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April 16, 2014

April 16, 2014

**Recent Homework Questions About Trigonometry**

Post a New Question | Current Questions

**trig**

POOP
*Thursday, November 21, 2013 at 6:03pm*

**Trig - Law of sines and cosines**

Thanks a lot
*Thursday, November 21, 2013 at 4:04pm*

**Trig-Medians and law of cosines and sines**

thanks a lot man!
*Wednesday, November 20, 2013 at 3:59pm*

**Trig-Medians and law of cosines and sines**

median length x = BC length median hits BC at D triangle ABC 4^2 = 3^2 + x^2 - 6 x cos B triangle ABD x^2 = 3^2 + (x/2)^2 - 3 x cos B ================================= 16 = 9 + x^2 - 6x cos B x^2 = 9 +x^2/4 -3 x cos B x^2 -6 x cos B -7 = 0 3x^2/4 +3 x cos B - 9 = 0 x^2 - 6 x ...
*Wednesday, November 20, 2013 at 12:30pm*

**Trig-Medians and law of cosines and sines**

In triangle ABC, we have AB=3 and AC=4. Side BC and the median from A to BC have the same length. What is BC? Not making sense to me, I think the answer must be simple, but I don't know how to solve I applied the law of sines but to no avail. Help is appreciated, thanks.
*Wednesday, November 20, 2013 at 11:53am*

**Trig**

since sin(34) = cos(56) sin(48) = cos(42) the expression evaluates to 2
*Wednesday, November 20, 2013 at 12:11am*

**Trig**

sin^2(34)+sin^2(48)+sin^2(56)+sin^2(42)
*Tuesday, November 19, 2013 at 10:18pm*

**Trig - Law of sines and cosines**

I started with AB, the vertical diagonal of the square. I then drew square ALMN At M, I drew the base BC of the equilateral triangle perpendicular to AM, letting BM = MC = 2 Joi AB and AC for the triangle. Let the intersection of ML and AB be P, and the intersection of MN and ...
*Tuesday, November 19, 2013 at 9:37pm*

**Trig**

√(100+100cot^2 x) √(100(1+cot^2 x)) √(100csc^2 x) 10cscx
*Tuesday, November 19, 2013 at 9:24pm*

**Trig - Law of sines and cosines**

ABC is an equilateral triangle with side length 4. M is the midpoint of BC, and AM is a diagonal of square ALMN. Find the area of the region common to both ABC and ALMN. I drew the diagram but I don't know how to find the answer? I think it has something to do with the law...
*Tuesday, November 19, 2013 at 8:09pm*

**Trig**

sqrt(100+(10cotx)^2)
*Tuesday, November 19, 2013 at 8:02pm*

**Pre-Cal**

these are all done the same way. You have to solve a quadratic in some trig function. #1. 2sec^2 4x - 3sec 4x - 5 = 0 (2sec 4x - 5)(sec 4x + 1) = 0 So, sec 4x = 5/4 or sec 4x = -1/4 There are two angles for each solution in [0,360°) where this is true. But that means there...
*Tuesday, November 19, 2013 at 12:19am*

**Calculus**

1) The period of a trig. function y=sin kx is 2pi/k. Then period of y=sin^2(pi.x/a) should be 2pi/(pi/a)=2a, but somewhere it is given as a. Which is correct? 2) The period of r=sin^3(theta/3) is given as 3pi. How is it worked out? Is it because after theta=0, the function ...
*Monday, November 18, 2013 at 12:18am*

**Trig**

V1 + V2 = 1200[70o] + 900[205o] X = 1200*cos70 + 900*cos205 = -405.3 Y = 1200*sin70 + 900*sin205 = 747.3 tanAr = Y/X = 747,3/-405.3 = -1.84376 Ar = -61.53o = Reference angle. A = -61.53 + 180 = 118.5o Vr = Y/sinA = 747.3/sin118.5 = 850.3[118.5o] = Resultant vector.
*Saturday, November 16, 2013 at 6:16pm*

**Trig**

2 vectors act on a bolt. vector 1(1200/70 degrees) vector 2(900/205 degrees) then E = ? must show work!
*Friday, November 15, 2013 at 8:57am*

**Trig**

Dunno what E is, but sounds like you just want to add two vectors. Bad news is you gotta convert to rectangular coordinates to do that.
*Friday, November 15, 2013 at 6:27am*

**Trig**

the resultant velocity is 320.0 mph @ 103° so, now you know how far it goes in 4 hours.
*Friday, November 15, 2013 at 6:22am*

**Trig**

Plane leaves airport with heading of 110 degrees at 300 mph, the wing out of the southwest is 42 mph. locate the plane after 4 hours? find the distance back to airport and angle from due east?
*Friday, November 15, 2013 at 1:14am*

**Trig**

2 vectors act on a bolt. vector 1(1200/70 degrees) vector 2(900/205 degrees) then E = ? must show work!
*Friday, November 15, 2013 at 12:37am*

**Algebra 2 Trig - Probability Question**

help solving a probability problem: probability that a person taking a survey is male "given" that he preferred a European automobile if 112 males prefer european autos out of 188, and 216 of a total of 452 individuals were surveyed were male. Additional information...
*Thursday, November 14, 2013 at 9:13am*

**Trig**

The magnitude of the resultant R is R^2 = 650^2 + 300^2 R = appr 715.89 if Ø is the angle it makes with the longer vector, then tanØ = 300/650 = ... Ø = appr 24.8°
*Wednesday, November 13, 2013 at 8:56pm*

**Trig**

2 vectors at a Right angle to each other are pulling on a bolt. vector 1 = 650; vector 2 = 300 then R = ?
*Wednesday, November 13, 2013 at 7:35pm*

**Trigonometry - Identities and proof**

Two of the standard conversion formulas are: sinA - sinB = 2sin((A-B)/2) cos( (A+B)/2) and cosA - cosB = - 2sin( (A-B)/2) sin( (A+B)/2) see: near bottom of page under: Sum-to-Product Formulas http://www.sosmath.com/trig/Trig5/trig5/trig5.html RS = -(sinx - siny)/(cosx...
*Wednesday, November 13, 2013 at 11:59am*

**Trigonometry - Identities and proof**

Show that cot((x+y)/2) = - (sin x - sin y)/(cos x - cos y) for all values of x and y for which both sides are defined. I tried manipulating both sides in terms of trig identities but I don't really have a solution....help would be appreciated, thanks.
*Wednesday, November 13, 2013 at 11:23am*

**Math**

absolute means global. It's possible to have several relative extrema, and some, all, or none may be absolute extrema. A parabola has one relative extremum, which is absolute. A cubic such as y=(x-1)(x-2)(x-3) has two relative extrema, but no absolutes at all, since the ...
*Sunday, November 10, 2013 at 4:38pm*

**trig**

¤©¤·¤¾¤·¤°¤·¤©
*Friday, November 8, 2013 at 1:53am*

**math iv**

it doesn't infer anything, though you may infer that it's a problem involving the first quadrant. Or a problem on acute triangles Or principal values of trig functions . . .
*Thursday, November 7, 2013 at 12:00pm*

**Math-Trigonometry**

Show that if A, B, and C are the angles of an acute triangle, then tan A + tan B + tan C = tan A tan B tan C. I tried drawing perpendiculars and stuff but it doesn't seem to work? For me, the trig identities don't seem to plug in as well. Help is appreciated, thanks.
*Wednesday, November 6, 2013 at 5:21pm*

**Math - Trigonometry**

If sin theta +cos theta =1.2, then what is sin^3 theta + cos^3 theta? Hmm...I don't understand how to proceed. I know I must apply a trig Identity, but which one? Thanks in advance
*Tuesday, November 5, 2013 at 10:26pm*

**Trigonometry**

I will read that as sin (x/7) = -√3/2 since √3/2 is one of the trig values I recognize So x/7 must be in quadrants III or IV I know sin 60° = +√3/2 so my angle in "standard position " is 60° x/7 = 180+60 = 240 x = 1680° or x/7 = 360-60...
*Monday, November 4, 2013 at 3:12pm*

**MATH ANALYSIS & TRIG**

as usual, draw a diagram. Al's distance a is 6+10t Barb's distance b is 2+5t d^2 = (6+10t)^2 + (2+5t)^2 = 5(25t^2 + 28t + 8) Now just set dd/dt=0 to find the minimum.
*Wednesday, October 30, 2013 at 5:45am*

**MATH ANALYSIS & TRIG**

At noon, Al is 6km north of point O traveling south at 10km/h. Also at noon, Barb is 2km east of point ) traveling east at 5km/h. A) express the distance "d" between Al and Barb as a function of time "t" hours after noon. B) Find the time at which the ...
*Tuesday, October 29, 2013 at 11:25pm*

**trig**

ij
*Tuesday, October 22, 2013 at 3:22pm*

**Math - Trigonometry**

doesn't matter. All of the trig functions are positive in exactly 2 of the 4 quadrants.
*Tuesday, October 15, 2013 at 4:07pm*

**Math - Trigonometry**

degrees or radians or grads, it's still 2. What you call the angle does not matter. If you plan to succeed in trig, get used to thinking in radians. Once you get past the introductory material, most problems will deal with radians.
*Tuesday, October 15, 2013 at 11:51am*

**pre-calc/ trig**

yes it is.
*Tuesday, October 15, 2013 at 10:21am*

**pre-calc/ trig**

Sinx=0
*Tuesday, October 15, 2013 at 10:07am*

**Trig**

hey ted did you ever figure that out because if so tell me tell me now plz i need this very much
*Saturday, October 12, 2013 at 11:14am*

**Calculus - please help!**

we have 4x^2 + 9y^2 = 36 The line x=3 lies at one end of the ellipse, so shells are probably the way to go. The volume v is thus v = ∫[-3,3] 2πrh dx where r = 3-x and h = y = √(36-4x^2)/3 v = 2π/3 ∫[-3,3] (3-x)√(36-4x^2) dx Now just break it ...
*Friday, October 11, 2013 at 12:09pm*

**Math - Trig**

Did you make sure your calculator was set to Radians ? (press DRG until you see RAD displayed) Here are my keystrokes .5 x 12 - 2 = sin = x 20 + 40 = and you should get 24.86° (Of course you can see that since sin(????) ranges from -1 to +1, the temperature must range from...
*Thursday, October 10, 2013 at 9:07am*

**Math - Trig**

Still don't get it, so yes need more help.
*Thursday, October 10, 2013 at 5:49am*

**Math - Trig**

assuming θ is the month, just plug in θ=12. Come on back if you don't like the answer.
*Thursday, October 10, 2013 at 4:33am*

**Math - Trig**

The equation y = 20sin(0.5θ - 2) + 40 models the monthly temp for a certain city. Use the equation to predict the temperature in the city during December. Thanks
*Thursday, October 10, 2013 at 12:46am*

**Trig**

we know we want y = a sin(bx+c) + d diameter is 2, so radius=1, and we have y = a sin(bx+c) + 1 Starting at t=0, the bead starts going downward, so y = -sin(bx+c)+1 If we make it so the axle is at (0,1) at t=0, we need to shift the graph to the right by 1, since the bead is on...
*Monday, October 7, 2013 at 4:23am*

**Trig**

a child puts beads on one spoke of a bicycle wheel. the tire has a diameter of 2ft. if the child rides so that the tire makes one full rotation every 15 sec, and the beads begin in the horizontal outward position, find an equation that models the position of the beads at time t.
*Monday, October 7, 2013 at 4:11am*

**MathsSs triG**

ThanQ steve
*Thursday, October 3, 2013 at 5:49pm*

**MathsSs triG**

sin(x-360)sin(90-x)tan(-x)/cos(90+x) = sin(x)cos(x)(-sin(x)/cos(x))/(-sin(x)) = sin(x) x = 30 8/sin^2A - 4/1+cosA 8/(1-cos^2A) - 4/(1+cosA) (8 - 4(1-cosA))/(1-cos^2A) (8-4+4cosA)/(1-cos^2A) 4(1+cosA)/(1-cos^2A) 4/(1-cosA) undefined when cosA = 1 So, on 0<A<360 it is ...
*Thursday, October 3, 2013 at 5:28pm*

**MathsSs triG**

Consider sin(x-360)sin(90-x)tan(-x)/cos(90+x) 1.A.SIMPLIFY sin(x-360)sin(90-x)tan(-x)/cos(90+x) to a single trigonometric ratio B.hence or otherwise without using a calculator,solve for X if 0<X<360. sin(x-360)sin(90-x)tan(-x)/cos(90+x) =0,5 2.A.prove that 8/sin^2A - 4/1...
*Thursday, October 3, 2013 at 4:22pm*

**Derivatives**

H(x) = sin2x cos2x = 1/2 sin4x That help? Using the product rule, it's a bit more work, but you can get the same answer: dH/dx = 2cos2x cos2x - 2sin2x sin2x = 2(cos^2 2x - sin^2 2x) = 2cos4x Don't forget your trig just because you're taking calculus! #2 df/dx = (u...
*Thursday, October 3, 2013 at 4:58am*

**Go with Bosnian - Trig**

sorry Kiki don't know how I missed that 2 in front of 2x^2 Go with Bosnian's solution or, how about this... 2x^2 + 2x > 4 x^2 + x - 2 > 0 (x+2)(x-1) > 0 (x+2>0 and x-1<0) OR ( x+2 < 0 and x-1 > 0 ) ( x>-2 and x<1) OR x< -2 and x> 1 -2 &...
*Sunday, September 29, 2013 at 4:44pm*

**Trig**

2x^2+2x>4 Divide both sides by 2 x ^ 2 + x > 2 ________________________________________ take one half of thje coefficient of x and square it. in this case ( 1 / 2 ) ^ 2 = 1 / 4 ________________________________________ Add 1 / 4 to both sides x ^ 2 + x + 1...
*Sunday, September 29, 2013 at 4:24pm*

**not Trig**

x^2 + 2x > 4 add 1 to both sides x^2 + 2x + 1 > 5 (x+1)^2 > 5 x+1>√5 or -x-1 > √5 x > √5 - 1 OR -x > √5 + 1 x > √5-1 OR x < -√5-1
*Sunday, September 29, 2013 at 2:59pm*

**Trig**

2x^2+2x>4
*Sunday, September 29, 2013 at 2:17pm*

**math - trig**

sinA = a/c = cosB cosA = b/c = sinB tanA = a/b = cotB plug and chug.
*Wednesday, September 25, 2013 at 5:47am*

**math - trig**

Suppose ABC is a right triangle with sides of lengths a, b, and c and right angle at C. Find the unknown side length using the Pythagorean Theorem and find the following trigonometric functions of the indicated angle. Given: a = 4 and b = 7 Find: sin A, cot B, sec A, tan B
*Wednesday, September 25, 2013 at 4:09am*

**Trig**

(x+1)(x-4) >= (x-2)^2 x^2-3x-4 >= x^2-4x+4 x >= 8 for the graphs, go to wolframalpha.com and enter plot y=(x+1)(x-4) , y=(x-2)^2, x=6..10 or, enter (x+1)(x-4)-(x-2)^2 and see where it's positive
*Monday, September 23, 2013 at 3:15pm*

**Trig**

(x+1)(x-4) is greater than or equal to (x-2)^2
*Monday, September 23, 2013 at 2:57pm*

**Physics**

SAme solution, but you should have the two solutions. it should be 45+45-37.5 degrees (if you think on the trig equations you get, you see why).
*Wednesday, September 18, 2013 at 5:48pm*

**Trig**

looks like "homework dumping" to me I have two of them for you. Please indicate where you are having difficulties with these standard type of trig questions.
*Wednesday, September 18, 2013 at 12:23am*

**Trig**

make quick sketches of right-angled triangles for each of the two cases, and use Pythagoras to find the missing sides. first triangle: since sinA = 12/13 , y = 12 and r = 13 x^2 + y^2 = r^2 x^2 + 144 = 169 x^2 = 25 x = ±5, but we are in quad II, so x = -5 then tanA = - ...
*Wednesday, September 18, 2013 at 12:22am*

**Trig**

Well, why not follow that hint sin(-15) = sin(30 - 45) = sin30cos45 - cos30sin45 = .... You should know the trig functions of those special angles.
*Wednesday, September 18, 2013 at 12:13am*

**Trig**

Rewrite the expression (tan A)(cot A) in terms of a single trigonometric ratio.
*Tuesday, September 17, 2013 at 11:24pm*

**Trig**

Use the equation mg sin A = umg cos A to determine the angle at which a waxed wood block on an inclined plane of wet snow begins to slide. Assume the coefficient of friction, u, is 0.17.
*Tuesday, September 17, 2013 at 11:24pm*

**Trig**

Find the exact value of sin π/12.
*Tuesday, September 17, 2013 at 11:23pm*

**Trig**

Evaluate and simplify cos(-15º). Hint: Substitute 30º-45º for -15º.
*Tuesday, September 17, 2013 at 11:23pm*

**Trig**

Suppose sin A = 12/13 with 90º≤A≤180º. Suppose also that sin B = 7/25 with -90º≤B≤0º. Find tan (A - B).
*Tuesday, September 17, 2013 at 11:22pm*

**Trig**

Suppose sinA = 12/13 with 90º≤A≤180º. Suppose also that sinB = -7/25 with -90º≤B≤0º. Find cos(A - B).
*Tuesday, September 17, 2013 at 11:22pm*

**Trig**

Let sin A = 12/13 with 90º≤A≤180º and tan B = -4/3 with 270º≤B≤360º. Find tan (A + B).
*Tuesday, September 17, 2013 at 11:22pm*

**trig**

A plane flies in a direction of N70°E for 80 km and then on a bearing of S10°W for 150 km. a How far is the plane from its starting point? b What direction is the plane from its starting point?
*Sunday, September 15, 2013 at 8:01am*

**Trig**

sec^2θ = 1+tan^2θ = 17 since secθ < 0 in QIII, secθ = -√17
*Friday, September 13, 2013 at 5:03am*

**calculus (integral)**

use partial fractions to get 13/27 (1/(x-3) - (x+6)/(x^2+3x+9)) 13/27 (1/(x-3) - 1/2 (2x+3)/(x^2+3x+9) - (9/2)/(x^2+3x+9)) 13/27 (1/(x-3) - 1/2 (2x+3)/(x^2+3x+9) - (9/2)/(x+3/2)^2 + 27/4) Now you can use substitutions u = x-3 v = x^2+3x+9 w = x + 3/2 to come up with 13/27 (du/...
*Friday, September 13, 2013 at 5:01am*

**Trig**

suppose tan theta equal 4 and theta lies in the third quadrant. find the values of sec theta
*Friday, September 13, 2013 at 2:03am*

**trig**

sketch right-angled triangles to match sinA = 12/13 and tanB = -4/3 from the first , y=12, r=13, so x = -5 and tanA = -12/5 tan(A-b) = (tanA - tanB)/(1 + tanAtanB) = (-12/5 + 4/3)/(1 + (-12/5)(-4/3)) = -16/63
*Friday, September 13, 2013 at 12:09am*

**trig**

tan 5π/12 or tan 75° I will use degrees ... tan 75 = tan(30 + 45) = (tan30+tan45)/( 1 - tan30tan45) = (1/√3 + 1)/( 1 - 1/√3) , multiply top and bottom by √3 = (1 + √3)/√3 - 1)
*Friday, September 13, 2013 at 12:01am*

**trig**

Let sin A = 12/13 with 90º≤A≤180º and tan B = -4/3 with 270º≤B≤360º. Find tan (A - B).
*Thursday, September 12, 2013 at 11:51pm*

**trig**

let's test it by applying the sin(A - B) = sinAcosB - cosAsinB relation sin(x - 3π/2) = sinx cos3π/2 - cosx sin3π/2 = (sinx) (0) - (cosx)(-1) = cosx yup, it is true.
*Thursday, September 12, 2013 at 11:49pm*

**trig**

Evaluate and simplify tan 5π/12.
*Thursday, September 12, 2013 at 11:47pm*

**trig**

This identity means that translating a basic sine graph 3π/2 units to the right produces a basic cosine graph. True or False? sin(X-(3(pi)/2)) = cosX
*Thursday, September 12, 2013 at 11:43pm*

**trig**

well, sine is zero at zero radians, and at PI radians e^x=0 because e^x cannot be zero, then e^x=PI take the ln of each side x=ln(PI)
*Wednesday, September 11, 2013 at 8:49pm*

**trig**

see other post.
*Wednesday, September 11, 2013 at 8:47pm*

**trig**

How do you solve sin(e^x)=0 for x? The question is "find the smallest number x such that sin(e^x)=0".... but I can't even get AN answer lol! Much appreciated!
*Wednesday, September 11, 2013 at 8:29pm*

**trig**

How do you solve sin(e^x)=0 for x? The question is "find the smallest number x such that sin(e^x)=0".... but I can't even get AN answer lol! Much appreciated!
*Wednesday, September 11, 2013 at 8:29pm*

**trig**

Just noticed this was for trig, not calculus. Algebraically, note that a(y) is a parabola, whose vertex occurs where y=60/4 = 15.
*Thursday, August 29, 2013 at 11:34pm*

**trig**

If the sides are x and y, with x parallel to the barn, x+2y = 60, so x = 60-2y the area is a = xy = (60-2y)y = 60y-2y^2 da/dy = 60 - 4y max a when y=15,x=30
*Thursday, August 29, 2013 at 11:33pm*

**trig**

if 60 m of fencing is available for a rectangular garden, one side of which is against a barn, what areb the dimensions of the garden that will give the maximum area?
*Thursday, August 29, 2013 at 9:44pm*

**Grammar**

with = not algebra for = not trig However, "for" version is a little awkward. "I want to take algebra instead of trigonometry" would be better.
*Thursday, August 29, 2013 at 2:38pm*

**trig**

obviously, it's (2*9-3(-5) , 2*2-3(-2)) = (?,?)
*Wednesday, August 28, 2013 at 3:44pm*

**trig**

u = (9, 2) v = (-5, -2) 2u - 3v = ?
*Wednesday, August 28, 2013 at 3:07pm*

**Math**

You should memorize the following diagrams: 1. the 30°-60°-90° triangle with matching sides 1 -- √3 -- 2 (angles in radians, π/6 π/3 π/2) 2. the 45° -- 45° -- 90° triangle with corresponding sides 1 -- 1 -- √2 So sin π/4 ...
*Thursday, August 22, 2013 at 5:25pm*

**Trig hep, please? Thank you!**

I will try the others soon.
*Sunday, August 18, 2013 at 11:24am*

**Trig hep, please? Thank you!**

thanks! what do the question marks stand for. and also CAST?
*Sunday, August 18, 2013 at 11:24am*

**Trig hep, please? Thank you!**

1. sin 195° = - sin15° , by CAST we know cos 30° = 1 - 2 sin^2 15° √3/2 = 1 - 2sin^2 15° 2sin^2 15 = 1 - √3/2 = (2-√3)/2 sin^2 15° = (2-√3)/4 sin 15 = √((2-√3)) /2 so sin 195 = -sin15 = -√((2-√3)) /2 3. ...
*Sunday, August 18, 2013 at 12:00am*

**Trig hep, please? Thank you!**

1. Find the exact value of sin(195(degrees)) 2. If cot2(delta)=5/12 with 0(<or =)2(delta)pi, find cos(delta), sin(delta) , tan(delta). 3.find the exact value of sin2(x) if cos(x)= 4/5. (X is in quadrant 1) 4. Find the exact value of tan2(x) if sin(x)=5/13. ((X) in quadrant ...
*Saturday, August 17, 2013 at 10:51pm*

**Trigonometry desperate help, clueless girl here**

#2 cos2x - 3sinx cos2x = 0 (cos2x)(1-3sinx) = 0 as you know, if the product of two numbers is zero, one or the other must be zero. So, cos2x = 0 or 1-3sinx = 0 cos2x=0 means x is pi/4,3pi/4,5pi/4,7pi/4 1-3sinx=0 means x = arcsin(1/3) = .3398 But, you need all angles between 0 ...
*Saturday, August 17, 2013 at 12:37am*

**trig help much appreciated! :))**

alright, I will. Thanks
*Friday, August 16, 2013 at 6:37pm*

**trig help much appreciated! :))**

no ideas on any of these? Also you have some major typos and/or formatting issues. I'll do a couple. Maybe you can clean up the post and indicate where you get stuck on the others. #2 cos 2x – 3sin x cos 2x = 0 (1-3sinx)cos2x = 0 so, either cos2x=0 ==> x = pi/4 or ...
*Friday, August 16, 2013 at 4:58pm*

**trig help much appreciated! :))**

1. Find the complete exact solution of sin x = . 2. Solve cos 2x – 3sin x cos 2x = 0 for the principal value(s) to two decimal places. 3. Solve tan2 x + tan x – 1 = 0 for the principal value(s) to two decimal places. 4. Prove that tan2 – 1 + cos2 = ...
*Friday, August 16, 2013 at 4:53pm*

**trig**

This is very cleverly solved at http://answers.yahoo.com/question/index?qid=20100617064137AAccXtF as a simple web search revealed.
*Friday, August 16, 2013 at 4:53pm*

**trig**

prove that :cosA-sinA+1/cosA+sinA-1=cosecA+cotA
*Friday, August 16, 2013 at 1:16pm*

**trig**

only true if theta = 45 degrees as far as I know identity: cos 2t = cos^2 t - sin^2 t on the right we have .5 [ cos t/ sin t - sin t/cos t ] = .5 [ cos^2 t - sin^2 t] /(sin t cos t) that is only equal to the left side if sin t cos t = .5 which is only true if t = 45 degrees or...
*Wednesday, August 14, 2013 at 3:50pm*

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