 5, 2013 at 10:26pm

Trigonometry
I will read that as sin (x/7) = -√3/2 since √3/2 is one of the trig values I recognize So x/7 must be in quadrants III or IV I know sin 60° = +√3/2 so my angle in "standard position " is 60° x/7 = 180+60 = 240 x = 1680° or x/7 = 360-60...
Monday, November 4, 2013 at 3:12pm

MATH ANALYSIS & TRIG
as usual, draw a diagram. Al's distance a is 6+10t Barb's distance b is 2+5t d^2 = (6+10t)^2 + (2+5t)^2 = 5(25t^2 + 28t + 8) Now just set dd/dt=0 to find the minimum.
Wednesday, October 30, 2013 at 5:45am

MATH ANALYSIS & TRIG
At noon, Al is 6km north of point O traveling south at 10km/h. Also at noon, Barb is 2km east of point ) traveling east at 5km/h. A) express the distance "d" between Al and Barb as a function of time "t" hours after noon. B) Find the time at which the ...
Tuesday, October 29, 2013 at 11:25pm

trig
ij
Tuesday, October 22, 2013 at 3:22pm

Math - Trigonometry
doesn't matter. All of the trig functions are positive in exactly 2 of the 4 quadrants.
Tuesday, October 15, 2013 at 4:07pm

Math - Trigonometry
degrees or radians or grads, it's still 2. What you call the angle does not matter. If you plan to succeed in trig, get used to thinking in radians. Once you get past the introductory material, most problems will deal with radians.
Tuesday, October 15, 2013 at 11:51am

pre-calc/ trig
yes it is.
Tuesday, October 15, 2013 at 10:21am

pre-calc/ trig
Sinx=0
Tuesday, October 15, 2013 at 10:07am

Trig
hey ted did you ever figure that out because if so tell me tell me now plz i need this very much
Saturday, October 12, 2013 at 11:14am

Calculus - please help!
we have 4x^2 + 9y^2 = 36 The line x=3 lies at one end of the ellipse, so shells are probably the way to go. The volume v is thus v = ∫[-3,3] 2πrh dx where r = 3-x and h = y = √(36-4x^2)/3 v = 2π/3 ∫[-3,3] (3-x)√(36-4x^2) dx Now just break it ...
Friday, October 11, 2013 at 12:09pm

Math - Trig
Did you make sure your calculator was set to Radians ? (press DRG until you see RAD displayed) Here are my keystrokes .5 x 12 - 2 = sin = x 20 + 40 = and you should get 24.86° (Of course you can see that since sin(????) ranges from -1 to +1, the temperature must range from...
Thursday, October 10, 2013 at 9:07am

Math - Trig
Still don't get it, so yes need more help.
Thursday, October 10, 2013 at 5:49am

Math - Trig
assuming θ is the month, just plug in θ=12. Come on back if you don't like the answer.
Thursday, October 10, 2013 at 4:33am

Math - Trig
The equation y = 20sin(0.5θ - 2) + 40 models the monthly temp for a certain city. Use the equation to predict the temperature in the city during December. Thanks
Thursday, October 10, 2013 at 12:46am

Trig
we know we want y = a sin(bx+c) + d diameter is 2, so radius=1, and we have y = a sin(bx+c) + 1 Starting at t=0, the bead starts going downward, so y = -sin(bx+c)+1 If we make it so the axle is at (0,1) at t=0, we need to shift the graph to the right by 1, since the bead is on...
Monday, October 7, 2013 at 4:23am

Trig
a child puts beads on one spoke of a bicycle wheel. the tire has a diameter of 2ft. if the child rides so that the tire makes one full rotation every 15 sec, and the beads begin in the horizontal outward position, find an equation that models the position of the beads at time t.
Monday, October 7, 2013 at 4:11am

MathsSs triG
ThanQ steve
Thursday, October 3, 2013 at 5:49pm

MathsSs triG
sin(x-360)sin(90-x)tan(-x)/cos(90+x) = sin(x)cos(x)(-sin(x)/cos(x))/(-sin(x)) = sin(x) x = 30 8/sin^2A - 4/1+cosA 8/(1-cos^2A) - 4/(1+cosA) (8 - 4(1-cosA))/(1-cos^2A) (8-4+4cosA)/(1-cos^2A) 4(1+cosA)/(1-cos^2A) 4/(1-cosA) undefined when cosA = 1 So, on 0<A<360 it is ...
Thursday, October 3, 2013 at 5:28pm

MathsSs triG
Consider sin(x-360)sin(90-x)tan(-x)/cos(90+x) 1.A.SIMPLIFY sin(x-360)sin(90-x)tan(-x)/cos(90+x) to a single trigonometric ratio B.hence or otherwise without using a calculator,solve for X if 0<X<360. sin(x-360)sin(90-x)tan(-x)/cos(90+x) =0,5 2.A.prove that 8/sin^2A - 4/1...
Thursday, October 3, 2013 at 4:22pm

Derivatives
H(x) = sin2x cos2x = 1/2 sin4x That help? Using the product rule, it's a bit more work, but you can get the same answer: dH/dx = 2cos2x cos2x - 2sin2x sin2x = 2(cos^2 2x - sin^2 2x) = 2cos4x Don't forget your trig just because you're taking calculus! #2 df/dx = (u...
Thursday, October 3, 2013 at 4:58am

Go with Bosnian - Trig
sorry Kiki don't know how I missed that 2 in front of 2x^2 Go with Bosnian's solution or, how about this... 2x^2 + 2x > 4 x^2 + x - 2 > 0 (x+2)(x-1) > 0 (x+2>0 and x-1<0) OR ( x+2 < 0 and x-1 > 0 ) ( x>-2 and x<1) OR x< -2 and x> 1 -2 &...
Sunday, September 29, 2013 at 4:44pm

Trig
2x^2+2x>4 Divide both sides by 2 x ^ 2 + x > 2 ________________________________________​ take one half of thje coefficient of x and square it. in this case ( 1 / 2 ) ^ 2 = 1 / 4 ________________________________________​ Add 1 / 4 to both sides x ^ 2 + x + 1...
Sunday, September 29, 2013 at 4:24pm

not Trig
x^2 + 2x > 4 add 1 to both sides x^2 + 2x + 1 > 5 (x+1)^2 > 5 x+1>√5 or -x-1 > √5 x > √5 - 1 OR -x > √5 + 1 x > √5-1 OR x < -√5-1
Sunday, September 29, 2013 at 2:59pm

Trig
2x^2+2x>4
Sunday, September 29, 2013 at 2:17pm

math - trig
sinA = a/c = cosB cosA = b/c = sinB tanA = a/b = cotB plug and chug.
Wednesday, September 25, 2013 at 5:47am

math - trig
Suppose ABC is a right triangle with sides of lengths a, b, and c and right angle at C. Find the unknown side length using the Pythagorean Theorem and find the following trigonometric functions of the indicated angle. Given: a = 4 and b = 7 Find: sin A, cot B, sec A, tan B
Wednesday, September 25, 2013 at 4:09am

Trig
(x+1)(x-4) >= (x-2)^2 x^2-3x-4 >= x^2-4x+4 x >= 8 for the graphs, go to wolframalpha.com and enter plot y=(x+1)(x-4) , y=(x-2)^2, x=6..10 or, enter (x+1)(x-4)-(x-2)^2 and see where it's positive
Monday, September 23, 2013 at 3:15pm

Trig
(x+1)(x-4) is greater than or equal to (x-2)^2
Monday, September 23, 2013 at 2:57pm

Physics
SAme solution, but you should have the two solutions. it should be 45+45-37.5 degrees (if you think on the trig equations you get, you see why).
Wednesday, September 18, 2013 at 5:48pm

Trig
looks like "homework dumping" to me I have two of them for you. Please indicate where you are having difficulties with these standard type of trig questions.
Wednesday, September 18, 2013 at 12:23am

Trig
make quick sketches of right-angled triangles for each of the two cases, and use Pythagoras to find the missing sides. first triangle: since sinA = 12/13 , y = 12 and r = 13 x^2 + y^2 = r^2 x^2 + 144 = 169 x^2 = 25 x = ±5, but we are in quad II, so x = -5 then tanA = - ...
Wednesday, September 18, 2013 at 12:22am

Trig
Well, why not follow that hint sin(-15) = sin(30 - 45) = sin30cos45 - cos30sin45 = .... You should know the trig functions of those special angles.
Wednesday, September 18, 2013 at 12:13am

Trig
Rewrite the expression (tan A)(cot A) in terms of a single trigonometric ratio.
Tuesday, September 17, 2013 at 11:24pm

Trig
Use the equation mg sin A = umg cos A to determine the angle at which a waxed wood block on an inclined plane of wet snow begins to slide. Assume the coefficient of friction, u, is 0.17.
Tuesday, September 17, 2013 at 11:24pm

Trig
Find the exact value of sin π/12.
Tuesday, September 17, 2013 at 11:23pm

Trig
Evaluate and simplify cos(-15º). Hint: Substitute 30º-45º for -15º.
Tuesday, September 17, 2013 at 11:23pm

Trig
Suppose sin A = 12/13 with 90º≤A≤180º. Suppose also that sin B = 7/25 with -90º≤B≤0º. Find tan (A - B).
Tuesday, September 17, 2013 at 11:22pm

Trig
Suppose sinA = 12/13 with 90º≤A≤180º. Suppose also that sinB = -7/25 with -90º≤B≤0º. Find cos(A - B).
Tuesday, September 17, 2013 at 11:22pm

Trig
Let sin A = 12/13 with 90º≤A≤180º and tan B = -4/3 with 270º≤B≤360º. Find tan (A + B).
Tuesday, September 17, 2013 at 11:22pm

trig
A plane flies in a direction of N70°E for 80 km and then on a bearing of S10°W for 150 km. a How far is the plane from its starting point? b What direction is the plane from its starting point?
Sunday, September 15, 2013 at 8:01am

Trig
sec^2θ = 1+tan^2θ = 17 since secθ < 0 in QIII, secθ = -√17
Friday, September 13, 2013 at 5:03am

calculus (integral)
use partial fractions to get 13/27 (1/(x-3) - (x+6)/(x^2+3x+9)) 13/27 (1/(x-3) - 1/2 (2x+3)/(x^2+3x+9) - (9/2)/(x^2+3x+9)) 13/27 (1/(x-3) - 1/2 (2x+3)/(x^2+3x+9) - (9/2)/(x+3/2)^2 + 27/4) Now you can use substitutions u = x-3 v = x^2+3x+9 w = x + 3/2 to come up with 13/27 (du/...
Friday, September 13, 2013 at 5:01am

Trig
suppose tan theta equal 4 and theta lies in the third quadrant. find the values of sec theta
Friday, September 13, 2013 at 2:03am

trig
sketch right-angled triangles to match sinA = 12/13 and tanB = -4/3 from the first , y=12, r=13, so x = -5 and tanA = -12/5 tan(A-b) = (tanA - tanB)/(1 + tanAtanB) = (-12/5 + 4/3)/(1 + (-12/5)(-4/3)) = -16/63
Friday, September 13, 2013 at 12:09am

trig
tan 5π/12 or tan 75° I will use degrees ... tan 75 = tan(30 + 45) = (tan30+tan45)/( 1 - tan30tan45) = (1/√3 + 1)/( 1 - 1/√3) , multiply top and bottom by √3 = (1 + √3)/√3 - 1)
Friday, September 13, 2013 at 12:01am

trig
Let sin A = 12/13 with 90º≤A≤180º and tan B = -4/3 with 270º≤B≤360º. Find tan (A - B).
Thursday, September 12, 2013 at 11:51pm

trig
let's test it by applying the sin(A - B) = sinAcosB - cosAsinB relation sin(x - 3π/2) = sinx cos3π/2 - cosx sin3π/2 = (sinx) (0) - (cosx)(-1) = cosx yup, it is true.
Thursday, September 12, 2013 at 11:49pm

trig
Evaluate and simplify tan 5π/12.
Thursday, September 12, 2013 at 11:47pm

trig
This identity means that translating a basic sine graph 3π/2 units to the right produces a basic cosine graph. True or False? sin(X-(3(pi)/2)) = cosX
Thursday, September 12, 2013 at 11:43pm

trig
well, sine is zero at zero radians, and at PI radians e^x=0 because e^x cannot be zero, then e^x=PI take the ln of each side x=ln(PI)
Wednesday, September 11, 2013 at 8:49pm

trig
see other post.
Wednesday, September 11, 2013 at 8:47pm

trig
How do you solve sin(e^x)=0 for x? The question is "find the smallest number x such that sin(e^x)=0".... but I can't even get AN answer lol! Much appreciated!
Wednesday, September 11, 2013 at 8:29pm

trig
How do you solve sin(e^x)=0 for x? The question is "find the smallest number x such that sin(e^x)=0".... but I can't even get AN answer lol! Much appreciated!
Wednesday, September 11, 2013 at 8:29pm

trig
Just noticed this was for trig, not calculus. Algebraically, note that a(y) is a parabola, whose vertex occurs where y=60/4 = 15.
Thursday, August 29, 2013 at 11:34pm

trig
If the sides are x and y, with x parallel to the barn, x+2y = 60, so x = 60-2y the area is a = xy = (60-2y)y = 60y-2y^2 da/dy = 60 - 4y max a when y=15,x=30
Thursday, August 29, 2013 at 11:33pm

trig
if 60 m of fencing is available for a rectangular garden, one side of which is against a barn, what areb the dimensions of the garden that will give the maximum area?
Thursday, August 29, 2013 at 9:44pm

Grammar
with = not algebra for = not trig However, "for" version is a little awkward. "I want to take algebra instead of trigonometry" would be better.
Thursday, August 29, 2013 at 2:38pm

trig
obviously, it's (2*9-3(-5) , 2*2-3(-2)) = (?,?)
Wednesday, August 28, 2013 at 3:44pm

trig
u = (9, 2) v = (-5, -2) 2u - 3v = ?
Wednesday, August 28, 2013 at 3:07pm

Math
You should memorize the following diagrams: 1. the 30°-60°-90° triangle with matching sides 1 -- √3 -- 2 (angles in radians, π/6 π/3 π/2) 2. the 45° -- 45° -- 90° triangle with corresponding sides 1 -- 1 -- √2 So sin π/4 ...
Thursday, August 22, 2013 at 5:25pm

Trig hep, please? Thank you!
I will try the others soon.
Sunday, August 18, 2013 at 11:24am

Trig hep, please? Thank you!
thanks! what do the question marks stand for. and also CAST?
Sunday, August 18, 2013 at 11:24am

Trig hep, please? Thank you!
1. sin 195° = - sin15° , by CAST we know cos 30° = 1 - 2 sin^2 15° √3/2 = 1 - 2sin^2 15° 2sin^2 15 = 1 - √3/2 = (2-√3)/2 sin^2 15° = (2-√3)/4 sin 15 = √((2-√3)) /2 so sin 195 = -sin15 = -√((2-√3)) /2 3. ...
Sunday, August 18, 2013 at 12:00am

Trig hep, please? Thank you!
1. Find the exact value of sin(195(degrees)) 2. If cot2(delta)=5/12 with 0(<or =)2(delta)pi, find cos(delta), sin(delta) , tan(delta). 3.find the exact value of sin2(x) if cos(x)= 4/5. (X is in quadrant 1) 4. Find the exact value of tan2(x) if sin(x)=5/13. ((X) in quadrant ...
Saturday, August 17, 2013 at 10:51pm

Trigonometry desperate help, clueless girl here
#2 cos2x - 3sinx cos2x = 0 (cos2x)(1-3sinx) = 0 as you know, if the product of two numbers is zero, one or the other must be zero. So, cos2x = 0 or 1-3sinx = 0 cos2x=0 means x is pi/4,3pi/4,5pi/4,7pi/4 1-3sinx=0 means x = arcsin(1/3) = .3398 But, you need all angles between 0 ...
Saturday, August 17, 2013 at 12:37am

trig help much appreciated! :))
alright, I will. Thanks
Friday, August 16, 2013 at 6:37pm

trig help much appreciated! :))
no ideas on any of these? Also you have some major typos and/or formatting issues. I'll do a couple. Maybe you can clean up the post and indicate where you get stuck on the others. #2 cos 2x – 3sin x cos 2x = 0 (1-3sinx)cos2x = 0 so, either cos2x=0 ==> x = pi/4 or ...
Friday, August 16, 2013 at 4:58pm

trig help much appreciated! :))
1. Find the complete exact solution of sin x = . 2. Solve cos 2x – 3sin x cos 2x = 0 for the principal value(s) to two decimal places. 3. Solve tan2 x + tan x – 1 = 0 for the principal value(s) to two decimal places. 4. Prove that tan2  – 1 + cos2  = ...
Friday, August 16, 2013 at 4:53pm

trig
This is very cleverly solved at http://answers.yahoo.com/question/index?​qid=20100617064137AAccXtF as a simple web search revealed.
Friday, August 16, 2013 at 4:53pm

trig
prove that :cosA-sinA+1/cosA+sinA-1=cosecA+cotA
Friday, August 16, 2013 at 1:16pm

trig
only true if theta = 45 degrees as far as I know identity: cos 2t = cos^2 t - sin^2 t on the right we have .5 [ cos t/ sin t - sin t/cos t ] = .5 [ cos^2 t - sin^2 t] /(sin t cos t) that is only equal to the left side if sin t cos t = .5 which is only true if t = 45 degrees or...
Wednesday, August 14, 2013 at 3:50pm

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