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April 19, 2014

Homework Help: Math: Trigonometry

Recent Homework Questions About Trigonometry

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trig
the other long side is 10sqrt2 sinTheta=-10/10sqrt2=-.707 tantheta=-10/10=-1 you do the rest, if necessary, I can check them.
Saturday, April 27, 2013 at 2:34pm

trig
solve the point given below is on the terminal side of an angle0. find the exact value of the six trigonometric functions of 0. (10,-10)
Saturday, April 27, 2013 at 2:27pm

trig
how many solutions (answers) does this equation has? sinx=sin1
Saturday, April 27, 2013 at 1:53pm

trig
how to do this? how much does this equals? 2cos160*cos140*cos100
Saturday, April 27, 2013 at 1:49pm

Height and distance
http://www.mathsisfun.com/algebra/trig-c​osine-law.html
Friday, April 26, 2013 at 11:45am

graphing trig functions
How to find general solutions
Friday, April 26, 2013 at 11:12am

trig
FIRE
Thursday, April 25, 2013 at 11:38am

trig
pi = 180°, so pi/5 = 36° so, what is 11pi/5?
Wednesday, April 24, 2013 at 2:16pm

trig
Convert 11pi/5 radians into degrees.
Wednesday, April 24, 2013 at 1:45pm

trig
Convert 11pi/5 radians into degrees.
Wednesday, April 24, 2013 at 1:45pm

Trig
let the parabola have equation y = 3(x-p)^2 + q (1,10) lies on it, so 10 = 3(1-p)^2 + q also the vertex (p,q) lies on y = 3x+1 thus: q = 3p + 1 sub back into equation above 10 = 3(1-p)^2 + 3p + 1 10 = 3 - 6p + 3p^2 + 3p + 1 3p^2 -3p -6 = 0 p^2 - p - 2 = 0 (p-2)(p+1) = 0 p = 2 ...
Wednesday, April 24, 2013 at 8:03am

Trig
A parabola has its vertex on the graph of the line y=3x+1 and passes through (1,10). If it is the same size,? shape, and direction as the graph of y=3x^2, find the equation (s) of all possible parabolas Thanks!!!
Wednesday, April 24, 2013 at 6:39am

Trig
2 (r cos T)(r sin T) = 1 r^2 cos T sin T = (1/2)
Monday, April 22, 2013 at 7:11pm

Trig
Convert 2xy = 1 to polar form.
Monday, April 22, 2013 at 7:04pm

Trig
Rewrite -2x^2 + sqrt(3)xy - y^2 +2 = 0 with the xy term eliminated. Round terms to nearest hundredth.
Monday, April 22, 2013 at 6:39pm

trig
A kite is in the shape of a quadrilateral with two pair of congruent adjacent sides. The lengths of two sides are 20.0 inches and the lengths of the other two sides are 35.0 inches. The two shorter sides meet at an angke of 115 degrees. A. Find the length of the diagonal ...
Sunday, April 21, 2013 at 6:09pm

trig
using the law of cosines, the distance d is d^2 = 16^2 + 18^2 - 2(16)(18)cos110° d = 27.87 ft
Sunday, April 21, 2013 at 5:47pm

trig
If we draw a triangle WBE with the obvious labels for angles and sides, we have angles W=53 degrees E=109 degrees B=18 degrees Now, using the law of sines w/sin109 = 135/sin18 w=413 miles
Sunday, April 21, 2013 at 5:44pm

Trig
250 @ N35E = (143.4,204.8) 25 @ E = (25,0) add them to get (168.4,204.8) = 265.1 @ N39.5E
Sunday, April 21, 2013 at 5:34pm

Advanced Functions/Trig
sin A sin B = ½ cos(A−B) − ½ cos(A+B) sin(x+y)sin(x-y) = 1/2 [cos(2y) - cos(2x)] 1/2 (2cos^2(y)-1 - 2cos^2(x)+1) cos^2(y) - cos^2(x)
Sunday, April 21, 2013 at 4:54pm

trig
Two tracking stations are on the equator 135 miles apart. A weather balloon is located on a bearing of N 37 degrees E from the weastern station and on a bearing of N 19 degrees E from the eastern station. How far is the balloon from the weastern station?
Sunday, April 21, 2013 at 3:32pm

trig
A pole is braced by two wires that extended from the top of the pole to the ground. The lengths of the wires are 16 feet and 18 feet and the measure of the angle between the wires is 110 degrees. Find, to the nearest foot, the distance between the points at which the wires are...
Sunday, April 21, 2013 at 2:45pm

Trig
A plane has a speed of 250mph and flies with a bearing of N35degE. The wind is blowing from the west at 25 mph with a heading of due east. What is the actual speed and heading of the plane? Round to one decimal place.
Sunday, April 21, 2013 at 1:05pm

Trig
Thank you <3 you are a lifesaver!
Sunday, April 21, 2013 at 12:29pm

Advanced Functions/Trig
Prove the following: sin(x+y)sin(x-y) = cos^2 (y) - cos^2 (x)
Sunday, April 21, 2013 at 11:47am

Trig
looks good to me. Yep, just do a scalar multiply on u,v and add the i,j values separately.
Sunday, April 21, 2013 at 6:18am

Trig
I would tack on the wind vector to the end of the plane vector at the point P so the final end point is Q Join Q to the origin O vector OQ has our required direction and its magnitude represents the actual speed By simple geometry angle P = 125° so by the cosine law: OP^2...
Saturday, April 20, 2013 at 10:02pm

Trig
Whoops I forgot to include the diagram...here it is: i.imgur [dot] com/RRY5pox.png
Saturday, April 20, 2013 at 4:51pm

Trig
This has to do with vectors...I'm wondering if I set up the diagram correctly? A plane has a speed of 250mph and flies with a bearing of N35degE. The wind is blowing from the west at 25 mph with a heading of due east. What is the actual speed and heading of the plane?
Saturday, April 20, 2013 at 4:50pm

Trig
This has to do with dot products (I can't make the arrow above the u, i, or j). Given u = 4i - 6j and v = -3i + 5j, find 4u-(1/2)v Am I just supposed to plug in what u and v are equal to in 4u-(1/2)v? If so, is the answer (35/2i - 53/2j)
Saturday, April 20, 2013 at 4:15pm

Trig
sec(A/2) = 1.4275 cos(A/2) = 1/1.4275 = 0.70053 A/2 = 45.53 A = 91.06o
Friday, April 19, 2013 at 8:34pm

trig
sort of :/ how did you get from 1/sin=3 to 3sin+1 ?
Thursday, April 18, 2013 at 12:06pm

Trig
sin x = 10.5/12 so, x = ?
Thursday, April 18, 2013 at 5:41am

trig
2sin+1/sin=3 2sin^2 - 3sin + 1 = 0 (2sin-1)(sin-1) = 0 sinx = 1/2 or 1 now it's a cinch, right?
Thursday, April 18, 2013 at 5:40am

trig
Solve exactly over the indicated interval: 2sinx+cscx=3, 0<theta<360
Thursday, April 18, 2013 at 12:03am

Trig
A 12 foot ladder leans against a building. The top of the ladder leans against the wall 10.5 feet from the ground. What is the angle formed by the ground and the ladder? Assume its a right triangle.
Wednesday, April 17, 2013 at 11:52pm

Trig
solve over the indicated interval expressed in degrees to 2 decimal places. sec(theta/2)=1.4275, 0<theta<360degrees
Wednesday, April 17, 2013 at 11:25pm

trig
solve over the indicated interval expressed in degrees to 2 decimal places. sec(theta/2)=1.4275, 0<theta<360degrees
Wednesday, April 17, 2013 at 11:24pm

trig
if cotØ =0 then tanØ is undefined and we know that the tangent is undefined at π/2 and 3π/2
Wednesday, April 17, 2013 at 10:48pm

trig
solve exactly over the indicated interval: cot(theta)=0, all real numbers
Wednesday, April 17, 2013 at 9:33pm

Trig
Nope. As usual, there is more than one way to do it. I picked my way to avoid actually having to find the value of x.
Wednesday, April 17, 2013 at 10:38am

trig
values of cosines lie between -1 and +1 so minimum value of 2cosx + 2cosy = 2(-1) + 2(-1) = -4 when both x and y = 3π/2
Wednesday, April 17, 2013 at 8:42am

trig
Find minimum value of 2cos x + 2cos y
Wednesday, April 17, 2013 at 8:31am

trig
Find minimum value of 2cos x + 2cos y
Wednesday, April 17, 2013 at 8:28am

trig
Find minimum value of 2cos x + 2cos y
Wednesday, April 17, 2013 at 8:27am

Trig
Using law of sines, this is what I got: x/sin37° = 100/sin10° x = 346.6 ft h/sin47° = 346.6/sin90° h = 253.5 ft Is this wrong...?
Tuesday, April 16, 2013 at 6:34pm

Math-Trig
Find all angles in degrees that satisfy each equation. tan(a)+rad3=0
Tuesday, April 16, 2013 at 5:32pm

Trig
as usual, draw a diagram. If the height is h, and the surveyor started out at a distance x from the tree, h/x = tan47° h/(x+100) = tan37° equating the values for x, h/tan47° = h/tan37° - 100 h/1.072 = h/.7535 - 100 h = 253.6 ft
Tuesday, April 16, 2013 at 5:12pm

Trig
From a certain distance from the base of a Giant Sequoia tree a surveyor determines that the angle of elevation to the top of the tree is 47deg. The surveyor then walks 100 feet away from the tree and determines that the angle of elevation to the top of the tree is now 37deg. ...
Tuesday, April 16, 2013 at 5:06pm

Trig
the graph is rotated through an angle θ where cot2θ = (A-C)/B = -1/√3 so θ = -π/6 So, we need new x'-y' axes where x = (x'(√3/2)-y'(-1/2)) y = (x'(-1/2)+y'(√3/2)) expand the expressions (yes, you need to pay careful...
Tuesday, April 16, 2013 at 2:28pm

Trig
First off, you can see that it's 2nd-degree in both x and y, so there is no simple y=f(x) expression. It's the equation of a rotated parabola, so there will be a top branch and a bottom branch. Anyway, express it as a quadratic in y: x^2 - 4xy + 4y^2 - 8y + 2 = 0 4y^2...
Tuesday, April 16, 2013 at 2:15pm

Trig
Rewrite 6x^2 - sqrt(3)xy + 5y^2 - 9 = 0 with the xy term eliminated.
Tuesday, April 16, 2013 at 12:12pm

Trig
Convert this equation so that it could be entered into a calculator in the standard y= format. x^2 - 4xy + 4y^2 - 8y + 2 = 0
Tuesday, April 16, 2013 at 12:11pm

Trig
sec 9x = 2 cos 9x = 1/2 so 9x = π/3 or 5π/3 but period of cos 9x is 2π/9 so adding / subtracting any multiples of 2π/9 to an existing answer will yield a new answer general solution: x = π/3 + 2nπ/9 , 5π/3 + 2nπ/9
Monday, April 15, 2013 at 9:42pm

Trig
Solve the multiple-angle equation. (Enter your answers as a comma-separated list. Let n be any integer. Enter your response in radians.) sec 9x = 2
Monday, April 15, 2013 at 8:47pm

math
In right triangle ABC, a = 52 ft and c = 63 ft. Which of the following trig ratios of angle B, would combine the given sides?
Sunday, April 14, 2013 at 4:57pm

Maths-trig
clear the fraction and recall that sin^2+cos^2=1 sin 2x = 2 sinx cosx and you're done
Friday, April 12, 2013 at 3:18pm

Maths-trig
Prov tht 1-sin2X/sinX-cosX=sinX-cosX
Friday, April 12, 2013 at 3:11pm

trig
Mathematicians should not write navigation problems: an airplane has an airspeed of 430 miles per hour ***ON*** a ****HEADING**** of 135 degrees. the wind velocity is 35 miles per hour in the direction N30degreesE. Find the resultant speed and direction of the plane. Usually ...
Friday, April 12, 2013 at 1:44pm

trig
an airplane has an airspeed of 430 miles per hour at a bearing of 135 degrees. the wind velocity is 35 miles per hour in the direction N30degreesE. Find the resultant speed and direction of the plane.
Friday, April 12, 2013 at 1:25pm

trig
What happens when you try to elevate sin-1(3/2)?
Thursday, April 11, 2013 at 5:19pm

trig
A point on the terminal side of an angle theta is given. Find the exact value of the six trigonometric functions of theta: (2,-3)
Thursday, April 11, 2013 at 3:06pm

Precalc with Trig
There isn't a typo this is the equation my professor gave us, thanks for the help!
Thursday, April 11, 2013 at 2:57pm

Precalc with Trig
cos^2 - cos - 6 = 0 (cos-3)(cos+2) = 0 cos = 3 or -2 Bzzzt. |cos| <= 1, so not possible. If there's a typo, fix it and apply the same steps.
Thursday, April 11, 2013 at 2:52pm

Precalc with Trig
What value(s) of x from 0 to 2pi solve the following equation: cos squared x - cos x - 6 = 0
Thursday, April 11, 2013 at 2:04pm

trig
analyze given equation for following characteristics. (amplitude period, phrase shift) y= 3sin (2x-(3.14/2))
Thursday, April 11, 2013 at 10:15am

Math
sec(alpha) = -4(√(5))/5 find exact values at alpha for the remaining five trig functions
Wednesday, April 10, 2013 at 5:46pm

trig
9cosx + 4cos(2x) + 1 = 0 9cosx + 4(2cos^2 x - 1) + 1 = 0 8cos^2 x + 9cosx - 3 = 0 now just solve the quadratic equation for cosx
Wednesday, April 10, 2013 at 3:39pm

Trig
secx + tanx = 1 1 + sinx = cosx 1 + 2sinx + sin^2 x = cos^2 x 1 + 2sinx + sin^2 x = 1 - sin^2 x 2sin^2 x + 2sinx = 0 2sinx(sinx+1)=0 so, sinx=0 or sinx = -1 x=0,π,3π/2 but secx,tanx are undefined at x=3π/2, secπ + tanπ = -1, not 1, so x=0 is the only ...
Wednesday, April 10, 2013 at 3:31pm

trig
Solve rounding to 2 decimal places. 9cosx + 4cos(2x) + 1 = 0 over 0° ≤ x < 360° Thanks to anyone that can help! I'm totally stumped.
Wednesday, April 10, 2013 at 2:40pm

Trig
Solve exactly over 0 ≤ θ < 2π secx + tanx = 1
Wednesday, April 10, 2013 at 2:24pm

trig
since sin^2 + cos^2 = 1, sin^2 = 1 - cos^2 = 1-(1/a^2) so sin = √(1 - 1/a^2) = 1/a √(a^2-1) to see this, draw your triangle and label the adjacent leg=1 and the hypotenuse=a. Then the other leg is √(a^2-1) and the sin is just √(a^2-1)/a. Since sin > ...
Wednesday, April 10, 2013 at 12:47pm

trig
GIVEN THE ANGEL ALFA WITH COS ALFA = -1/a. IF ALFA IN QUADRANT TWO, THE VALUE OF SIN ALFA IS....
Wednesday, April 10, 2013 at 12:42pm

trig
that's fer shure
Wednesday, April 10, 2013 at 10:49am

trig
i need to know how to do evaerything because i am stupid
Wednesday, April 10, 2013 at 9:39am

applied calculus
3. a) Write out the limit definition for the derivative of y = xx. Attempt to solve it. b) Write out the limit definition for the derivative of the inverse trig function from question 2. Attempt to solve it. c) Discuss the value of implicit differentiation. (Use questions 1 ...
Tuesday, April 9, 2013 at 9:26pm

applied calculus
Let y=〖tan〗^(-1) x Rewrite it as an equation involving a trig function. Use implicit differentiation to determine an expression for y.
Tuesday, April 9, 2013 at 9:25pm

Physics
Please help! I've tried solving this with trig but I can't seem to figure out what I've been doing wrong. Two positive point charges, each of which has a charge of 2.3 × 10−9 C, are located at y = +0.60 m and y = −0.60 m. A)Find the magnitude of ...
Tuesday, April 9, 2013 at 7:25pm

Applied calculas
3. a) Write out the limit definition for the derivative of y = xx. Attempt to solve it. b) Write out the limit definition for the derivative of the inverse trig function from question 2. Attempt to solve it. c) Discuss the value of implicit differentiation. (Use questions 1 ...
Monday, April 8, 2013 at 5:32pm

Applied calculas
Let y=〖tan〗^(-1) x Rewrite it as an equation involving a trig function. Use implicit differentiation to determine an expression for y.
Monday, April 8, 2013 at 5:31pm

trig
I see. A trailing ,numer; works. For asin(.8), asin(.8)*180/%pi, numer; produces 53.13... I also found that writing numbers with a trailing .0 forces approximation. And for the built-in constants, float() forces approximation. For example: a: 4.0/5.0$ disp(a)$ b: asin(a)$ ...
Monday, April 8, 2013 at 4:13pm

Trig
treat the 2Ø as one angle , let's say 2Ø = A then sin A = -.7843 so we know that A is in III or IV , like you had I then find the "angle in standard position" by finding sin^-1 (+.7843) which is 51.67° so in III, A = 180 + 51.66 = 231.66° in...
Monday, April 8, 2013 at 3:50pm

Trig
well, you are correct, in that 2θ = 231.655° or 308.345° That makes θ = 115.83° or 154.17° However, since we're dealing with 2θ, that gives us answers in the 0<θ<180 range. To get all the values up to 360, we need to add 360 to 2&#...
Monday, April 8, 2013 at 3:43pm

trig
using online maxima, I found also that acos(.8)*180/%pi; does what you show, but acos(.8)*180/%pi,numer; produces 36.86... as the answer in degrees. Maybe wxMaxima has something similar
Monday, April 8, 2013 at 3:32pm

Trig
Solve the equation on 0° ≤ θ < 360° and express in degrees to two decimal places. sin(2θ) = -0.7843 I've gotten 308.345° (QIV) and 231.655° (QIII). I'm unsure how to get the answer though? The final answer is: 115.83°, 295.83°...
Monday, April 8, 2013 at 3:14pm

trig
Using b+0 doesn't phase it. The problem is getting wxMaxima to make 4/5 into a decimal. If I use a: 0.8$ b: asin(a)$ print("the angle is ", b)$ Then it works. It gives me the angle is 0.92729521800161 But if I try to convert that to degrees: c: b*180/%pi$ print(&...
Monday, April 8, 2013 at 2:15pm

trig
Hmm. don't know wxMaxima, but have you tried something like print("the angle is ", b+0)$ so it has to evaluate b.
Monday, April 8, 2013 at 1:54pm

trig
I am trying to get wxMaxima to give me arcsin(4/5), either in radians or degrees. I am using a: 4/5$ b:asin(a)$ print("the angle is ", b)$ But it returns the angle is asin(4/5) How do I get wxMaxima to give me the angle? I want .927 radians or 53.1 degrees.
Monday, April 8, 2013 at 1:06pm

trig
we know sin^2 41° + cos^2 41° = 1 m^2 = 1 - cos^2 41° m = ±√(1 - cos^2 41°)
Monday, April 8, 2013 at 9:47am

trig
if sin 41'=m. write m in terms of cos 41'
Monday, April 8, 2013 at 9:24am

Trig
given: sinØ = 3/4, and Ø is in quadrant II then by Pythagoras, y= 3, x = ? , r = 4 x^2 + 9 = 16 x = ±√5 , but in II x = -√7 so cosØ = -√7/4 sin 2Ø = 2sinØcosØ = 2(3/4)(-√7/4) = -6√7/16 = -3√7/...
Monday, April 8, 2013 at 7:59am

Trig
Find the exact values of sin 2 theta, cos 2 theta and tan 2thea using the double angle formula sin theta = 3/4 , pi/2 < theta < pi
Monday, April 8, 2013 at 2:31am

trig
Please clarify before I attempt this: is it 3cos (2Ø) ... or is it 3cos^2 Ø .... ?
Sunday, April 7, 2013 at 11:17pm

trig
i nead to find to the nearest degree all values of theta in the interval 0 x 360 that satisfy the equation 3 cos 2 theta+ sin theta-1=0
Sunday, April 7, 2013 at 10:33pm

Precalc with Trig
No, I don't just my professor trying to trick us, thanks for the help!
Sunday, April 7, 2013 at 5:23pm

Trig
tanØ = sinØ/cosØ = 0 in sinØ/cosØ = 0 , that is only possible if sinØ = 0 and you should know that the basic sine curve has zeros at 0 , π and 2π since the period of sinØ = π, a new answer can be obtained by ...
Sunday, April 7, 2013 at 3:53pm

Precalc with Trig
cos^2 Ø - sin^2 Ø = 6 cos 2Ø = 6 no solution, the cosine of anything cannot be greater than 1 do you have a typo?
Sunday, April 7, 2013 at 3:49pm

Precalculus with Trig
make a sketch of a rightangled triangle label the base angle Ø, then opposite side is x hypotenuse is 1 then the adjacent side is √(1 - x^2) so cos(arcsin(x)) = √(1-x^2)1
Sunday, April 7, 2013 at 3:47pm

Trig
Solve the trig equation exactly over the indicated interval. tanθ = 0, all real numbers I know the answer is πn, but I don't understand how they got this. I get that tanθ = 0 on π and 2π on the unit circle, so did they just put n on the end of &#...
Sunday, April 7, 2013 at 1:48pm

Precalc with Trig
What value(s) of theta (0 less than or equal to theta less than or equal to 2pi) solve the following equation: cos squared theta - cos theta - 6 = 0
Sunday, April 7, 2013 at 1:42pm

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