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April 19, 2014

April 19, 2014

**Recent Homework Questions About Trigonometry**

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**trig**

5 csc^2 Ø - 3cot^2 Ø = 2csc^2 Ø + 3 LS = 5/sin^2 Ø - 3cos^2 Ø/sin^2 Ø = (5 - 3cos^2 Ø)/sin^2 Ø = (5 - 3(1 - sin^2 Ø) )/sin^2 Ø = ( 2 + 3sin^2 Ø)/sin^2 Ø = 2/sin^2 Ø + 3sin^2 Ø/sin...
*Friday, July 13, 2012 at 11:30pm*

**trig**

establish identity 5csc^2theta-3cot^2theta=2csc^2theta+3
*Friday, July 13, 2012 at 10:00pm*

**trig**

cos^2t+2cos(t)+1=0 (cost + 1)^2 = 0 cost + 1 = 0 cost = -1 t = 270°
*Thursday, July 12, 2012 at 1:05pm*

**trig**

sin 4t = √3/2 you know sin 60° = √3/2, so 4t=60° or 120° t=15° or t=30° Now, to get all solutions in [0,360), add multiples of 2pi/4 = 90°: t = 15° + 90n° = 15,105,195,285 t = 30° + 90n° = 30,120,210,300
*Thursday, July 12, 2012 at 11:45am*

**trig**

Solve the equation on the interval [0,360) cos^2t+2cos(t)+1=0
*Thursday, July 12, 2012 at 10:51am*

**trig**

Find all solutions in the interval [0,360) sin(4t)=ã3/2
*Thursday, July 12, 2012 at 10:49am*

**TRIG**

And it is not even impish
*Wednesday, July 11, 2012 at 12:29pm*

**algebra, not trig**

CAPS LUK ST00PID, as do the contractions and typos. Assuming his present salary is the base salary, 24 * 1.10^2 = 29.04 Where can I get a job with an annual 10% raise?
*Wednesday, July 11, 2012 at 11:04am*

**TRIG**

cos theta = + OR - 1 / sqrt ( 1 + tan ^ 2 theta ) In Quadrant II, cosine are negative so : cos theta = - 1 / sqrt ( 1 + tan ^ 2 theta ) cos theta = - 1 / sqrt [ 1 + ( 3 / 4 ) ^ 2 theta ] cos theta = - 1 / sqrt ( 1 + 9 / 16 ) cos theta = - 1 / sqrt ( 16 / 16 + 9 / 16 ) cos ...
*Wednesday, July 11, 2012 at 10:48am*

**Airplane trig question**

the speed is irrelevant, since it does not change. So the relative times are the same as the relative distances. 90 min at 131° moves the plane 90(cos41°,-sin41°) = (67.92,-59.04) 30 min at 41° moves it an additional 30(cos41°,sin41°) = (22.64,19.68) So...
*Wednesday, July 11, 2012 at 10:46am*

**TRIG**

That was a good atempt bob but the options i have here is -2/35, 30.78, -24/35, AND 18
*Wednesday, July 11, 2012 at 10:34am*

**TRIG**

A WORKER PRESENT SALARY IS 24 PER ANNUM HIS ANUAL INCREMENT IS 10% OF HIS BASIC SALARY WAT WUD BE HIS ANNUAL SALARY AT THE BEGINNIN OF THE THIRD YEAR
*Wednesday, July 11, 2012 at 10:27am*

**TRIG**

see other post.
*Wednesday, July 11, 2012 at 10:22am*

**TRIG**

if tan is 3/4, then the right triangle is 3,4,5 so sin is 3/5, second quadrant.
*Wednesday, July 11, 2012 at 10:22am*

**TRIG**

Given that tan¤=3/4 and ¤ is in the second quadrant find sin2¤ (¤=theta)
*Wednesday, July 11, 2012 at 10:17am*

**TRIG**

Given that tan¤=3/4 and ¤ is in the second quadrant find sin2¤ (¤=theta)
*Wednesday, July 11, 2012 at 10:17am*

**trig**

rewrite it in the most general form y = -2 cos π(x - 3/π) amplitude = 2 period = 2π/π = 2 phase shift = 3/π units to the right
*Wednesday, July 11, 2012 at 7:14am*

**Trig**

sqrt ( 3 ) * csc ( 2 theta ) = - 2 Divide both sides by sqrt ( 3 ) csc ( 2 theta ) = - 2 / sqrt ( 3 ) Take the inverse cosecant of both sides. 2 theta = - pi / 3 and 2 theta = 4 pi / 3 [ Becouse csc ( pi / 3 ) = 2 / sqrt ( 3 ) , and csc ( 4 pi / 3 = 2 / sqrt ( 3 ) ] Divide ...
*Wednesday, July 11, 2012 at 2:09am*

**trig**

determine the amplitude,period and phase shift of y=-2cos(pix-3)
*Wednesday, July 11, 2012 at 12:19am*

**Airplane trig question**

An airplane, flying at a speed of 420 miles per hour, flies from point A in the direction 131 degrees for 90 min. and then flies in the direction 41 degrees for 30 min. in what direction does the plane need to fly in order to get back to point A?
*Tuesday, July 10, 2012 at 7:34pm*

**Trig**

find all solutions to the equation √3 csc(2theta)=-2 Would the answer be π/6 + 2πn, 5π/6 +2πn or π/6 + πn, 5π/6 +πn? or neither?
*Tuesday, July 10, 2012 at 7:30pm*

**trig**

sin t /(1-cos t) - (1+cos t)/sin t sin t (1+cos t)/((1-cos t)(1+cos t)) - (1+cos t)/sin t sin t (1+cos t)/(1-cos^2 t) - (1+cos t)/sin t sin t (1+cos t)/sin^2 t - sin t (1+cos t)/sin^2 t 0
*Tuesday, July 10, 2012 at 5:35pm*

**trig**

sin t /1-cos t - 1+ cos t/sin t= 0
*Tuesday, July 10, 2012 at 5:24pm*

**trig**

Prove sin 1 cos 1 cos sin 0 1. Show each step of your proof. 2. Provide written justification for each step of your proof. C. If you use sources, include all in-text citations and references in
*Tuesday, July 10, 2012 at 5:22pm*

**Trig**

Given y=2-4cos(3x-pi/4), find each of the following, giving the general representation of the location of all minimums. Also provide the graph of two full cycles, labeling everything. Domain = Range = Amplitude = Period = Ph. Shft. = Interval for one Complete cycle = ...
*Monday, July 9, 2012 at 4:24pm*

**Trig/Precal**

add the vectors. flying: 90sin310 E + 90 cos310 N windblowing: 40sin(332-180)E+40cos(332-180)N now add them result: East(90sin310+40sin152)+North(90cos310+40cos152) figure those, and you are near the end. Post your questions if you get lost from here.
*Friday, July 6, 2012 at 8:41pm*

**Trig/Precal**

An airplane flies on a compass heading of 90° at 310 mph. The wind affecting the plane is blowing from 332° at 40 mph. What is the true course and ground speed of the airplane?
*Friday, July 6, 2012 at 7:37pm*

**Alg/Trig**

(x-2)^2 + (y-3)^2 = 4.5^2
*Friday, July 6, 2012 at 1:17pm*

**Alg/Trig**

Write an equation of a circle with center (2, 3) and radius 4.5.
*Friday, July 6, 2012 at 1:13pm*

**trig HELP**

at each end, you have a triangle with base 15, and a height 15tan40°=12.58 so, each end has an area of 1/2 * 15 * 12.58 = 94.35 each half of the roof is 60 x 15sec40° = 60x19.58 = 1174.87
*Thursday, July 5, 2012 at 3:43pm*

**trig HELP**

A barn is 30ft. wide by 60ft long; the rafters make an angle of 40degrees with the horizontal. Find the area of each of the two gable ends and the area of the roof. Please Include Solution And If Possible An Illustration. Thank You In Advance :)
*Thursday, July 5, 2012 at 9:51am*

**Algebra & Trig**

Was there a picture given? I don't know if I can answer the question without a picture.
*Thursday, July 5, 2012 at 2:05am*

**Algebra & Trig**

under the influence of gravity? consider the maximum height, velocity vertical is zero. Vf=vi+gt 0=12sinTheta-9.8t t=12sinTheta/9.8 time in the air is twice that time. no, horizontal distance 13=12cosTheta*t=12cosTheta*2*12sinTheta/9.8 13=12^2/9.8 *2 cosTheta*sinTheta ...
*Wednesday, July 4, 2012 at 7:45pm*

**Algebra & Trig**

at a throwing speed pf 12.2 meters per seconds,what is the angle that produce a distance of 13 meters
*Wednesday, July 4, 2012 at 6:24pm*

**Algebra & Trig**

calculate the velocity of 10 meters per second with an angle of 30 degrees
*Wednesday, July 4, 2012 at 6:20pm*

**trig**

in general the period of y = tan (kØ) is π/k so for y = tan (2x) the period is π/2
*Wednesday, July 4, 2012 at 11:52am*

**trig**

determine the period of y = tan 2x
*Wednesday, July 4, 2012 at 10:47am*

**TRIG**

Hey, I ca help you with the first one as this is the only one I understand so far. Since you have pi*x in the tan, you have to take the pi and divide by the k of the function, which in this case is pi. This gets you a perio of one, meaning that there is a point at every single...
*Tuesday, July 3, 2012 at 1:38pm*

**trig**

Connecting the center of the circle with the 2 ends of the chord makes an isc.. triangle. Half of that is a right triangle. The angle at the center of the circle is 39.1Âº and 1/2 the chord is 59.2 cm. 59.2 = r*sin(12/2) solve for r.
*Tuesday, July 3, 2012 at 9:59am*

**trig**

Find the radius of a circle which a 59-foot chord subtends an angle of 12degrees at the center. I Just Want To See The Illustration. I Can Handle The Rest. Thank You In Advance =)
*Tuesday, July 3, 2012 at 9:51am*

**trig**

57/X = sin 24deg, so you are correct.
*Saturday, June 30, 2012 at 5:49am*

**trig**

using a right triangle with one angle 24 degrees and the opposite side being 57 feet, write an expression to fine the hypotenuse, X? X=57/sin(24) is this correct?
*Friday, June 29, 2012 at 11:47pm*

**trig**

8/cos40=10.44 inches check that
*Friday, June 29, 2012 at 8:37pm*

**trig**

a circular chimney pipe with an 8-inch diameter passes vertically from a heater through a metal plate in the roof. the plate has an elliptical hole that measures 8 inches along one axis. if the roof is slanted at an angel of 40 degrees with the horizontal, what is the length ...
*Friday, June 29, 2012 at 6:22pm*

**trig**

tanA = 2 in QIII means sinA = -2/√5 cosA = -1/√5 cosA/(sin^3A + cos^3A) = -1/√5 / (-8/5√5 - 1/5√5) = -1/√5 / (-9/5√5) = -1/√5 * -5√5/9 = 5/9 that's PI, not PIE!
*Friday, June 29, 2012 at 4:31pm*

**trig**

If tan A = 2 and A belongs to [pie,3pie/2] then the expression cos A divided by sin cubed A +cos cubed A is equal to what?
*Friday, June 29, 2012 at 12:41pm*

**trig**

yes, you should get an angle of 30.1°
*Friday, June 29, 2012 at 10:24am*

**trig**

the horizontal speed of an airplane is 173 miles per hour. the magnitude of the planes velocity is 200 miles per hour. what angle is the plane climbing? My answer is Cos-1 173/200. is this correct
*Friday, June 29, 2012 at 9:38am*

**trig**

looks like height = .5n + 800 , where n is the number of weeks
*Thursday, June 28, 2012 at 10:27am*

**trig**

A certain species of tree grows an average of 0.5 cm per week. Write an equation for the sequence that represents the weekly height of this tree in centimeters if the measurements begin when the tree is 800 centimeters tall?
*Thursday, June 28, 2012 at 8:10am*

**trig**

to prove (sec8A-1)/(sec4A-1) = tan8A/tan2A Not sure how to do this with sec and tan only, but using sin and cos, we have sec8A-1 = (1-cos8A)/cos8A = (1-(2cos^2(4A)-1))/cos8A = 2(1-cos^2(4A))/cos8A sec4A-1 = (1-cos4A)/cos4A so, the LS is 2cos4A(1-cos^2(4A))/(2cos^2(4A)-1)*(1-co...
*Wednesday, June 27, 2012 at 2:20pm*

**trig**

sec8A-1/sec4A-1=tan8A/tan2A
*Wednesday, June 27, 2012 at 12:53pm*

**Trig**

If the height is h, and P is x m from the base, h/x = tan 27°10' = .5132 h/(x-23) = tan 51°30' = 1.2572 rearranging things a bit, h = .5132x = 1.2572(x-23) x = 38.8651 so, h = 19.9456, or 19.9m
*Tuesday, June 26, 2012 at 11:32pm*

**Trig**

From a point P on level ground, the angle of elevation of the top of a tower is 27°10'. From a point 23.0 meters closer to the tower and on the same line with P and the base of the tower, the angle of elevation of the top is 51°30'. Approximate the height of ...
*Tuesday, June 26, 2012 at 11:14pm*

**trig**

I will try to guess what you mean .... If π/3 is a rotation angle on a unit circle (60°) then 14π/3 = 4 + 2π/3 or 4 rotations + 2π/3 this would put you into quadrant II (120°) a point with angle π/3 and a point with angle 2π/3 has the same...
*Sunday, June 24, 2012 at 8:21am*

**trig**

explain how you will find the coordinate of P(14pi/3) if you knew the coordinate of P(pi/3)
*Sunday, June 24, 2012 at 3:09am*

**trig**

given cot-A square root of 5, find cot-B at the 4th quadrant.
*Saturday, June 23, 2012 at 9:10pm*

**Trig**

Sketch the graph through at least one complete cycle. Label the axes on the coordinate plane. Use your graph to predict the height of point P above the surface of the water when the stopwatch shows 7 seconds. Round to the nearest tenth of a foot. (Enter only the number.)
*Thursday, June 21, 2012 at 6:11pm*

**Trig**

P was at its maximum after 5 seconds. Use this information to write the value of the last constant, h. (If the value is a fraction, use a slash to separate the numerator from the denominator without spaces, such as 3/4.)
*Thursday, June 21, 2012 at 6:09pm*

**Trig**

6 + sin k x 6 feet
*Thursday, June 21, 2012 at 5:17pm*

**Trig**

Since the center is 6 feet above the water, the horizontal line of symmetry of the sine curve will be 6 units above the x-axis. This information helps you find the value of vertical shift. Write its value.
*Thursday, June 21, 2012 at 5:15pm*

**Trig**

sin theta = 4.7/25 so theta = sin^-1 (4.7/25) 10.8 degrees
*Thursday, June 21, 2012 at 5:12pm*

**Trig**

Jessica attains a height of 4.7 feet above the launch and landing ramps after 1 second. Her initial velocity is 25 feet per second. To find the angle of her launch, which equation can you use with the given information to solve for θ? (Answer: 1 or 2). Use the equation ...
*Thursday, June 21, 2012 at 5:02pm*

**trig**

We form a rt triangle: Y = 150 Ft. = Ver. side. X = Hor. side = Dist. from bottom of tower to point where wire attaches to gnd, Z = hyp. = Length of wire. Z = Y/sin40 = 150 / sin40 = 233.4 Ft. X = Z*cos40 = 233.4*cos40178.8 Ft.
*Wednesday, June 20, 2012 at 9:01pm*

**trig**

A guy wire runs from the ground to a cell tower. The wire is attached to the cell tower 150 feet above the ground. The angle formed between the wire and the ground is 40° (see figure). (Round your answers to one decimal place.)
*Wednesday, June 20, 2012 at 3:46pm*

**trig**

After you make your sketch you should see that tanØ = 5/14 Ø = 19.65°
*Wednesday, June 20, 2012 at 2:00pm*

**trig**

Two forces of 5 lb. and 14 lb. act on a body at right angles to each other. Find the angle their resultant force makes with the force of 14 lb
*Wednesday, June 20, 2012 at 1:50pm*

**geometry**

looks like a straight-forward right-angled triangle trig question. tan75° = x/145 x = 145tan75 = 541.15
*Wednesday, June 20, 2012 at 1:28pm*

**trig**

tan = rise/run = 55/2640 = 1/48
*Tuesday, June 19, 2012 at 2:42pm*

**trig**

A roadway rises 55ft in horizontal distance of 1/2 mile (1mile=5280ft) Find the tangent of the angle that it makes with the horizontal.
*Tuesday, June 19, 2012 at 12:26pm*

**math**

Looks like "homework or assignment dumping" to me Exactly what problems do you have with this list of trig questions? I will do this one.... show me some attempts at the others LS = cosx(2tan^2 x + 5tanx + 2) = 2cosx sin^2 x/cos^2 x + 5cosxsinx/cosx + 2cosx = 2sin^2 ...
*Monday, June 18, 2012 at 9:45am*

**Algebra**

Using similar triangles and ratios, we can find the height of the tower. h/60 = 3/2 h = 90 then by Pythagoras: let the length of the wire be x x^2 = 60^2+90^2 x = √11700 = appr 108.2 ft or using trig, let the angle formed by Ø tanØ = 3/2 Ø = 56.31&...
*Friday, June 15, 2012 at 11:45pm*

**trig**

Use the laws of scalar multiplication of vectors: A=(x,y) aA=(ax,ay) u=(10,5) -u=(-10,-5) -3u=(-30,-15)
*Friday, June 15, 2012 at 2:26pm*

**trig**

Give the component form of the resultant vector in the following. NOTE: Answer must be typed in using the following format -- including the parentheses: (#,#). u =(10, 5) -3u=
*Friday, June 15, 2012 at 11:15am*

**trig**

s = rθ = 18(44*pi/180) = 13.82ft in one arc It makes 10 (5 back and forth) swings every minute. Total travel: 138 ft
*Thursday, June 14, 2012 at 1:06pm*

**trig**

A pendulum is 18 feet long. Its central angle is 44º. The pendulum makes one back and forth swing every 12 seconds. Each minute, the pendulum swings _____ feet. (Answer to the nearest foot.)
*Thursday, June 14, 2012 at 12:59pm*

**trig**

the angle is appr 15 degrees You don't state where the sides A and H are, but once you have the angle I am sure you can find the other side.
*Wednesday, June 13, 2012 at 5:01pm*

**trig**

Use a table of trigonometric values to find the angle è in the right triangle in the following problem. Round to the nearest degree, if necessary. cos è =0.9659 A=? H=20
*Wednesday, June 13, 2012 at 4:44pm*

**Trig**

If csc Ø = 13/12, the sinØ = 12/13 so we are dealing with the 12-5-13 right-angled triangle in II then cosØ = -5/13 sin 2Ø = 2sinØcosØ = 2(12/13)(-5/13) = -120/169 cos 2Ø = cos^2 Ø - sin^2 Ø = 25/169 - 144/169 = -...
*Wednesday, June 13, 2012 at 7:58am*

**Trig**

Find sin2theta, cos2theta if csctheta=13/12 and lies in quadrant II. Thank you!
*Wednesday, June 13, 2012 at 2:06am*

**trig**

hjk.hjk.hkl;
*Tuesday, June 12, 2012 at 9:06am*

**Trig**

base of triangle --- x inches side of triangle ---- 4x inches 4x + 4x + x = 16 9x = 16 x = 16/9 the base is 16/9 inches long base = 16/9 each side = 64/9 check: 16/9+64/9+64/9 = 144/9 = 16
*Monday, June 11, 2012 at 8:28pm*

**Trig **

The length of the base of an isosceles triangle is one fourth the length of one of its legs. If the perimeter of the triangle is 16 inches, what is the length of the base?
*Monday, June 11, 2012 at 6:15pm*

**trig/math**

Determine the period, amplitude and phase shift for each given function: A)y = -4 cos 3x + 5 B)y = 2/3 sin (30x-90)-10 c)y = -0.38 tan (x/3+pi/3) d)y = pi cos(2x)+ pi
*Monday, June 11, 2012 at 5:17pm*

**Math/Trig**

sin(-θ) = -sinθ sin(θ±2kπ) = sinθ so, sin(-13π/6) = sin(-π/6) = -1/2 tan(θ±kπ) = tanθ -- same for cot cot(11π/6) = cot(-π/6) = -cot(π/6) = -√3 cot(-14π/4) = cot(2π/4) = cot(π/2...
*Monday, June 11, 2012 at 11:29am*

**Math/Trig**

How would I evaluate these trig functions without using a calculator? U: sin(-13π/6) cot 11π/6 cot(-14π/4) sec 23π/6 Thanks in advance ^^; and if you'd tell me step by step on how to do problems like these I'd be grateful :D
*Monday, June 11, 2012 at 5:03am*

**Calculus**

integral of cscx^(2/3)(cot^3)x i know that cot^2x is csc^2(x)-1, but i just don't understand how to solve the cscx^(2/3), any help? i also know that its trig integrals/substitution...
*Sunday, June 10, 2012 at 12:05pm*

**trig**

I answered that question for you last night. http://www.jiskha.com/display.cgi?id=1338941277 What didn't you like about the answer?
*Wednesday, June 6, 2012 at 1:59pm*

** trig**

A tunnel for a new highway is to be cut through a mountain that is 260 feet high. At a distance of 200 feet from the base of the mountain, the angle of elevation is 36 degrees. From a distance of 150 feet on the other side of the mountain, the angle of elevation is 47 degrees...
*Wednesday, June 6, 2012 at 1:36pm*

**Coterminal Angles**

lol this was one of my final questions for trig/precal
*Tuesday, June 5, 2012 at 10:50pm*

**trig**

I drew triangle ABC where BC is a horizontal (the road) and angle B is 36° , and is on the left side of the diagram. Angle C = 47° , on the right side D is on BC and is the altitude of the mountain of 260 feet I let BD = 200+x , and DC = 150+y in the left triangle, tan...
*Tuesday, June 5, 2012 at 8:33pm*

** trig**

A tunnel for a new highway is to be cut through a mountain that is 260 feet high. At a distance of 200 feet from the base of the mountain, the angle of elevation is 36 degrees. From a distance of 150 feet on the other side of the mountain, the angle of elevation is 47 degrees...
*Tuesday, June 5, 2012 at 8:07pm*

**trig.**

It is not Sin 90° = 1 sin 90 radians = appr .89699...
*Wednesday, May 30, 2012 at 11:45pm*

**trig.**

How is it that sin90 is negative?
*Wednesday, May 30, 2012 at 9:35pm*

**trig**

I read that as sin (x/2) = √2 - sin(x/2) 2sin(x/2) = √2 sin(x/2) = √2/2 x/2 = 45° or 135° x = 90° or x = 270°
*Wednesday, May 30, 2012 at 7:42am*

**trig**

I have a suspicion that you meant 2tanx/(3 - tan^2x) = 1 then 2tanx = 3 - tan^2x tan^2x + 2tanx - 3 = 0 (tanx + 3)(tanx - 1) = 0 tanx = -3 or tanx = 1 if tanx = -3, x must be in II or IV by the CAST rule x = 108.4° or 288.4° if x = 1, then x must be in I or III x = 45&...
*Wednesday, May 30, 2012 at 7:39am*

**Trig**

For terms from 1 to 40 you have an arithmetic series, where a = 25, d=2 and n = 40 sum of those seats = (40/2)(2(25) + 39(2) ) = 2560 then there are 35 rows of 125 seats each or 4375 seats so we have 6935 seats
*Wednesday, May 30, 2012 at 7:18am*

**trig**

solve equation for the exact solutions if possible put answer in degrees. 2tanx/3-tan^2x=1
*Wednesday, May 30, 2012 at 7:04am*

**trig**

solve equation for exact solution if possible leave answer in degree sinx/2=square root of 2- sinx/2
*Wednesday, May 30, 2012 at 6:59am*

**Trig**

The first 40 rows of seating in a certain section of a stadium have 25 seats, 28 seats, 31 seats, and so on. The 41st through the 75th rows each contain 125 seats. Find the total number of seats in this stadium section.
*Wednesday, May 30, 2012 at 12:18am*

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