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April 20, 2014

April 20, 2014

**Recent Homework Questions About Trigonometry**

Post a New Question | Current Questions

**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*

**trig**

PROVE: Cos 2theta = .5 (cot theta - tan theta) Please help me
*Wednesday, August 14, 2013 at 3:32pm*

**Trig**

same as the others .... cscØ = -4 so sinØ = -1/4 we also know that tanØ>0 so the sine is negative and the tangent is positive ---> Ø must be in III since sinØ = -1/4, in III, y = -1, r = 4 by Pythagoras and the CAST rule x = -√15 ...
*Wednesday, August 7, 2013 at 8:12am*

**Trig**

If cosØ = 5/13 and Ø is in IV , then using the standard definition of cos Ø = x/r we have x=5, r=13, then 5^2 + y^2 = 13^2 y^2 = 144 and in IV .... a) y = -12 b) sinØ = -12/13 c) secØ = 13/5 d) tanØ = y/x = -12/5
*Wednesday, August 7, 2013 at 8:06am*

**Trig**

a. If you are only given the value of ƒÆ , you cannot deduce the value of the radius r. b. ƒÆ = pi - 0.213 rad = 2.929 rad csc ƒÆ = 4.739 c. cos ƒÆ = -0.977 d. cot ƒÆ = 1/(tan ƒÆ )= - (8/(ã3)
*Wednesday, August 7, 2013 at 7:12am*

**Trig**

Given that csc θ = -4 and tan θ > 0, find the exact value of a. x b. sin θ c. cos θ d. r e. sec θ f. cot θ
*Wednesday, August 7, 2013 at 5:20am*

**Trig**

Given that tan θ = - (√3/8) and θ is in QII, find the exact value of a. r = b. csc θ c. cos θ d. cot θ
*Wednesday, August 7, 2013 at 5:20am*

**Trig**

Given that cos θ = 5/13 and θ is in QIV, find the exact value of a. y b. sin θ c. sec θ d. tan θ
*Wednesday, August 7, 2013 at 5:19am*

**trig**

sorry about the double-post proof of above: tan200(cot10-tan10) = tan20(1/tan10 - tan10) , 9since by CAST tan200 = tan200) = tan20( 1 - tan^2 10)/tan10 = 2tan10/(1 - tan^2 10) * ( 1 - tan^2 10)/tan10 = 2
*Friday, August 2, 2013 at 7:59am*

**trig**

keystrokes I used on my SHARP scientific calculator I assume your units are degrees, so make sure your calc is set to DEG 1÷ tan 10 - tan 10 = x ta200 = you should get 2
*Friday, August 2, 2013 at 7:50am*

**trig**

keystrokes I used on my SHARP scientific calculator I assume your units are degrees, so make sure your calc is set to DEG 1÷ tan 10 - tan 10 = x ta200 = you should get 2
*Friday, August 2, 2013 at 7:43am*

**trig**

tan200(cot10-tan10)
*Friday, August 2, 2013 at 6:00am*

**trig**

tan200(cot10-tan10)
*Friday, August 2, 2013 at 5:57am*

**trig**

cosx + (1/2)(1/cosx) = -3/2 times 2 2cosx + 1/cosx = -3 times cosx 2cos^2 x + 1 = -3cosx 2cos^2 x + 3cosx + 1 = 0 (2cosx + 1)(cosx + 1) = 0 cosx = -1/2 or cosx = -1 x = 120° or x = 240° or x = 180° in radians: x = 2π/3 , 4π/3 , π BUT, you want the ...
*Wednesday, July 31, 2013 at 8:49pm*

**trig**

Find all solutions of cos (x) + 1/2 sec (x) = -3/2 in the interval (2pi, 4pi) (Leave your answers in exact form and enter them as a comma-separated list.)
*Wednesday, July 31, 2013 at 6:03pm*

**trig**

label the parallelogram ABCD, where AD = 12, BC = 22 angle B = 45° and angle C = 65° draw a line DE || AB to meet BC at E Thus, BE = 12, and EC = 10 label AB = DE = x and DC = y clearly, angle DEC = 45° In triangle DEC, sin45/y = sin65/x = sin70/10 sin45/y = sin70/...
*Wednesday, July 31, 2013 at 7:27am*

**trig**

the base of a trapezoid are 22 and 12 respectively. the angle at the extremities of one base are 65 degree and 45 degree respectively. find two legs?
*Wednesday, July 31, 2013 at 6:10am*

**trig**

PQ = [(-2+1), (1-8)] = [-1 , -7] 3PQ = [-3,-21) or -3i - 21j magnitude = √(9 + 441) = √450 = 15√2
*Sunday, July 28, 2013 at 1:00pm*

**trig**

Given that P=(-1,8) and Q=(-2,1), find the component form and magnitude of the vector 3 PQ.
*Sunday, July 28, 2013 at 12:45pm*

**Calculus HELP!**

cosh(at) = d/dt (1/a sinh(at)) sinh(at) = d/dt (1/a cosh(at)) So, Since L{f(at)} = 1/|a| F(s/a) L{cosh(at)} = sL{1/a sinh(t)} = s/a L{sinh(t)} L{sinh(at)} = sL{1/a cosh(at)} = s/a L{cosh(at)} - s/a So, L{cosh(at)} = s/a (s/a L{cosh(at)} - s/a L{cosh(at)}(1-s^2/a^2) = -s/a = (s...
*Sunday, July 28, 2013 at 5:51am*

**Trig.**

0 greater or equal to t greater or equal to 1.6
*Sunday, July 28, 2013 at 2:29am*

**math**

Haw can solev this qen trig application measruments width of lake?
*Tuesday, July 23, 2013 at 8:21am*

**math03**

Trig aplication meseramants width of lake
*Tuesday, July 23, 2013 at 8:08am*

**trig**

I feel the question required not the ht of tree but distance between bird's eye and mouse which will be=hypoteneuse =30/cos65. May please check.
*Tuesday, July 23, 2013 at 4:45am*

**trig**

csc x - sin x = 1/sinx - sinx = (1 - sin^2 x)/sinx = cos^2 x /sinx = (cosx/sinx) /sinx = cotx cscx
*Monday, July 22, 2013 at 8:43pm*

**trig**

we know that tan 2A = 2tanA/(1 - tan^2 A) so tan 210 = 2tan105/(1 - tan^2 105) you should know that tan 210 = tan 30° = 1/√3 let tan 105 = x then 1/√3 = 2x/(1-x^2) 2√3x = 1 - x^2 x^2 + 2√3 - 1 = 0 x = (-2√3 ± √16)/2 = -√3 &...
*Monday, July 22, 2013 at 8:34pm*

**Jackie, Karen, Lois Sharla ? trig**

Why are you switching names ? Please stick with one name.
*Monday, July 22, 2013 at 8:20pm*

**trig**

The o has a diagonal line thru it if that helps. I don't know about the p and that's why I'm confused.
*Monday, July 22, 2013 at 8:17pm*

**trig**

no idea what "p" and "o" represent, or which trig function is required.
*Monday, July 22, 2013 at 8:16pm*

**trig**

did you make a sketch of the right-angled triangle? Remember the line of sight is parallel to the ground, and alternate angles are equal. I have tan 65° = h/30 , where h is the height of the tree h = 30tan65 = 64.3 metres
*Monday, July 22, 2013 at 8:12pm*

**trig**

Which expression is equivalent to csc x - sin x?
*Monday, July 22, 2013 at 8:10pm*

**trig**

use a half-angle identity to find the exact value of tan 105 degrees.
*Monday, July 22, 2013 at 8:10pm*

**trig**

Write the standard form of the equation where p=π and o=7π/12
*Monday, July 22, 2013 at 8:09pm*

**trig**

A bird at the top of a tree looks down at a field mouse with an angle of depression of 65 degrees. If the field mouse is 30 meters from the base of the tree, find the distance from which the field mouse to the bird's eyes. Round the answer to the nearest tenth.
*Monday, July 22, 2013 at 8:08pm*

**Trig**

17sec2 θ − 15tanθsecθ − 15 = 0 17/cos^2 Ø - 15(sinØ/cosØ)(1/cosØ;) - 15 = 0 times cos^2 Ø 17 - 15sinØ - 15cos^2 Ø = 0 17 - 15sinØ - 15(1 - sin^2 Ø) = 0 15sin^2 Ø - 15sin&...
*Monday, July 22, 2013 at 5:01pm*

**Trig**

17sec^2θ - 15secθ tanθ - 15 = 0 15secθ tanθ = 17sec^2θ - 15 225 sec^2θ tan^2θ = 289sec^4θ - 510sec^2θ + 225 225sec^4θ - 225sec^2θ = 289sec^4θ - 510sec^2θ + 225 64sec^4θ - 285sec^2θ + 225 = 0 That&#...
*Monday, July 22, 2013 at 4:52pm*

**Trig**

Find all solutions in the interval 0 degrees<θ<360 degrees. If rounding necessary, round to the nearest tenth of a degree. 17sec2 θ − 15tanθsecθ − 15 = 0
*Monday, July 22, 2013 at 3:55pm*

**trig**

If you are using the parallelogram method where the resultant (r) is the diagonal, you should have r^2 = 12^2 + 17^2 - 2(12)(17)cos 120° = 637 r = √637 = appr 25.24 knots let Ø be the angle between the resultant r and the first tug boat, then sinØ/17 = ...
*Sunday, July 21, 2013 at 11:38pm*

**trig**

A barge is pulled by two tugboats. The first tugboat is traveling at a speed of 12 knots with heading 140°, and the second tugboat is traveling at a speed of 17 knots with heading 200°. Find the resulting speed and direction of the barge. (Round your answers to the ...
*Sunday, July 21, 2013 at 10:58pm*

**trig**

set 2pi/k = 16 and solve for k. Because the period is 16. Normally, sin(x) has period 2pi. Review your text and reread my first line.
*Sunday, July 21, 2013 at 4:57am*

**trig**

general shape d = a sin kt we know a = 15 and if it takes 4 sec to complete 5 cycles then 1 cycle or 1 period = 4/5 seconds to find k, 2π/k= 4/5 4k = 10π k = 5π/2 equation: d = 15 sin (5π/2) t
*Saturday, July 20, 2013 at 10:42pm*

**trig**

why pi?
*Saturday, July 20, 2013 at 3:30pm*

**trig**

An object suspended from a spring is oscillating up and down. The distance from the high point to the low point is 30 cm, and the objects take 4 sec to complete 5 cycles.For the first few cycles the distance from the mean position, d(t), in cm with respect to the time t sec is...
*Saturday, July 20, 2013 at 3:27pm*

**trig**

since sin(kt) has period 2pi/k, we have y = 7sin(pi/8 t) + 8.5 with appropriate shift depending on where the wheel is at time 0. slower turning means longer period, so the coefficient of t will decrease.
*Saturday, July 20, 2013 at 3:24pm*

**physics**

c = cliff height h = lighthouse hight a = distance of A from cliff b = distance of B from cliff c/a = tan 23° (c+h)/a = tan 40° (c+h)/b = tan 17° So, clearing fractions and plugging in the trig values, we have c = .4245a c+h = .8391a c+h = .3057b b = a+1320 h = 313...
*Saturday, July 20, 2013 at 3:20pm*

**trig**

1. A Ferris wheel with a radius of 7m makes one complete revolution every 16 s. The bottom of the wheel is 1.5m above ground. a)Find the equation of the graph b)predict how the graph and the equation will change if the Ferris wheel turns more slowly c) test your predictions ...
*Saturday, July 20, 2013 at 3:13pm*

**trig**

Tan (7pi/4)+ Tan (5pi/4) =Tan (pi+3pi/4)+ Tan (pi+pi/4) =-Tan (pi/4)+ Tan (pi/4) = 0
*Friday, July 19, 2013 at 3:35am*

**trig**

Tan7pi/4=Tan(pi+3pi/4) lies in IVQdt and is =-Tan pi/4. Tan 5pi/4=tan (pi+pi/4) lis in IIIQdt and is=tan pi/4, hence your ans is correct.
*Friday, July 19, 2013 at 12:12am*

**trig**

tan(7π/4) + tan(5π/4) Does this = 0
*Thursday, July 18, 2013 at 10:49pm*

**trig**

C = 2 pi r, so the circumference of the tire is 2 pi * 1.1 = 2.2 pi 2.2 pi ft/rev * 4 rev/s = 8.8 pi ft/s
*Thursday, July 18, 2013 at 5:58pm*

**trig **

each wheel of a bicycle is of radius is 1.1 ft. if the wheels are making 4 revoultions per second how fast is the bicycle moving
*Thursday, July 18, 2013 at 5:05pm*

**Calculus**

Find the positive value of x which satisfies x = 4.3cos(x). Give the answer to six places of accuracy. Remember to calculate the trig functions in radian mode
*Wednesday, July 17, 2013 at 10:47am*

**trig**

sec^2 + csc^2 = 1/cos^2 + 1/sin^2 = (sin^2+cos^2)/(sin^2 cos^2) = 1/(sin^2 cos^2) csc^2 sec = 1/(sin^2 cos) = cos/(sin^2 cos^2) so , doing the division, we end up with just cos x
*Tuesday, July 16, 2013 at 2:26pm*

**trig**

Simplify: csc^2 x sec x / sec^2 x + csc^2 x
*Tuesday, July 16, 2013 at 12:36pm*

**college trig**

arc/circumference = angle/360° 37.3/2πr = 108.03/360 2πr(108.03) = 37.3(360) r = 19.783 mm
*Monday, July 15, 2013 at 1:55pm*

**college trig**

find the radius of a pulley if rotating the pulley 108.03 degrees raises the pulley 37.3mm
*Monday, July 15, 2013 at 1:34pm*

**trig**

but but but ... you should have used my hints. csc^2(-x) = (-cscx)^2 = csc^2(x) sin^2+cos^2 = 1 divide by sin^2 to get 1+cot^2 = csc^2 cot^2 - csc^2 = -1 sin^2+cos^2=1 will help solve many problems.
*Monday, July 15, 2013 at 5:01am*

**trig**

works for me. the co-functions are the functions of the complementary angles. So, by definition, tan(π/2-x) = cot(x). Your proof works as well, though.
*Monday, July 15, 2013 at 4:56am*

**trig**

tan (pi/2 - x) = sin (pi/2 - x) / cos (pi/2 - x) But sin (pi/2 - x) = cos x and cos (pi/2 - x) = sin x <=> tan (pi/2 - x) = cos x / sin x = cotan x <=> tan (pi/2 - x) * tan x = cotan x * tan x = (cos x / sin x) * (sin x / cos x) = 1
*Sunday, July 14, 2013 at 8:57pm*

**trig**

simplify the expression. tan(π/2-x)tanx
*Sunday, July 14, 2013 at 8:48pm*

**trig**

Parentheses are assumed missing. Implied parentheses are ALWAYS required in numerator and denominator of fractions. (cos^2x+sin^2x)/(cot^2x-csc^2x) This problem can be solved by converting all functions in terms of sine and cosine according to the standard definitions. (cos^2x...
*Sunday, July 14, 2013 at 7:21pm*

**trig**

simplify the expression. cos^2x+sin^2x/cot^2x-csc^2x
*Sunday, July 14, 2013 at 6:22pm*

**trig**

But my teacher gave the answer on the review as -1.
*Sunday, July 14, 2013 at 5:17pm*

**trig**

csc(-x) = -csc(x) sin^2+cos^2 = 1
*Sunday, July 14, 2013 at 4:56pm*

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