How can I find 2 angles, t, 0<=t<=2pi, such that tan(t)=-2?
921171
If tan( t ) = -2 then t = atan( -2 ) = -1.107 + multiples of π.\n\nIn the range from 0 to 2π the required angles are\n\n-1.107 + π and -1.107 + 2π, or\n\n2.034 and 5.176
449610
First, you need a protractor. Create a right triangle ABC with C as the right triangle. Let angle B be your angle. Make rough measurements of the three sides of the triangle. now:\nsin B = AC/AB\n\nOnce you know its sine, use these formulae for the other trigonometric functions:\ncos B = sqrt(1 - sin² B)\ntan B = sin B/cos B\n\nSpecial cases are:\nsin 0 = 0\nsin 90 = 1\nsin 30 = 1/2\nsin 60 = sqrt3/2\nsin 45 = sqrt2/2\nFrom these, you can compute all angles.\nremember,\nsin B = sin (B - 360n) where n is an integer\nsin B = cos (90 - B)\ncos B = sin (90 - B)\ncos (-B) = cos B\nsin (-B) = -sin B\n\nexample:\ncalculate sin, cos and tan of -150.\nfirst, sin -150 = -sin 150\n-sin 150 = -cos(90 - 150)\n= -cos (-60)\n= - cos 60 = -sin(90 - 60) = - sin 30\n= -1/2\nsin -150 = -1/2\ncos -150 = cos 150 = sin(90 - 150) \n= sin (-60) = -sin 60\n= -sqrt3/2\ntan -150 = sin -150/cos -150\n= (-1/2)/(-sqrt3/2)\n= sqrt3/3\n\nJust use these sign rules:\nif angle is in Q1 = sin cos and tan are +\nif angle is in Q2 = sin is +,cos is -,tan is -\nif angle is in Q3 = sin is -,cos is -,tan is +\nis angle is in Q4 = sin is -,cos is +,tan is -\n\n^_^
835259
I'll give you a hint and let you try to figure out the rest yourself (unless someone else tells you the exact answer). =)\n\nUse your trig identities:\n\ntan A = sinA/cosA\n\ncot A = 1/tanA or cot A = cosA/sinA
72362
I forgot how to do those, but it's some sort of trig identity isn't it?  Like\n\nsqrt (1-tan [x])\n\nProbably not though.
890950
Did they tell you that\nIf x is very close to 0, then sin(x)=tan(x)=x;\n(sin(x/2))^2 / x^2 = (x/2)^2/x^2=1/4\n(tan(2x))^3 / x^3 = (2x)^3/x^3=8
714467
(2 tan x ) / (3 - 2 tan squared x) = 1\n\nMultiply both sides by  (3 - 2 tan squared x)  and get\n\n2 tan x = 3 - 2 tan squared x\n\nLet u= tan x and the equation becomes\n\n2 u= 3 - 2 u^2\n2u^2 +2u -3 = 0\nu^2 + u -1.5=0\n\ncompleting the square you get\n(u  + 0.5)^2 -1.5 - 0.25 = 0\n(u  + 0.5)^2 -1.75= 0\n(u  + 0.5)^2 =1.75\nu+0.5 = square root (1.75)\nu +0.5=1.32287566\nu=1.32287566-0.5\nu=0.82287566\n\ntan x = 0.82287566\n\nx=arctan ( 0.82287566)\nx=0.688534708 radians \n\nas the tan shape repeats \n\nx = 0.688534708 radians AND (0.688534708 +pi) radians \nx = 0.688534708  and 3.83012736 radians
274244
Put Y=tan x and solve the equation for Y.\nThis gives two solutions Y=tan x = 1,7808 and Y =tan x = -0,2808.\nThis leads to x = 60.68 degrees and x = -15.68 degrees.\n\nAdd 360 degrees to the second solution and transform to degrees and minutes for the final answer.
746463
I'm guessing the second one should be h/(x+300), not h/x + 300.\nLets make h the subject of each.\nh = x tan 31\nh = (x+300) tan 24.\nEquate those:\nx tan 31 = (x+300) tan 24\nx tan 31 = x tan 24 + 300 tan 24\nx(tan 31 - tan 24) = 300 tan 24\nx = 300 tan 24 / (tan 31 - tan 24).\nYou can then work out h = x tan 31.\nYou should get x = 858.2, h = 515.7, approx.
404800
sec = (1/cos); cot = (1/tan) = (cos/sin)\n\ncos(-pi/3) = 1/2 \n\n1/1/2 = 2 = sec(- pi/3)\n\ncos(-5pi/4) = -sqrt(2)/2 ;      sin(-5pi/4) = sqrt(2)/2\n\n(-sqrt(2)/2)/(sqrt(2)/2) = -1 = cot(- 5pi/4)\n\n2 - (-1) = 3 There's your answer.
830888
OK, first let's try to figure out the arctan part.  Arctan(-1) means the angle whose tangent is -1, right?  And the tangent is the sine divided by the cosine.  So you're looking for angles whose sines and cosines are equal, but with opposite signs.  They point out to the right and down, or out to the left and up, at 45-degree angles.  That means the angles are 135 degrees and 315 degrees, or 3pi/4 and 7pi/4 radians.\n\nNow, the question asked you to find the sines of these two angles.  The sine of 3pi/4 is sqrt(2)/2, and the sine of 7pi/4 is -sqrt(2)/2.  So your answer is +/- sqrt(2)/2.
710047
Since theta is between 90 and 180 degrees, it is in the II quadrant. Tan = opp/adj. The y value (opp) is +3, and the x value (adj) is -4.  By the Pythagorean theorem, hyp^2 = opp^2 + adj^2. So the hypotenuse  is the sqrt of 3^2 +4^2 and is equal to 5. Cos is add/hyp, and is -4/5. So the answer is -4/5.
303020
The half-angle formula for tangent is:\n\ntan(x/2) = sin(x)/(1+cos(x))\n\npi/8 is not a very easy angle to work with, but pi/4 is. So let x = pi/4 in that half-angle formula it'll give you tan(pi/8).
930129
r = √1² + (-1)² = √2\nθ = arctan(1/-1) = -π/4 ± n2π, where n is any integer.
1153228
If the two triangles are similar, then all the angles are the same.  So tan(Vert/Horiz) = the top left angle.  Tan 1.5/1.6 = 43.15   Tan 43.15 = Vert/Horiz\n\n1.5/1.6 = x / 4\n\nx = 4.267 m
943371
f(x)=a+tan(bx)\n\n(-7/2) < b < (7/2)\n\nf(0)=-5\nf(π/7)=sqrt(3)-5\n\nf(0)=-5=a+tan(0)=a\na=5\nf(-π/7)=-5+sqrt(3)=a+tan(bπ/7)\nsqrt(3)=tan(b*π/7)\narctan sqrt 3=60°=π/3\nb*π/7=π/3\nb/7=1/3\nb=7/3  so \nf(x)=-5 +tan(7x/3)
18861
tanA + tanB + tanC = tanA tanB tanC is true for any non-right angle triangle\n\nrefer the link below . adequately solved
324333
Look at this link.\n\nUse the first two "sum to product" identities.\n\nOn the right you will have:\n\n= (2 / 2) (sin a / cos a) (cos b / sin b)\n\n= 1 (tan a) (cot b)\n\na = (x + y) / 2,  b = (x - y) / 2\n\nso\n\n= (tan (x+y)/2) (cot (x-y)/2)\n\nOh, darn!  You're right, the signs are reversed from this.  I wonder if the problem was given to you incorrectly.  I don't see how you could go from our result to the problem the way it was given.  Ugh!\n\nI'm sorry, but I can't go any further with this.  Good luck, and let us know if the problem was stated wrong, please.