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April 18, 2014

April 18, 2014

**Recent Homework Questions About Trigonometry**

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**trig**

Four wires support a 40-meter radio tower. Two wires are attached to the top and two wires are attached to the center of the tower. The wires are anchored to the ground 30-meters from the base of the tower.
*Tuesday, February 26, 2013 at 10:33am*

**trig**

b=a(tanB) where a is your altitude, B is your angle of depression and b is your distance to airport. b=30,000(tan(72)) b=92,330.51
*Monday, February 25, 2013 at 8:32pm*

**Math**

secθ = 1/cosθ If the very names of the 6 basic trig functions are unfamiliar to you, you have some major catching up to do.
*Monday, February 25, 2013 at 5:36pm*

**geometry(triangle)**

There's probably a good geometric way to do it, but here's a trig way, using the law of cosines. AB^2 = 6^2+15^2 - 2 * 6 * 15 * (-1/2) = 351 AC^2 = PC^2 + 15^2 - 2*PC*15(-1/2) BC^2 = PC^2 + 6^2 - 2*PC*6(-1/2) AC^2-BC^2 = AB^2 = 351, so 351 = 189 + 9PC 9PC = 162
*Monday, February 25, 2013 at 12:05pm*

**trig**

sec0+ten0-1/tan0+sec0+1=1+sin0/cos0
*Sunday, February 24, 2013 at 4:02am*

**Trig identities:(**

?? Im sorry i dont understand what u did but im sure tht this is an identity we have to prove tht the left side equals the right side:)
*Saturday, February 23, 2013 at 8:17pm*

**Trig identities:(**

I don't think it is an identity. let A=90 deg sin(360)-cos(360)=0-1 sin(180)-cos(180)=0+1 The sides are not equal.
*Saturday, February 23, 2013 at 7:52pm*

**Trig identities:(**

Sin4A - cos4A = sin2A - cos2A Can i just square root the left side??:O
*Saturday, February 23, 2013 at 7:43pm*

**Math**

Tht will take awhile wont it? Do u think i can master it by monday:( i also have cast rule, ambiguous case, sine and cosine law, primary trig ratios to study tho:( do u think its possible:(?
*Saturday, February 23, 2013 at 7:28pm*

**Math**

OK, but the effort is the same. cos/(1+sin) Now, you know that a^2-b^2 = (1+a)(1-a), and you knwo that cos^2 = 1-sin^2, so that should lead you to simplify the denominator by multiplying top and bottom by (1-sin) cos/(1+sin) * (1-sin)/(1-sin) = cos(1-sin)/(1-sin^2) = cos(1-sin...
*Saturday, February 23, 2013 at 7:19pm*

**Math**

coem on. This is trig. You shoudl know algebra I a/b * b * 1/a = 1 Just multiply the fractions and cancel factors!
*Saturday, February 23, 2013 at 7:08pm*

**Math**

Just cross-multiply to clear the fractions. You get cos*cos = (1-sin)(1+sin) If that doesn't look true, you do need to work some more of these. Dozens of them. sin^2 + cos^2 = 1 is one of the most useful identities you have for trig problems.
*Saturday, February 23, 2013 at 7:07pm*

**trig sub question**

thank you Steve
*Saturday, February 23, 2013 at 8:57am*

**trig sub question**

You need 3x = 2tanθ, so 9x^2 = 4tan^2 θ 9x^2+4 = 4tan^2 θ + 4 = 4sec^2 θ
*Saturday, February 23, 2013 at 5:07am*

**trig**

sec^2(x)-1 = tan^2(x) csc^2(x)-1 = cot^2(x) since cot = 1/tan, it follows immediately
*Saturday, February 23, 2013 at 4:58am*

**trig**

(sec^2x-1)(csc^2x-1)=1 prove the following identity ive been stuck on this for hours please help
*Friday, February 22, 2013 at 7:53pm*

**trig sub question**

Hello, I have a question concerning trigonometric substitution. let's say we have integral of dx/sqrt(9x^2 + 4), so after doing a few steps we have: 2/3 integral of sec0/sqrt((2tan0)^2 + 4*) d0 (the * is for later on) the next step turns into: 2/3 integral of sec0/(sqrt(...
*Friday, February 22, 2013 at 6:30pm*

**trig**

since A+B+C = 180, we can see that C = 78° Now, use the law of sines: c/sinC = a/sinA
*Friday, February 22, 2013 at 6:25pm*

**trig**

What is the length of side c to the nearest whole number if side a 105 and angle A 65 degrees and angle B 37 degrees?
*Friday, February 22, 2013 at 5:49pm*

**trig**

tan x = opposite/adjacent cot x = adjacent/opposite they are reciprocals of each other. tan x = 1/cot x Can you finish from here?
*Thursday, February 21, 2013 at 7:26pm*

**trig**

If cot x = .78, what is tan x?
*Thursday, February 21, 2013 at 7:24pm*

**trig**

use your product-to-sum formulas: sinAcosB = 1/2 (sin(A+B) + sin(A-B)) sin4x cos3x = 1/2 (sin7x + sin(x))
*Thursday, February 21, 2013 at 5:48pm*

**trig**

Express sin4xcos3x as a sum or differences of sines and cosines
*Thursday, February 21, 2013 at 4:04pm*

**trig needs help**

Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question.
*Thursday, February 21, 2013 at 11:36am*

**trig needs help**

Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question.
*Thursday, February 21, 2013 at 11:22am*

**trig needs help**

Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question.
*Thursday, February 21, 2013 at 10:39am*

**Trig**

We need to find the arclength of a central angle of 22° let's use a ratio arc/(2π(30)) = 22/360 arc = 22(60π)/360 = appr 11.5 inches/sec
*Wednesday, February 20, 2013 at 10:33am*

**Trig **

A pendulum swings through an angle of 22 degrees each second. If the pendulum is 30 inches long, how far does its tip move each second?
*Wednesday, February 20, 2013 at 10:25am*

**Trig**

No, they are not 11π/12 is in quadrant II =π/12 is in IV
*Tuesday, February 19, 2013 at 9:02pm*

**Trig**

Whoops nevermind. I was thinking if they were 11pi/6 and -pi/6.
*Tuesday, February 19, 2013 at 8:56pm*

**Trig**

Is the angle 11pi/12 on the unit circle the same as the angle -pi/12? I'm thinking it would be, but I'm not sure.
*Tuesday, February 19, 2013 at 8:51pm*

**trig**

An object Is propelled upward at an angle θ, 45° < θ<90°, to the horizontal with an initial velocity of (Vo) feet per second from the base of a plane that makes an angle of 45° with the horizontal. If air resistance is ignored, the distance R it ...
*Tuesday, February 19, 2013 at 8:35pm*

**trig**

To determine the distance to an oil platform in the Pacific Ocean, from both ends of a beach, a surveyor measures the angle to the platform from each end of the beach. The angle made with the shoreline from one end of the beach is 83 degrees, from the other end 78.6 degrees. ...
*Tuesday, February 19, 2013 at 7:50pm*

**trig.**

thank you!
*Monday, February 18, 2013 at 8:38pm*

**trig.**

solution sets? Is there an equals sign anywhere? If all this is equal to zero.. 4x(2x^2+x-1)=4x(2x-1)(x+1)=0 x=0, x=1/2, x=-1 check those
*Monday, February 18, 2013 at 8:31pm*

**trig.**

8x^3+4x^2-4x solution sets
*Monday, February 18, 2013 at 8:26pm*

**Trig**

using the identity cos 2A = cos^2 A - sin^2 A = 1 - 2sin^2 x = 2cos^2 x -1 LS = cos 4x + cos 2x = cos^2 2x - sin^2 2x + cos 2x = (1 - sin^2 (2x) ) - sin^2 (2x) + (1 - 2sin^2 x) = 1 - 2sin^2 (2x) + 1 - 2sin^2 x = 2 - 2sin^2 (2x) - 2sin^2 x = RS
*Monday, February 18, 2013 at 12:32am*

**trig**

cos^(2) 20 degrees + sin^(2) 20 degrees +pi/2 = (cos^(2) 20 degrees + sin^(2) 20 degrees) +pi/2 = 1 + π/2
*Sunday, February 17, 2013 at 11:46pm*

**Trig**

Verify the identity. cos 4x + cos 2x = 2 - 2 sin^2(2x) - 2 sin^2 x
*Sunday, February 17, 2013 at 11:33pm*

**Trig**

sides of triangle are (2+2.5),(2+3),(2.5+3) Start with law of cosines c^2=a^2+b^2-2abCosC label the sides a,b, c solve for angle C Then, use the law of sines a/SinA=c/SinC solve for angle A then use the fact that the sum of the angles is 180 deg, find angle B. check angle B ...
*Sunday, February 17, 2013 at 9:40pm*

**Trig**

Three circles with radii of 4, 5, and 6 cm, respectively, are tangent to each other externally. Find the angles of the triangle whose vertexes are the centers of the circles.
*Sunday, February 17, 2013 at 9:36pm*

**trig**

cos^(2) 20 degrees + sin^(2) 20 degrees +pi/2
*Sunday, February 17, 2013 at 8:59pm*

**Trig**

if the sides have length a and b, a^2 = 6^2 + 3.5^2 - 2(6)(3.5)cos42 b^2 = 6^2 + 3.5^2 + 2(6)(3.5)cos42
*Sunday, February 17, 2013 at 8:37pm*

**Trig**

The diagonals of a parallelogram intersect at a 42◦ angle and have lengths of 12 and 7 cm. Find the lengths of the sides of the parallelogram. (Hint: The diagonals bisect each other.)
*Sunday, February 17, 2013 at 8:17pm*

**Trig Identities**

OHHH!!! Ok thankyou soooo much I greatly appreciate your help and one compliment...Your way better than my teacher...once again thankyou sooo much I think I get the jist of it now:)
*Sunday, February 17, 2013 at 12:48pm*

**small correction - Trig Identities**

My first line should have been: There is NO one correct and foolproof way to prove identities.
*Sunday, February 17, 2013 at 10:25am*

**Trig Identities**

There is on one correct and foolproof way to prove identities. There are some general rules you might follow 1. look for obvious relations , like sin^2 x + cos^2 x = 1 or 1 + tan^2 x = sec^2 x -- make yourself a summary of these collected from your text or notebooks 2. usually...
*Sunday, February 17, 2013 at 10:24am*

**Trig Identities**

Proving identities: 1) 1+ 1/tan^2x = 1/sin^2x 2) 2sin^2 x-1 = sin^2x - cos^2x 3) 1/cosx - cosx = sin x tan x 4) sin x + tan x =tan x (1+cos x) 5) 1/1-sin^2x= 1+tan^2 x How in the world do I prove this...please help... I appreciateyour time thankyou soo much!!
*Sunday, February 17, 2013 at 10:06am*

**geometry**

first off, trig functions are just numbers. they have no units (such as meters). Maybe those little m's are meant to be ° symbols. If sin(x) = 8/32, x = 14.48° did you visit the trig site I provided? http://www.rapidtables.com/calc/math
*Friday, February 15, 2013 at 4:04pm*

**geomtrey**

Handy trig calculators can be found at http://www.rapidtables.com/calc/math
*Friday, February 15, 2013 at 2:31pm*

**geomtrey**

the question probably says tanØ = 38 and sinØ = 8/32 a statement like sin = 8/32 is a "mathematical sin" sin is a trig operator and needs an argument that is like saying √ = 12 , another meaningless statement. anyway..... if tanØ = 38 , we ...
*Friday, February 15, 2013 at 2:28pm*

**geometry**

You are just looking for the hypotenuse of a right-angled triangle with height 27 and base angle of 60° using fundamental trig .... sin60° = 27/h, where h is the hypotenuse h = 27/sin60 = 31.1769 ft = 31 ft, 2 inches time =distance/rate = 31.1769/75 = .416 minutes ...
*Friday, February 15, 2013 at 1:05pm*

**Geometry**

Handy trig calculators can be found at http://www.rapidtables.com/calc/math just plug in your value for 22 degrees, get the sine, and then multiply by 11. Hard to believe that someone with access to a computer can't figure out how to get some calculations made. Heck, ...
*Friday, February 15, 2013 at 12:32pm*

**trig**

This is just a straightforward law of cosines problem. We want to find a, given A,b,c. A=58°, c=7.5, b=8.6 a^2 = b^2+c^2 - 2bc cosA plug and chug. . .
*Thursday, February 14, 2013 at 4:34pm*

**trig**

a lighthouse is located at point A. a ship travels from point B to point C. At point B,, the distance between the ship and the lighthouse is 7.5km. At point C the distance between the ship and the lighthouse is 8.6km. Angle BAC is 58 degrees. Determine the distance between B ...
*Thursday, February 14, 2013 at 2:16pm*

**Trig**

let's use a sine function .... amplitude = 20 period = 2π/k = 2 2k = 2π k = π so far we have height = 20 sin π(t + c) + d , assuming we have a phase shift and a vertical shift clearly, the whole sine curve has to shifted upwards by 24 units, so that the...
*Wednesday, February 13, 2013 at 9:41pm*

**Trig**

A Ferris wheel is 40 meters in diameter and boarded from a platform that is 4 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 2 minutes. How many minutes of the ride are ...
*Wednesday, February 13, 2013 at 8:30pm*

**MATHS**

Find the lengths of the missing sides in the triangle. Write your answers as integers or as decimals rounded to the nearest tenth. The diagram is not drawn to scale. triangle has sides 7, Y and X and 45 degree angle can someone show me how to do this? im supposed to use trig ...
*Wednesday, February 13, 2013 at 9:40am*

**trig**

y = 3cos(3x + pi) y = 3cos 3(x + π/3) amplitude : 3 period = 2π/3 phase shift: π/3 to the left
*Wednesday, February 13, 2013 at 9:26am*

**trig**

Given the function y = 3cos(3x + pi), identify: Amplitude (if applicable, give the answer in fractional form) Period (in radians as a multiple of pi - note: do not write "rad" or "radians" in your answer) Phase Shift (if the shift is right, enter + and if ...
*Wednesday, February 13, 2013 at 12:51am*

**Trig/precalc**

cos(x) cos(7pi/4) - sin(x) sin(7pi/4) + cos(x) cos(7pi/4) + sin(x) sin(7pi/4) = 1 2cos(x) cos(7pi/4) = 1 2cos(x) * 1/√2 = 1 cos(x) = 1/√2 x = pi/4 or 7pi/4
*Tuesday, February 12, 2013 at 11:03pm*

**Trig/precalc**

Cos[x + (7 Pi)/4] + Cos[x - (7 Pi)/4] = 1
*Tuesday, February 12, 2013 at 10:22pm*

**Trig**

Cos x
*Tuesday, February 12, 2013 at 5:47pm*

**trig**

Perhaps you recognized the 30-6-90 triangle (sinØ=1/2 , so Ø=30°) so just use a simple ratio: h/2 = 8/1 h = 16 or sinØ - .5 Ø = 30° then sin30 = 8/h h = 8/sin30 = 8/(1/2) = 16
*Tuesday, February 12, 2013 at 11:45am*

**trig**

ABC- with angles AB, and C and sides AB,BC, and AC, angle B is right 90degree angle, if sin of angle A is 0.5, side BC 8in., what is length of AC
*Tuesday, February 12, 2013 at 11:01am*

**trig**

For height at lower angle: tan 15°10' = height1/140 height1 = 140tan15°10' in the same way: height2 = 140tan29°30' rise of the balloon in that change of angle = 140tan29°30' - 140tan15°10' = 41.256 I am sure your units are not correct. ...
*Monday, February 11, 2013 at 10:36pm*

**trig**

As a hot-air balloon rises vertically, its angle of elevation from a point P on level ground d = 140 kilometers from the point Q directly underneath the balloon changes from 15°10' to 29°30' (see the figure). Approximately how far does the balloon rise during ...
*Monday, February 11, 2013 at 9:37pm*

**maths (trig)**

tangent of theta is opposite / adjacent. So your theta (angle) here is 50, your opposite is 27, and we don't know the adjacent. So do simple algebra & bring x to the other side by multiplying so you have tan50 * x = 27, now divide tan50 * x by tan50 to get x by itself. Now...
*Monday, February 11, 2013 at 2:05am*

**maths (trig)**

how would you work out tan 50=27/x please help
*Monday, February 11, 2013 at 1:56am*

**trig**

Did you mean solve cos(2Ø) = π/4 ?
*Sunday, February 10, 2013 at 9:34pm*

**trig**

what is the exact value of this equation. cos(20) if it equals pi/4
*Sunday, February 10, 2013 at 8:02pm*

**trig**

Two people decide to estimate the height of a flagpole. One person positions himself due north of the pole and the other person stands due east of the pole. If the two people are the same distance from the pole and a = 30 feet from each other, find the height of the pole if ...
*Sunday, February 10, 2013 at 12:23pm*

**Trigonometry**

You have no equations to solve, you probably meant "simplify the expressions" 1. sin^2 x (1/sin^2 x - 1) = 1 - sin^2 x = cos^2 x 2. (cosx/sinx) (1/cosx) = 1/sinx = cscx 3. recall the property of complementary trig ratios, that is... sin(π/2 - x) = cosx and cos...
*Sunday, February 10, 2013 at 7:28am*

**maths**

Think of a right triangle with the bottom leg 1 and the side leg 73. It is very tall and skinny. The base angle is very close to 90 degrees. If that angle is x, then tan(x) = 73. Go back to basics and start looking at right triangles to see how the trig functions are defined. ...
*Friday, February 8, 2013 at 4:38pm*

**maths**

so the answer is 89.22? sorry I hope I don't sound dumb but I don't understand trig AT ALL.
*Friday, February 8, 2013 at 4:34pm*

**maths**

just use your calculator for the arctan function. Or, go to http://www.rapidtables.com/calc/math/Arctan_Calculator.htm and enter 73 and get arctan(73) = 89.22° so, tan(89.22°) = 73 just think of the trig functions as another set of functions and inverses If I ...
*Friday, February 8, 2013 at 4:30pm*

**math-trig**

sec^2(t) = 1+tan^2(t) That help? If not, add some parentheses so we know just what is being discussed.
*Friday, February 8, 2013 at 2:31pm*

**math-trig**

tan(t)-sec^2(t)/tan(t)
*Friday, February 8, 2013 at 12:51pm*

**Trig**

Troll
*Thursday, February 7, 2013 at 9:35pm*

**trig**

4cos^{2}2x-1=cos4x
*Thursday, February 7, 2013 at 6:32am*

**trig**

First observation point A second observation point B top of tower P bottom of tower Q In triangle ABP, angle A = 70 --- given angle PBA = 95 --- exterior angle to 85° angle APB = 15° AB = 55m --- given by the sine law BP/sin70 = 55/sin15 BP = 55sin70/sin15 in triangle ...
*Tuesday, February 5, 2013 at 9:21am*

**trig**

a tower that is a 200 meters is leaning to one side. from a certain point on that side, the angle of elevation to the top of the tower is 70 degree. From a point 55 meters closer to the tower, the angle of elevation is 85 degree. Determine the acute angle from the horizontal ...
*Tuesday, February 5, 2013 at 8:33am*

**Precalc**

Your lack of brackets make your equations totally ambiguous. Secondly your trig ratios have no argument, sin and cos by themselves are meaningless that's like saying 5 + √ I happen to know the second one , so (1+ sinØ)/(1 - sinØ) = (secØ + tan&...
*Monday, February 4, 2013 at 7:03pm*

**trig**

1/tanβ + tanβ (1+tan^2 β)/tanβ sec^2 β/tan since sec^2 β = 1+tan^2 β
*Monday, February 4, 2013 at 4:58pm*

**trig**

1/tan beta +tan beta=sec^2 beta/tan beta
*Monday, February 4, 2013 at 4:50pm*

**trig Elev. & Depress part ii**

as usual, draw a diagram. If the tree has height h, (h-4.5)/55 = tan61°
*Monday, February 4, 2013 at 4:12pm*

**trig Elev. & Depress part ii**

You are 55ft from a tree. The angle of elevation from your eyes, which are 4.5 ft off the ground,to the top of the tree is 61 degrees. To the nearest foot, how tall is the tree?
*Monday, February 4, 2013 at 4:09pm*

**trig**

tan 8.4 = h/3.4 h =3.4 tan 8.4° = .......
*Friday, February 1, 2013 at 2:26pm*

**trig**

A mountain road makes an angle θ = 8.4° with the horizontal direction. If the road has a total length of 3.4 km, how much does it climb? That is, find h.
*Friday, February 1, 2013 at 1:57pm*

**trig**

cos(π/2 - Ø) = sinØ , complementary angle property then cos(+pi/2-theta)/csctheta+cos^2theta = cos(π/2-Ø) * 1/cscØ + cos^2 Ø = sinØ ( sinØ) + cos^2 Ø = sin^2 Ø + cos^2 Ø = 1
*Thursday, January 31, 2013 at 11:46pm*

**trig**

cos(+pi/2-theta)/csctheta+cos^2theta
*Thursday, January 31, 2013 at 10:49pm*

**nicole**

1. determine the exact value.. cot 7pi/6 and sec(-210degrees) 2. determine approximate measure of all angels that satisfy following, and draw a sketch to show the quadrants involved cosTHETA = -0.77, -pi lessOr same than theta less than pi and cscTHETA=9.5, -270Degrees less ...
*Wednesday, January 30, 2013 at 3:22pm*

**geometry**

basic trig ... tan 56° = height/7 height = 7tan56° = ......
*Tuesday, January 29, 2013 at 11:29pm*

**trig**

I don't know if you are supposed to do this by simply substiting, or actually solve the equation. I will solve it 2sinx + 4cos 2x = 3 2sinx + 4(1 - 2sin^2 x) -3=0 2sinx + 4 -8sin^2 x - 3 = 0 8sin^2 x - 2sinx -1 = 0 sinx = (2 ± √36)/16 = 1/2 or -1/4 x = 30°...
*Tuesday, January 29, 2013 at 10:41pm*

**trig**

To the nearest degree, all of the following angles are solutions of the equation 2sin x + 4 cos 2x =3 except: (1) 40 degrees (2) 150 degrees (3) 166 degrees (4) 194 degrees
*Tuesday, January 29, 2013 at 9:52pm*

**Trigonometry**

What is -6i in standard form? It is originally in trig form.
*Tuesday, January 29, 2013 at 12:13pm*

**math (trig.)**

You will need a phase shift your amplitude is correct at 17.5 your vertical shift of 17.5 is also correct period = 2π/k = 8 8k = 2π k = 2π/8 = π/4 so your k for the period is correct so let's adjust: height = 17.5 sin (π/4)(t + d) + 17.5 , where d ...
*Monday, January 28, 2013 at 8:25pm*

**math (trig.)**

I answered 17.5*sin((pi/4)*t)+17.5 but it's wrong I'm not sure what I'm doing incorrectly
*Monday, January 28, 2013 at 7:30pm*

**math (trig.)**

A ferris wheel is 35 meters in diameter and boarded at ground level. The wheel makes one full rotation every 8 minutes, and at time (t=0) you are at the 3 o'clock position and descending. Let f(t) denote your height (in meters) above ground at t minutes. Find a formula for...
*Monday, January 28, 2013 at 7:24pm*

**Trig**

not true, if you meant: tan(x/2) =( plus/minus) sqrt(1-cos(x/(1+cos(x))) let x = 2 radians LS = tan 1 = appr 1.557 RS = ± √(1 - cos( 2/(1 + cos 2) let's look at cos (2/(1+ cos2) = cos(2/.58385) = cos(3.4255) = -.95996 so RS = ± √( 1 - (-.95996...
*Sunday, January 27, 2013 at 4:00pm*

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