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April 16, 2014

April 16, 2014

**Recent Homework Questions About Trigonometry**

Post a New Question | Current Questions

**Trig**

A 16 - foot ladder leaning against the side of a house reaches 12 feet up the side of the house. What angle does the ladder make with the ground
*Sunday, September 23, 2012 at 5:02pm*

**TRIG**

a) one rotation = 2π radians so 4400 rpm = 2π(4400) radians/min = 8800π rad/min = appr 27646 rad/min b) one rotation = 2π(8.5) inches = 17π inches so 4400 rotations would be 17π(4400) inches = 74800π so the linear speed of a tooth = 74800&#...
*Sunday, September 23, 2012 at 2:41pm*

**TRIG**

A circular power saw has an 8-1/2 inch diameter blade that rotates at 4400 revolutions per minute. a)Find the angular speed of the saw blade in radians per minute. b)Find the linear speed in feet per minute of one of the 24 cutting teeth as they contact the wood being cut. I&#...
*Sunday, September 23, 2012 at 12:26pm*

**trig**

Write 8y = 300 in its equivalent logarithmic form
*Friday, September 21, 2012 at 3:52pm*

**math/trig**

if the flagpole has height h, h/44 = tan 60.5° h = 77.77 now you want the angle x where 77.77/88 = tan x x = 41.5°
*Tuesday, September 18, 2012 at 10:59am*

**math/trig**

What does the ladder have to do with the flagpole?
*Monday, September 17, 2012 at 11:35pm*

**math/trig**

A flagpole casts a shadow 44 ft with the sun angle of elevation is 60.5 degrees. Find the height of the flagpole and at what angle is the shadow twice as long?
*Monday, September 17, 2012 at 11:18pm*

**math/trig**

A 16 ft ladder casts a shadow 44 ft with the sun angle of elevation is 60.5 degrees. Find the height of the flagpole and at what angle is the shadow twice as long?
*Monday, September 17, 2012 at 11:16pm*

**trig**

I think you mean find all angles theta between 0 and 2pi.
*Sunday, September 16, 2012 at 12:26pm*

**Definite Integral**

The way to get their answer is to complete the square and then use trig substitution. If you go to wolframalpha.com and type in ∫ (5x + 3) / (x^2 + 4x + 7) dx then click the "Show Steps" button you can see how they did it. If you want the final numeric answer, ...
*Saturday, September 15, 2012 at 11:18pm*

**trig**

construct your right-angled triangle by drawing a perpendicular to the x-axis from (1,-4) then r^2 = x^2 + y^2 = 1 + 16 r = √17 b) sinA = -4/√17 cscA = -√17/4 cosA = 1/√17 secA = √17 tanA = -4/1 = -4 cotA = -1/4
*Saturday, September 15, 2012 at 7:49pm*

**trig**

-20000/360 = -55.555555.. = -55 - 200/360 so we have 55 clockwise rotations and then another clockwise rotation of 160° OR 360-200 = 160° in the counterclockwise of positive rotation direction 160° is in the second quadrant, where the sine is positve and we are 20&...
*Saturday, September 15, 2012 at 7:43pm*

**trig**

(a) sketch an angle A in standard position whose terminal ray passses through the point (1,-4) (b) Find the exact value of the six trigonometric functions of A, (c) Let B be the reference angle of A. Find a point on the terminal rsy of B in standard position and add B to your ...
*Saturday, September 15, 2012 at 4:47pm*

**trig**

find the reference angle of -20000 and express the six trigonometric functions of -20000 in terms of the six trigonometric functions of its reference angle.
*Saturday, September 15, 2012 at 4:40pm*

**trig**

my question is find all angles of theta and 2pi whose reference angle is alpha = pi/12. Give exact answers. i cannot for the live of me find this
*Saturday, September 15, 2012 at 4:30pm*

**math**

first of all, to make things easier to type, let's say h = delta x so your question becomes lim [ sin(π/6 + h) - sin π/6 ]/h , as h ---> 0 This looks like you want the derivative of sin x, when x = π/6 if f(x) = sinx f'(x) = cosx f'(π/6) = ...
*Thursday, September 13, 2012 at 10:12am*

**Calculus**

One of the fundamental limits with trig expressions is Lim tanx/x = 1 , as x---> 0 so notice that you have exactly that pattern, so lim tan(4Ø)/(4Ø) =1 , assuming 4Ø ---> 0
*Wednesday, September 12, 2012 at 2:42pm*

**Math**

Actually it probably has something to do with trig identities...
*Tuesday, September 11, 2012 at 11:09pm*

**Algebra 2/Trig**

A 5'6" person is standing near a light post that is 18' above the ground. How long is the man's shadow when he is 5' from the base of the light post?
*Tuesday, September 11, 2012 at 8:47pm*

**trig**

53. the longest side is 55 cm. find the length of the shortest side
*Tuesday, September 11, 2012 at 7:59pm*

**Physics**

Picture a right angled triangle with one side AB being 25 x 125 metres long to the west; the other side BC being 19 x 66 metres long to the south. The displacement is the length of the hypotenuse AC; the direction being between 180 degrees and west. Sketch it out, then do the ...
*Tuesday, September 11, 2012 at 8:39am*

**Math - Trig**

find in degrees and radians the angle between the hour hand and the minute hand of a clock at half past three
*Sunday, September 9, 2012 at 8:44am*

**Calculus**

You're ok to this point: 10 sin^-1 (5x)(x) + 2√(1-25x^2) + C |1/5, 1/10 By this time you should realize that radians are the measure of choice for trig stuff. sin^-1(1/2) = pi/6 sin^-1(1) = pi/2 so you end up with [10(1/5 * pi/2) + 2√(1-1)] - [10(1/10 * pi/6...
*Friday, September 7, 2012 at 10:07am*

**Calculus**

think back, back, back to your days in trig. d^2 = 350^2 + 400^2 - 2*350*400*cos 50 d = 320.19
*Monday, September 3, 2012 at 12:42pm*

**calculus**

think back, back, back to your days in trig. If the tower has height h, tan 67 = h/50 h = 117.8 m
*Monday, September 3, 2012 at 12:36pm*

**math trig**

unless you can measure the angle, you have to use similar triangles. If you can measure the angle sighting from the tip of the stick's shadow to the top of the stick, then tanθ = opposite/adjacent The same angle is made from the pole's shadow to the top of the ...
*Friday, August 31, 2012 at 6:08pm*

**math trig**

I think i have to use soh cah toa instead of similar triangles
*Friday, August 31, 2012 at 4:46pm*

**math trig**

stick a stick vertically in the ground. measure the height and shadow of the stick. measure the shadow of the tree, and use the same ratio to calculate its height. extra credit: why does this work?
*Friday, August 31, 2012 at 4:17pm*

**math trig**

lets say you have a tree or a flagpole. Describe how you would measure this object using right angled triganometry
*Friday, August 31, 2012 at 11:35am*

**trig**

My calculator says sin(70°)=0.93969262
*Friday, August 31, 2012 at 8:06am*

**trig**

what is the value of sin 70
*Friday, August 31, 2012 at 1:51am*

**trig**

Sin (180 - A) = sin A. (angle in degrees) sin 120 = sin (180 - 60) = sin 60 Cos (180 - A) = - cos A cos 150 = cos (180 - 30) = - cos 30 tan (360 - A) = - tan A, tan 280 = tan (360 - 80) = - tan 80
*Friday, August 31, 2012 at 12:20am*

**trig**

how to prove ; tan 10 degrees + tan 70 degrees - tan 50 degrees = sqrt 3
*Friday, August 31, 2012 at 12:08am*

**Math**

velocity is v = dh/dt = -sin(2t) average v over an interval is ∫[1,3] dv/dt / (3-1) = 1/2 ∫[1,3] -sin(2t) dt = 1/2 (1/2 cos 2t) [1,3] = 1/4 (cos6-cos1) = 1/4 (.4198) = 0.105 Note that this is just the total displacement / total time h(pi)-h(0) = .5(1) - .5(1) = 0 ...
*Wednesday, August 29, 2012 at 2:19pm*

**trig**

pick any real number x > 0. x/2 < x so there is always a smaller positive real same for rational numbers pick any y such that x < y The difference is (y-x) x < x + (y-x)/2 < y
*Tuesday, August 28, 2012 at 3:09pm*

**trig**

prove that there exists no smallest possitive real number dose there exist a smallest positive rational number given a real number x, does there exist a smallest real y>x?
*Tuesday, August 28, 2012 at 2:40pm*

**physics-please help**

Now Sabrina, We did the previous problem for you. It is very similar to this problem. Please try it. It is really just trig.
*Monday, August 27, 2012 at 8:11am*

**Math (Trig/pre-calculus)**

nevermind, I figured it out, it is approximately 4.1 x10^13 km from the star
*Thursday, August 23, 2012 at 10:29pm*

**Math (Trig/pre-calculus)**

some stars are so far away that their position appear fixed as earth orbits the sun. other stars, however, appear over time to shift their positions relative to the background of "fixed" stars. suppose that the star shown below appears to shift through an arc of ...
*Thursday, August 23, 2012 at 9:52pm*

**Trig**

-17 pi/6
*Monday, August 20, 2012 at 11:37am*

**trig**

The Tangent Function: tanA = Y/X = 8/6.
*Wednesday, August 15, 2012 at 7:05pm*

**trig**

Imagine that you are sitting 6 feet away from a television that is hung on a wall. The top of the TV is 8 feet off the ground. Which function correctly represents the angle that you make with the top of your television?
*Wednesday, August 15, 2012 at 2:34pm*

**math**

how would I measure a distance or object using right angled trig?
*Tuesday, August 14, 2012 at 10:43pm*

**trig**

If Ø is your angle, and since r=√(a^2 + b^2) sinØ = b/r cscØ = r/b cosØ = a/r secØ = r/a tanØ = b/a cotØ = a/b These definitions should be in your textbook and are probably defined in terms of x, y and r. You should ...
*Monday, August 13, 2012 at 9:02pm*

**trig**

Let be an angle in standard position and the point (a, b) be the point of intersection of the terminal side of with the unit circle. State the unit circle definitions of the six trigonometric functions. cos = sec = sin = csc = tan = cot =
*Monday, August 13, 2012 at 8:53pm*

**trig**

sin -1 as principal values in [-π/2,π/2] sin -1-.5 = -π/6
*Monday, August 13, 2012 at 5:24pm*

**trig**

find the exact value of the expression , sin-1(-0.5)
*Monday, August 13, 2012 at 3:58pm*

**trig**

sin(x) has amplitude 1, period 2pi, no phase shift sin(x-pi/3) is shifted pi/3 to the right sin(2(x-pi/3)) has period pi, shifted right by pi/3 2sin(2(x-pi/3)) meets the requirements. Do the others in lime wise. Recall that tan(x) has period pi, not 2pi.
*Monday, August 13, 2012 at 12:15am*

**trig**

13
*Sunday, August 12, 2012 at 8:46pm*

**trig **

Construct the equations of the following trigonometric functions: A)A sine function with amplitude 2, period , phase shift /3 right B)A tangent function with a reflection in the y-axis, period ¾, translation up 5 units C)A cosine function with period 270°, ...
*Sunday, August 12, 2012 at 7:07pm*

**trig**

you know that sin(x) has maximum value at x = pi/2. So, x - pi/3 = pi/2, or x = 5pi/6. However, that is not in the interval [0,2]. So, try x - pi/3 = -3pi/2. x = -7pi/6. So, y does not achieve its maximum possible value over x in [0,2]. So, the max achieved in [0,2] occurs ...
*Sunday, August 12, 2012 at 1:58pm*

**trig**

For the function -2sin(x-(pi/3)) between x = 0 and x = 2 : (6 marks) For what value(s) of x does y have its maximum value? For what value(s) of x does y have its minimum value? For what value(s) of x does y=1?
*Sunday, August 12, 2012 at 1:44pm*

**trig**

Determine the period, amplitude and phase shift for each given function: A)y = -4 cos 3x + 5 B)y = 2/3 sin (30x-90degrees)-10 c)y = -0.38 tan (x/3+pi/3) d)y = pi cos(2x)+ pi
*Sunday, August 12, 2012 at 1:33pm*

**trig**

start-> programs-> Accessories-> Calculator-> View-> Scintific 17 sin
*Thursday, August 9, 2012 at 11:40pm*

**trig**

Please help to find the value of: Sin 17 degree.
*Thursday, August 9, 2012 at 10:33pm*

**trig**

One of the first things I usually do is test if the identity is true by picking any value of x. Since it is an identity it should work for all value of x let x=10 LS = (1/cos80 -1)/(1/cos40-1) = appr 15.58 RS = tan80/tan40 = appr 6.758 so , not true, no point trying to prove it.
*Thursday, August 9, 2012 at 12:39pm*

**trig**

(sec 8 x - 1)/(sec 4x -1) = tan 8x/tan 4x Prove it!
*Thursday, August 9, 2012 at 11:57am*

**algebra/trig**

1. Which numbers are NOT perfect squares? 2. √48-5√27+2√75 = √16√3 - 5√9√3 + 2√25√3 = 4√3 - 15√3 + 10√3 = -√3 3. I will assume you meant √(x+2) - 3 = 0 then √(x+2) = 3 square both sides x...
*Tuesday, August 7, 2012 at 10:46am*

**algebra/trig**

1. let y = x^3 + 5 inverse is x = y^3 + 5 y^3 = x-5 y = (x-5)^(1/3) yes , it is a function 2. (f∙ g)(2) = f(g(2)) = f(3) = 10 3. Come on, you can do this!
*Tuesday, August 7, 2012 at 10:39am*

**algebra/trig**

1. which number below is irrational? a)√4/9 b) √20 c)√121 Why is the number you chose irrational? 2. express in simplest form: √48-5√27+2√75 3. solve for x: (√x+2)-3=0
*Tuesday, August 7, 2012 at 9:59am*

**algebra/trig**

1. f(x)=x^3+5 does f(x) have an inverse? if so, find the inverse and decide if it is a function. 2. if f(x)= 3x+1 and g(x)= x^2-1, find (f∙ g)(2) 3. if f(x)= (2x)^2, find f(-4) thank you :)
*Tuesday, August 7, 2012 at 9:51am*

**mathematics**

Given y = 3cos2x you know it has amplitude of 3, so its minimum value is y = -3, since there is no y-translation. Unless you are specifically doing calculus, I'd surely use my knowledge of trig to answer this one. Good analysis, though, Bosnian!
*Saturday, August 4, 2012 at 12:11pm*

**Calculus 1 Integration Trig Inverse**

see answer in prior post
*Tuesday, July 31, 2012 at 11:03am*

**Caculus:Integration Inverse Trig**

for integrating 1/√(x^2-1) use x = secθ Then x√(x^2-1) = secθtanθ dx = secθtanθ dθ The integral of the first term then becomes ∫4 dθ = 4θ = 4 arcsec(x) for integrating 1/(1+x^2) use x = tanθ dx = sec^2θ dθ...
*Tuesday, July 31, 2012 at 11:02am*

**Calculus 1 Integration Trig Inverse**

integral of (4/(x times the square root of (x^2-1)) + (1+x+x^3)/(1+x^2)dx
*Monday, July 30, 2012 at 10:42pm*

**Caculus:Integration Inverse Trig**

integral of (4/(x times the square root of( x^2-1)) + (1+x+x^3)/(1+x^2)dx
*Monday, July 30, 2012 at 10:40pm*

**physics**

You can just use basic Trig a few times. First, find x1 and y1 (I’m defining these as the x and y components of the first triangle, the one where a is the hypotenuse. I drew them in green in the diagram above). x1 = 175*cos(30) y1 = 175*sin(30) You can do the same thing ...
*Sunday, July 29, 2012 at 1:50am*

**Radicals and Trig.**

Thank you! Next question. What if instead they gave me the hypotenuse? 2/(sqrt2? Still tan 30.
*Friday, July 27, 2012 at 10:50am*

**trig**

For the second, draw you triangle in quadrant II hypotenuse = 3 , opposite = 1 x^2 + 1 = 9 x = ±√8 or ±2√2 , but we are in II, so use x - -2√2 cosØ = -1/3 tanØ = - 1/2√2 ---> cotØ = -2√2 I don't know how ...
*Friday, July 27, 2012 at 8:53am*

**trig**

so cos would be root 15/ 4 and cot would be 1/ root 15? one more question, how do you know it is in the second quadrant?
*Friday, July 27, 2012 at 12:43am*

**trig**

given cosØ = 2/3 and cotØ > 0 so Ø must be in quadrant I make a sketch of a right -angled triangle with hypotenuse 3 and adjacent 2 then y^2 + 2^2 = 3^2 y = √5 and sin Ø - √5/3 tanØ = √5/2 do the second the same way (&...
*Thursday, July 26, 2012 at 11:48pm*

**trig**

find sin theta and tan theta if cos theta = 2/3 and cot theta is >0 and find cos theta and cot theta if sin theta = 1/3 and tan theta is <0 thanks! :)
*Thursday, July 26, 2012 at 11:23pm*

**Trig Functions**

cosine, sine, vary between 1, -1? So they cannot be in the correct answer.
*Thursday, July 26, 2012 at 5:11pm*

**Trig Functions**

Which of the following lists contains only functions with vertical asymptotes in their graphs? A. Cosine, sine, tangent, cotangent B. Tangent, secant, cosecant, cotangent C. Sine, tangent, secant, cosecant D. Cosine, sine, secant, cosecant
*Thursday, July 26, 2012 at 4:23pm*

**Radicals and Trig.**

Because that is the value of tan 30 . The short side length is half the hypotenuse for a 30-60-90 triangle. If the hypotenuse length is 1, the Second longest side length must be sqrt[3/4] That makes the tangent of 30 degrees 1/(sqrt3)
*Wednesday, July 25, 2012 at 10:08pm*

**Radicals and Trig.**

I am catching up over summer on math units that I have to rewrite. In my textbook it asks me to find the base of two triangles and add them together. the smallest angle of the larger triangle is 30 degrees. The opposite side to the angle is 40cm. it gives me Tan 30 = 40/x 1/...
*Wednesday, July 25, 2012 at 3:32pm*

**Trig**

LS = (1+ tanØ)^2 = 1 + 2tanØ + tan^2 Ø = 1 + 2tanØ + sec^2 Ø - 1 = 2sinØ/cosØ + 1/cos^2 Ø = (2sinØcosØ + 1)/cos^2 Ø = (sin^2 Ø + cos^2 Ø + 2sinØcosØ)/cos^2 Ø = (sin&...
*Tuesday, July 24, 2012 at 11:40am*

**Trig**

Claim: For all theta such that -pie/2<theta<pie/2 the following holds true: (1+tan(theta))^2=1/cos(theta)
*Tuesday, July 24, 2012 at 11:02am*

**Trig Help Please!!!**

I didn't know this, and I feel like I was cheated by my geomtery teacher.I feel like I should have known this I am a math professor with a Ph.D. but I'm comforted to hear that a lot of other people didn't know about it, either.
*Tuesday, July 24, 2012 at 1:41am*

**math**

One of the basic trig identities says: cos 2A = cos^2 A - sin^2 A
*Monday, July 23, 2012 at 1:08pm*

**Math/Trig**

Is the answer y = 21.15sin(pi/6 t + 3)+66.25?
*Saturday, July 21, 2012 at 2:26pm*

**Trig**

Did you make a sketch? if so , you will see that tan 1° = 12000/x , where x is the distance along the ground x = 12000/tan 1° = ..... "How far on the ground is the plane ? " sounds like a very strange question.
*Friday, July 20, 2012 at 6:43pm*

**Trig**

There is an airplane at an altitude of 12000 ft. The angle of depression is 1 degree. How far on the ground is the plane.
*Friday, July 20, 2012 at 2:36pm*

**trig**

tan(25+5)= tan25+5/1-tan25tan5 tan(30)= tan25+5/1-tan25tan5 1/root(3)= tan25+5/1-tan25tan5 ur answer is 1/root(3)
*Friday, July 20, 2012 at 6:27am*

**trig**

cos195
*Thursday, July 19, 2012 at 10:52pm*

**Trig**

looks like we have some "homework dumping" going on here. I will do one more, and let you looks at my methods. Then you tell me where you are having problems Again, you must mean (tanx + cotx)/cscx = secx LS = (sinx/cosx + cosx/sinx)(sinx) = (sin^2 x + cos^2 x)/(...
*Thursday, July 19, 2012 at 2:18pm*

**Trig**

You must mean (1-cosx)/sinx = sinx/(1+cosx) LS = (1-cosx)/sinx * (1+cosx)/(1+cosx) = 1 - cos^2 x)/(sinx(1+cosx) ) = sin^2 x/(sinx(1+cosx) ) = sinx/(1+cosx) = RS
*Thursday, July 19, 2012 at 2:15pm*

**Trig**

verify the following identity: 2sinxcos^3x+2sin^3cosx=sin(2x)
*Thursday, July 19, 2012 at 12:50pm*

**Trig**

verify the following identity: tanx+cotx/cscx=secx
*Thursday, July 19, 2012 at 12:49pm*

**Trig**

verify the following identity: cos(A-B)/cosAsinB=tanA+cotB
*Thursday, July 19, 2012 at 12:48pm*

**Trig**

verify the following identity: sin(x+y)*sin(x-y)=sin^2x-sin^2y
*Thursday, July 19, 2012 at 12:47pm*

**Trig**

Verify the following identity: 1-cosx/sinx=sinx/1+cosx
*Thursday, July 19, 2012 at 12:45pm*

**Trig/Physics**

Hi, thank you so much. Everything I tried hit all around the borders, but just was not working out correctly. I really appreciate your help.
*Thursday, July 19, 2012 at 3:50am*

**Trig/Physics**

total displacement north = 5 sin 30 + 10 cos20 = 2.5 + 9.397 = 11.897 miles total displacement east = 5 cos30 - 10 sin20 = 4.33 - 3.42 = 0.91 miles (a) sqrt[(11.897)^2 + (0.91)^2] = 11.93 miles (b) direction = tan^-1 0.91/11.897 E of N = 4.37 deg E of N
*Thursday, July 19, 2012 at 1:18am*

**Trig/Physics**

Hi, sorry it was 30 degrees North of East.
*Wednesday, July 18, 2012 at 11:38pm*

**Trig/Physics**

You left out an important number that is needed to compute the answer. It should follow the first word "direction". How many degrees is the first direction North of East?
*Wednesday, July 18, 2012 at 11:14pm*

**Trig/Physics**

Joe walks 5 miles in the direction degrees North of East, and then 10 miles in the direction 20 degrees West of North. (a) How far is Joe from his starting position? (b) In what direction is Joe from his starting position?
*Wednesday, July 18, 2012 at 10:01pm*

**Trigonometry**

Simplify the expression using trig identities: 1. (sin4x - cos4x)/(sin2x -cos2x) 2. (sinx(cotx)+cosx)/(2cotx)
*Sunday, July 15, 2012 at 6:29pm*

**Math Trig**

13. amplitude 3: 3cos(...) period p: 3cos(2pi(...)/p) ... translate by (-1,1): 3cos(2pi(x+1)/p) + 1 So, it appears to be (D), if by p you mean pi. 16: (0,4): (C) 20: (A) Your use of transition point appears unusual. In #13 the only possible interpretation, given the answer ...
*Saturday, July 14, 2012 at 12:05pm*

**Math Trig**

13. What is the equation of a cosine function with amplitude 3, transition point (−1, 1), and period p? A. y = p cos [3(x − 1)] − 1 B. y = 3 cos [2(x − 1)] + 1 C. y = 3 cos [p (x + 1)] − 1 D. y = 3 cos [2(x + 1)] + 1 16. What is the transition ...
*Saturday, July 14, 2012 at 6:34am*

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