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April 18, 2014

April 18, 2014

**Recent Homework Questions About Trigonometry**

Post a New Question | Current Questions

**Precalculus with Trig**

Rewrite the following expression as an algebraic expression in x: cos(arcsin(x))
*Sunday, April 7, 2013 at 1:38pm*

**11th grade math**

If you have trig as a resource, then the altitude h from F to DE is h = 10 sin 50° = 7.66 Then, the area a = 14*h
*Sunday, April 7, 2013 at 1:07pm*

**trig**

evidently 195 = P-45
*Saturday, April 6, 2013 at 4:00pm*

**trig**

If it is given that tan(A-B)=tanA-tanB/1+tanAtanB and tanP-1/1+tanP=tan195 find p
*Saturday, April 6, 2013 at 3:44pm*

**trig**

Indoor steps generally are steeper and have smaller landings. A staircase has a 9 inch vertical rise per step to get to the second floor of a house. If the angle of elevation is 36.8 degrees and there are 16 feet of horizontal space total for each of the landings combined, ...
*Wednesday, April 3, 2013 at 1:34pm*

**Trig**

csc^2 = 1+cot^2, so you have sin^2 b csc^2 b = 1
*Wednesday, April 3, 2013 at 11:04am*

**Trig**

Sin^2 b(1 +cot^2 b) =1
*Wednesday, April 3, 2013 at 10:14am*

**calculus**

converting your directions to standard trig notation (x-axis is zero, counterclockwise) S 30° W ----> 240° S 20° E ----> 290° 300 km/h in S30W ---> vector(400cos240,400sin240) 50 km/h in S20E --> vector (50cos290 , 50sin 290) resultant = (400cos240...
*Tuesday, April 2, 2013 at 4:28pm*

**trig**

I have a triangle with sides 60 and 400 and the contained angle is 70° let the resultant be R R^2= 60^2 + 400^2 - 2(60)(400) cos70° = 147183.0331 R = 383.64 mi/h finding the angle opposite the 70° sinØ/60 = sin70/R sinØ = .14696.. Ø = 8.45°...
*Monday, April 1, 2013 at 9:53pm*

**trig**

An airplane pilot wishes to maintain a true course in the direction 250° with a ground speed of 400 mi/hr when the wind is blowing directly north at 60 mi/hr. Approximate the required airspeed and compass heading
*Monday, April 1, 2013 at 9:33pm*

**Trig**

I will assume you mean 1/(1-sinx) + 1/(1+sinx) = 2 sec^2 x LS = (1 + sinx + 1 - sinx)/((1-sinx)(1+sinx)) = 2/(1-sin^2 x) = 2/cos^2 x = 2sec^2 x = RS
*Sunday, March 31, 2013 at 9:41pm*

**Trig**

Verify that each trigometric equation is an identity. 1/1-sinx + 1/1+sinx = 2 sec^2x
*Sunday, March 31, 2013 at 9:16pm*

**math**

sketch a right-angled triangle in quadrant I , labeling the two acute angles A and B then A+B = 90° or π/2 and B = (90°-A) or (π/2 - A) label the horizontal x and the vertical side y sinA = y/r cosB = y/r so sinA = cosB = cos(π/2 - A) or in the ...
*Sunday, March 31, 2013 at 8:44pm*

**Math: Solving Trig Equation**

there's no equation the expression is equivalent to (9tan^2-9sec^2)/tan and since sec^2-tan^2=1, that's just -9/tan
*Sunday, March 31, 2013 at 5:04am*

**Math: Solving Trig Equation**

9tan (x) - 9 sec^2 (x)/ tan (x)
*Sunday, March 31, 2013 at 1:16am*

**math**

1. sin (π/6) csc (π/6) = sin (π/6) * 1/sin (π/6) = 1 2. sec(π/3)cos(π/3) + tan(π/3)cot(π/3) I think for both questions, they want you to realize that if you multiply a trig function by its reciprocal function you get 1 = 1 + 1 = 2
*Saturday, March 30, 2013 at 2:26pm*

**Trig**

Thank you very much!
*Thursday, March 28, 2013 at 5:20pm*

**Trig**

note that the problem was sin^2(67.5)-cos^2(67.5) = -cos(135) my hint was not intended to make you forget what was originally asked. :-) And, I know lots of folks hate roots in the denominator, but I see no reason why √2/2 is better than 1/√2. Just mt preference
*Thursday, March 28, 2013 at 3:20pm*

**Trig**

How exactly do you get there though? I tried to follow a similar homework problem and got: = (cos^2)*(67.5deg) - (sin^2)*(67.5deg) = cos(2*67.5deg) = cos135deg = -√2/2
*Thursday, March 28, 2013 at 3:13pm*

**Trig**

well, cos 2x = cos^2 x - sin^2 x, so what do you think? If you end up with 1/√2 you are correct.
*Thursday, March 28, 2013 at 1:49pm*

**Trig**

Simplify (sin^2)*(67.5deg) - (cos^2)*(67.5deg) and then evaluate exactly.
*Thursday, March 28, 2013 at 1:46pm*

**Trig**

Since sin = -12/13 in QIV, cos = 5/13 sec x/2 = 1/cos(x/2) = √(2/(1+cosx)) = √(1/(1+5/13)) = √(2/(18/13)) = √(26/18) = √13/3
*Thursday, March 28, 2013 at 1:29pm*

**Trig**

Use a half-angle identity to find the exact value for sec(x/2) if sinx = -12/13 ; 3pi/2 < x < 2pi
*Thursday, March 28, 2013 at 12:27pm*

**trig**

swagger.
*Wednesday, March 27, 2013 at 9:13pm*

**Katlyn Krystien - Math**

Please stick to one name, don't switch names from one post to another. this question is similar to the one you posted as Katlyn since the radius is 7 m, a = 7 remember : period = 2π/k , where k is the coefficient of your trig function ... sin (kt) so 2π/k = 16 ...
*Monday, March 25, 2013 at 7:05pm*

**Trig**

draw the circle and the line at the given angle. The intersection will give you a point (x,y). in this case, the line is at an angle of 45 degrees, in QIII, where y=x. tan (-3pi/4) = (-1√2)/(-1/√2) = 1
*Monday, March 25, 2013 at 5:09pm*

**Trig **

How do I use the unit circle to evaluate tan(-3pi/4)
*Monday, March 25, 2013 at 5:03pm*

**trig**

Find the first two terms a_1 =5 and an=3a _n-1+2
*Sunday, March 24, 2013 at 9:50pm*

**Trig**

tanA = sinA/cosA so tanA / sinA = (sinA/cosA)/sinA = 1/cosA = secA
*Sunday, March 24, 2013 at 8:17pm*

**Trig**

prove that tanA divided by the sinA equals the secA
*Sunday, March 24, 2013 at 8:02pm*

**Trig needs help**

Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question.
*Saturday, March 23, 2013 at 1:14pm*

**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, March 21, 2013 at 1:25pm*

**mathematics**

I assume you want "exact" answers for these, the usual kind of request for these types of questions. All your given angles are combinations of the standard angles of 30, 45, 60 etc e.g. 75 = 45+30 15 = 45-30 105 = 60+45 You should also have memorized or have quick ...
*Thursday, March 21, 2013 at 9:49am*

**trig**

make a sketch labeling the height of the mountain as PQ, with P as the top of the mountain. Label the 2 points of observation as A and B, with B the closer point, In triang ABP, angle A = 35°15' angle ABP = 180 - 50°25' = 129°35' then angle APB = 15°...
*Wednesday, March 20, 2013 at 3:29pm*

**trig**

Sandra wants to find the height of a moutain. From her first location on the ground, she finds the angle of elevation to the top of the mountain to be 35 degrees 15 minutes. After moving 1000 meters closer to the moutain on level ground, she finds the elevation to be 50 ...
*Wednesday, March 20, 2013 at 1:28pm*

**Trig**

A is in QIII, so cosA = -2/3 tan(A/2) = (1-cosA)/sinA = (5/3)/(-√5/3) = -√5
*Wednesday, March 20, 2013 at 11:32am*

**Trig**

If 180<A<270 and sinA= -radical(5)/3, find tan(1/2)A.
*Tuesday, March 19, 2013 at 11:36pm*

**trig**

I keep trying to find the power reducing formula for sin^4(x), but I can't seem to get all the fractional parts correct. The answer I should be getting is: sin^4(x)=(1/8)cos4x-(1/2)cos2x+(3/8) I can only get this far knowing feeling confident. When I go further I start ...
*Tuesday, March 19, 2013 at 9:57pm*

**Math**

What is the inverse trig function cotangent of square root of 3/3? What is the value?
*Tuesday, March 19, 2013 at 5:19pm*

**Trig**

True or False: For a trignometric function, y = f(x), then x = f^-1(y). Explain your answer. Thanks.
*Tuesday, March 19, 2013 at 4:51pm*

**Trig**

True or False: For a one-to-one function, y = f(x),then x = F^-1(y). Explain your answer. Please help.
*Tuesday, March 19, 2013 at 4:46pm*

**Trig**

draw the triangles and you will see that tan(s) = 12/5 tan(t) = 4/3 Now just plug in your sum formula: tan(s+t) = (tan s + tan t)/(1 - tan s * tan t)
*Tuesday, March 19, 2013 at 3:28pm*

**Trig**

What is the exact value of tan(s+t) if sin(s+t)=56/65. And cos(s)=5/13 and sin(t)=4/5. Everything is in the first quadrant
*Tuesday, March 19, 2013 at 3:22pm*

**Trig**

A pulley with a diameter of 24 inches is driven by a belt which is moving 1045 ft/min. How many revolutions per minute are made by the pulley?
*Tuesday, March 19, 2013 at 3:16pm*

**Trig**

13sinx+5=0 sinx = -5/13 sinx is negative in QIII,QIV, but we want the QIII value. so, cosx = -12/13 sin2x = 2sinx cosx = 2(-5/13)(-12/13) = 120/169
*Monday, March 18, 2013 at 3:53pm*

**trig**

sin π/6 = 1/2 cos π/6 = √3/2 so, cos π/12 = √(1+√3/2) / 2 sin π/24 = √(1-cos π/12) / 2 = 1/2 √(1 - (√(1+√3/2) / 2)) = 1/2√2 √(2 - √(1+√3/2)) = 1/4 √(2√2 - √(1+√3...
*Monday, March 18, 2013 at 9:50am*

**Trig**

How do I find the exact value of sin (pi/24)?I am crying
*Sunday, March 17, 2013 at 6:03pm*

**trig**

Do I subtract ?
*Sunday, March 17, 2013 at 5:51pm*

**trig**

I really need help . I find this hard Steve .
*Sunday, March 17, 2013 at 5:43pm*

**trig**

I applied the half angle formula for sin and I got the square root of 2 minus the square root of 3 divided by 2 . and for cos I got thesqurof 2 plus thesqr of3divided by 2 .
*Sunday, March 17, 2013 at 5:40pm*

**trig**

I am not under standing could you show me ? to tell me ?
*Sunday, March 17, 2013 at 5:36pm*

**trig**

sin π/6 = 1/2 cos π/6 = √3/2 sin(x/2) = √(1-cosx) / 2 cos(x/2) = √(1+cosx) / 2 apply the half-angle formula twice to get sin π/24 = 1/2 √(2-√(2+√3))
*Sunday, March 17, 2013 at 5:24pm*

**trig**

How do I find the exact value of sin (pi/24)?
*Sunday, March 17, 2013 at 5:11pm*

**Trigonometry**

Hi, I really need help in understanding how to solve trig equations. How do you solve this equation? Solve the equation on the interval 0 < or equal to x < or equal to 2pi. sinx = (square root of 3)/2 I appreciate your help. Thank you!
*Sunday, March 17, 2013 at 2:18pm*

**trig**

h/100 = sin 40°
*Friday, March 15, 2013 at 10:30am*

**trig**

A pole is braced with a wire from the top of a pole to the ground. The wire is 100 feet long and makes an angle of 40° with the ground. Find the height of the pole.
*Friday, March 15, 2013 at 10:03am*

**trig**

Draw a diagram. If the nearer point is a distance x from the base of the mountain, then the height h can be figured from h/x = tan 43° h/(x+235) = tan 30° x = h/tan 43° x = h/tan 30° - 235 h/tan 43° = h/tan 30° - 235 h = 235/(cot30° - cot43°) h...
*Thursday, March 14, 2013 at 12:42pm*

**trig**

A road runs from the base of a mountain. From two points 235 meters apart on the road, the angles of elevation to the top of the mountain are 43 and 30. how high above the road is the mountaintop?
*Wednesday, March 13, 2013 at 10:45pm*

**trig**

A road runs from the base of a mountain. From two points 235 meters apart on the road, the angles of elevation to the top of the mountain are 43 and 30. how high above the road is the mountaintop?
*Wednesday, March 13, 2013 at 10:29pm*

**trig**

A road runs from the base of a mountain. From two points 235 meters apart on the road, the angles of elevation to the top of the mountain are 43 and 30. how high above the road is the mountaintop?
*Wednesday, March 13, 2013 at 10:29pm*

**trig**

2 csc 2θ cos 2θ = 2cot 2θ = 2/tan 2θ = 2(1-tan^2 θ)/(2tanθ) = 1/tanθ - tan^2θ/tanθ = cotθ - tanθ
*Wednesday, March 13, 2013 at 5:19pm*

**Trig**

tan34 = Y/X = Y/37. Y = 37*tan34 = 25 Ft. Ht. = Y + 5.3 = 25 + 5.3 = 30.3 Ft.
*Wednesday, March 13, 2013 at 4:39pm*

**trig**

no figure. Go figure.
*Wednesday, March 13, 2013 at 4:23pm*

**trig**

cos è = 0.9659 A = ? H = 20
*Wednesday, March 13, 2013 at 4:19pm*

**trig**

A math tutor told me to FOIL it. This is just a hint, or starting point because I don't know for sure.
*Tuesday, March 12, 2013 at 7:01pm*

**Trig**

so, what's the problem? Divide the equation in two and make new equations.
*Tuesday, March 12, 2013 at 3:43am*

**Trig**

a. Form a pair of simultaneous equations by letting y1 equal the left side and y2 equal the right side of sqrt 5-x=1. b.Repeat part (a) with the equivalent equation sqrt 5=x+1 c. Repeat part (a) with the equivalent equation sqrt 5-x-1=0
*Tuesday, March 12, 2013 at 1:56am*

**math**

My sketch looked like this: Side view, A vertical wall, T is the top and V is the bottom of the TV, Q is the point along a horizontal line level with her eyes, which I called P (we want that distance PQ) so we have TV = 30, and VQ = 6 Let PQ = x let angle TPV = A ---. that'...
*Tuesday, March 12, 2013 at 12:08am*

**Maths**

right-angled triangle trig . tan 28° = 76/x x = 76/tan28 = appr 142.9 m
*Monday, March 11, 2013 at 10:13am*

**Trig**

two sin x < 0 in QIII, QIV
*Monday, March 11, 2013 at 2:39am*

**Trig**

Whats the height of a flagpole if a student stand 37feet from it and determines the angle of elevation to be 34degrees and her eyes are 5.3 feet from the ground(round to the nearest whole number)
*Monday, March 11, 2013 at 2:06am*

**Trig**

the number of solutions of sin x= -sqrt3/2 for x between 0 and 2pi
*Monday, March 11, 2013 at 2:02am*

**Trig**

assuming the angle is to the top of the pole, the height h = 5.3 + 37 tan 34°
*Monday, March 11, 2013 at 1:16am*

**Trig**

the height of a flagpole if a student stands 37 feet from it and determines the angle of elevation to be 34 degrees and her eyes are 5.3 feet from the ground.
*Monday, March 11, 2013 at 12:46am*

**Trigonometry**

g(3) = 5(3)+1= 16 h(g(3)) = h(16) = 2(16-7) = 18 why do you call this trig ?
*Sunday, March 10, 2013 at 5:27pm*

**trig**

An airplane flying at 550 mph has a bearing of N 58 E. After flying for 1.5 hours, how far north and how far east has the plane traveled from its point of departure?
*Wednesday, March 6, 2013 at 7:54pm*

**trig**

182 * pi/180
*Wednesday, March 6, 2013 at 3:58pm*

**trig **

The radian measure of an angle of 182 degrees is ?
*Wednesday, March 6, 2013 at 3:43pm*

**trig**

cos(2x) = cos(x+x) = cosx cosx - sinx sinx = cos^2(x) - sin^2(x) = cos^2(x) - (1-cos^2(x)) = 2cos^2(x) - 1
*Monday, March 4, 2013 at 3:29pm*

**trig**

use an angel sum identity to verify cos2theta = -2cos^@-1
*Monday, March 4, 2013 at 3:11pm*

**Trig**

help please!: (sinx + sin2x +sin3x) / (cosx + cos2x + cos3x) = tan2x
*Sunday, March 3, 2013 at 9:05pm*

**trig**

cos(2x+4h)-cos(2x+2h)=?
*Sunday, March 3, 2013 at 8:23pm*

**trig**

cos(2x+4h)-cos(2x+2h)=?
*Sunday, March 3, 2013 at 8:23pm*

**trig**

cos(2x+4h)-cos(2x+2h)
*Sunday, March 3, 2013 at 8:22pm*

**Algebra and Trigonometry**

Trig - Damon, Sunday, March 3, 2013 at 6:24pm East speed = 100 + 200 sin 45 = 241 North speed = 200 cos 45 = 141 speed = sqrt( 141^2 + 241^2) = 279 mph tan east of north = 241/141 so angle east of north = 60 degrees
*Sunday, March 3, 2013 at 6:28pm*

**Trig**

East speed = 100 + 200 sin 45 = 241 North speed = 200 cos 45 = 141 speed = sqrt( 141^2 + 241^2) = 279 mph tan east of north = 241/141 so angle east of north = 60 degrees
*Sunday, March 3, 2013 at 6:24pm*

**Trig**

A plane if flying at 200 mph with a heading of 45 degrees and encounters a wind of 100 mph from the west. What is the resulting velocity and heading?
*Sunday, March 3, 2013 at 6:13pm*

**trig**

10sinB=sinB True if sinB=0 so, where does that happen?
*Friday, March 1, 2013 at 4:17pm*

**Trig needs help**

Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question.
*Friday, March 1, 2013 at 1:05pm*

**trig**

10sinB+5 = sinB+5
*Friday, March 1, 2013 at 12:51pm*

**trig**

make a sketch of your triangle since sinØ = opposite /hypotenuse = 5/13 opposite = 5 and hypotenuse = 13 by Pythagoras: x^2 + 5^2 = 13^2 x^2 +25 = 169 x^2 = 144 x = ± 12 , but in quadrant II, x = - 12 sinØ = 5/13 , cscØ = 13/5 cosØ = -12/13...
*Thursday, February 28, 2013 at 11:10pm*

**trig**

If sin theta is equal to 5/13 and theta is an angle in quadrant II find the value of cos theta, sec theta, tan theta, csc theta, cot theta.
*Thursday, February 28, 2013 at 10:44pm*

**Trig**

N32.47E
*Thursday, February 28, 2013 at 8:05pm*

**trig**

Incomplete.
*Thursday, February 28, 2013 at 3:40pm*

**trig**

well, what's arctan(-28/45?) -31.9 degrees. So, That's 2x You didn't specify quadrant, but tan x is negative in QII and QIV
*Thursday, February 28, 2013 at 10:16am*

**trig**

solve tan2x = -28/45 Thank you
*Thursday, February 28, 2013 at 10:11am*

**trig**

A population of wolves in a country is represented by the equation p(t)=80(0.98)^t, where t is the number of years since 1998. Predict the number of wolves in the population in the year 2008. how many years will it take for the population of wolves to reach 500?
*Tuesday, February 26, 2013 at 8:15pm*

**trig**

A population of wolves in a country is represented by the equation p(t)=80(0.98)^t, where t is the number of years since 1998. Predict the number of wolves in the population in the year 2008. how many years will it take for the population of wolves to reach 500?
*Tuesday, February 26, 2013 at 8:14pm*

**Trig**

You have two wires of length 50 and two wires of length 10√13 add them up
*Tuesday, February 26, 2013 at 10:42am*

**Trig**

Four wires support a 40-meter radio tower. Two wires are attached to the top and two are attached to the center of the tower. The wires are anchored to the ground 30-meters from the base of the tower. What is the total length of wire needed?
*Tuesday, February 26, 2013 at 10:36am*

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