Trigonometric identities?
924056
sin φ + sin φ tan φ + cos φ cot φ + cos φ = csc φ sec φ\n\nsin φ + sin φ (sin φ / cos φ) + cos φ (cos φ / sin φ) + cos φ = csc φ sec φ\nsin φ + sin²φ / cos φ + cos²φ / sin φ + cos φ = csc φ sec φ\n\ngetting the least common denominator which is sin φ cos φ\n\nthen \n\n(sin²φ cos φ + sin²φ sin φ + cos²φ cos φ + sin φ cos²φ) / sin φ cos φ = csc φ sec φ\n\ncombining similar terms\n\n(sin²φ cos φ + cos²φ cos φ) + (sin²φ sin φ + cos²φ sin φ) / sin φ cos φ = csc φ sec φ\n\nthen\n\n[cos φ(sin²φ + cos²φ) + sin φ(sin²φ + cos²φ)] / sin φ cos φ = csc φ sec φ\n\nsin²φ + cos²φ = 1\n\n(cos φ + sin φ) / sin φ cos φ = csc φ sec φ\n\n1/sin φ + 1/cos φ = csc φ sec φ\n\ncscφ + sec φ = cscφ sec φ\n\ni think you forgot to place a plus sign in between the csc φ and sec φ. i you do, then my solution is right. hope this will help.
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.
1132938
I can help you with the part of  "Identify the expression that completes the equasion so that it is an identity.\nsin^2x/(1-cosx)"\n\nYou got sin^2x/(1-cosx). Then to complete the equation you multiply[(1+cosx )/(1+cosx)] to the equation.  \n\nThen you get {[(sin^2x)(1+cosx)]/[(1-cosx)(1+cosx)]}...Then by factorization you get that {[(sin^2x)(1+cosx)]/(1-cos^2)]}....The (1-cos^2) is the same as to sin^2x.  \n\nThen you cancel the terms sin^2x/(1-cos^2)...It leaves the equation as following ==> (1+cosx)(1) = (1+cosx)\n\nHope someone helps you with the rest SOrry
287405
Use the identity cos 2x = 2 cos^2 x - 1\n\nThen you have 3(2cos^2 x -1) - 5cosx - 1 = 0\n\n6 cos^2 x - 3 - 5cosx - 1 = 0\n\n6 cos^2 x - 5 cosx - 4 = 0\n\n(3 cosx - 4)(2 cosx + 1) = 0\n\ncosx = 4/3 (impossible) or cosx = -1/2\n\nIn degrees, x = 120, 240, or if you need, 120+360n, 240+360n\nIn radians, x = 2pi/3 or 4pi/3 etc.\n\nHope that helps you see what to do.
676730
Opening brackets \nLHS=SinA+ SinA. TanA+ CosA+ CosA.CotA\n=SinA+Sin^2A/CosA+CosA+Cos^2A/SinA\nGrouping\n=(SinA+Cos^2A/SinA)+(CosA+Sin^2A?CosA)\n=(Sin^2A+Cos^2)/sinA+(Cos^2A+Sin^2A)/CosA\n=1/sinA+1/CosA\n=SecA+CosecA\n\nEasy Pesy
895946
There is a trig identity you need: cos(2x) = 2cos^2(x)-1.\nIf you want to derive it, you can use the cosine of sum formula:\ncos(x+x) = cos(x)cos(x)-sin(x)sin(x) and go from there...\n\nYou can apply this to your equation cosx+cos(2x)=0\n\nThis gives cos(x)+2cos^2(x)-1 = 0\n\nThen yes, you can substitute p for cos(x) giving\n\n2p^2 + p - 1 = 0\n\n(2p-1)(p+1) = 0\n\np = 1/2, p = -1\n\ncos(x) = 1/2, -1\n\nx = Arccos(1/2) or Arccos(-1)\n\nx = 60 degrees or 180 degrees
411302
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=1
935340
= cosx[cosxcosy+sinxsiny]+sinx[sinxcosy-cosxsiny]\n\n= (Cosx)^2cosy+cosxsinxsiny+(sinx)^2cosy-cosxsinxsiny\n\n= (Cosx)^2cosy + (Sinx)^2cosy\n\n= Cosy [ (cos)x^2 + (sinx)^2]\n\n= Cos(y)
677828
Yes use the trig identities:\n\ncsc x = 1/sin x\n\nSo for your first question\n\ncsc x = 3 sin x becomes 1 = 3 sin(x)^2\n\nso sin(x)^2 = 1/3  sin(x) = 1/rt(3) = 0.57735 and 35.26 degrees using the arcsin function.\n\nNote 1/cot x = tan x = sinx/cos x\n\nSo your second question becomes:\n\n2 sin x/cosx + 3sinx= 0  or 2sinx(1/cosx +3)=0\n\nThis is true if sinx = 0 or x= 0 or 180 degrees or k2pi, kpi where k=(0,1,2,3... \n\nor 1/cosx + 3 =0 which happens 1/cosx = - 3, means cosx = -1/3 means x = 109.5 degrees using arc cos function etc.
1021739
Since tanx = sinx/cosx,\n\ntan(A-B) = sin(A-B)/cos(A-B)\n\nNow, write out the expression for each term on the right side of this equation.\n\nWhen you finish that, divide each term by cosAcosB and you will get the result you are seeking.
756840
cosx=tanx\ncosx=sinx/cosx\ncos^2(x)=sinx\n1-sin^2(x)=sinx    {pythagoras}\nsin^2(x)+sinx-1=0\nuse quadratic formula\nsinx=(-1+or-sqrt(5))/2\n=0.6180339888\nor -1.6180339888\nsine has to lie between +/-1\nsinx= 0.6180339888\nx=arcsin(0.6180339888)\n  =38.17270763\n+ve value for sine lies in 1st and 2nd\nquadrants \ntherefore,x=38.17270763 degrees or\n141.8272924 degrees\n\nthese two values were checked in the \noriginal equation and were found \nto be correct\n\ni hope that this helps
982428
LHS\n= (sin^2B-tan^2B)/(1-sec^2B)\n= (sin^2B-tan^2B)/(-tan^2B)\n= (sin^2B/-tan^2B) + (-tan^2B/-tan^2B)\n[splitting the numerator up]\n= (-cos^2B) + 1\n[cosx= sinx/tanx]\n= 1 - cos^2B\n= sin^2B\n= RHS
419071
You can cheat and use a calculator to get the values, but I don't think this is what your teacher is looking for. \n\nFirst, use the relationship sin^2(x)+cos^2(x)=1 to find |sin(a)| and |cos(b)|. Pay close attention to the ranges of a and b to get the sign right-- sin(a) and cos(b) should both be negative. Doing this you should find:\n\nsin(a) = -12/13 \ncos(b) = (-2 sqrt(2))/3\n\nHere are two useful identities:\n\ncos(a-b) = cos(a)cos(b) + sin(a)sin(b)\ncos(2b) = 1 - 2sin^2(b)\n\nSince you now know cos(a),cos(b),sin(a) and sin(b), you can find that:\n\ncos(a-b)=(12-10sqrt(2))/39\ncos(2b)=7/9
1150095
its sooo easy  \nsinA-sinB = 2 sin[(A-B)/2] * cos[(A-B)/2]\ncosA - cosB = 2 cos[(A-B)/2] * cos[(A-B)/2]\nthus LHS = \n(2 sin[(A-B)/2] * cos[(A-B)/2]) / (2 cos[(A-B)/2] * cos[(A-B)/2])\n= sin[(A-B)/2] / cos[(A-B)/2]\n= tan[(A-B)/2 \n= tan (A/2-B/2)\n=RHS \n\nhence proved
333253
You can solve this using x, y, and r (the lengths of the 3 sides of the triangle made by the angle).  The trigonometric functions are originally defined as:\n\nsin = y/r\ncos = x/r\ntan = y/x\nsec = r/x\ncsc = r/y\ncot = x/y\n\nYou need to find x, y, and r for angles a and b.  Do this using the given sin and cot values and the equation\n\nx^2 + y^2 = r^2\n\nSo for a:\n\nsin a = 3/5\ny=3, r=5\n\nx^2 + 9 = 25\nx^2 = 16\nx=-4\n(this is negative because the angle is in the 2nd quadrant)\n\nFor b:\ncot b = -12/5\nx=-12, y=5\n(x is the negative one because this is in the 2nd quadrant)\n\n144 + 25 = r^2\n169 = r^2\nr=13\n\nNow you need to turn tan (a-b) into a form where you can plug in your x, y, and r values, and then plug and solve.  Remember that tan = y/x.\n\ntan (a-b) = (tan a + tan b) / (1 + tan a * tan b)\n=(-3/4 + -5/12) / (1 + (-3/4)(-5/12)\n=(-14/12)/(1 + 15/48)\n=(-7/6)/(63/48)\n=-8/9\n\nTherefore, your final answer is -8/9.\n\nHope this helps!
932836
Are you sure this is the correct identity?\n\n 1+ tanx.tan2x = 1 + (sinx/cosx)(sin2x/cos2x)\n                        = 1 + (sinx/cosx)(2sinxcosx/cos2x)\n                        = 1 + 2sinx.sinx/cos2x\n\n cos2x = 2cosx.cosx - 1  = 2cos^2x - 1        (Note:  ^2 = square)\n\n then..\n      1+tanx.tan2x = 1 + [2sin^2x/(2cos^2x - 1)]\n                            = (2cos^2x + 2sin^x -1) / (2cos^2x - 1)\n                            = [2(cos^2x + sin^2x) -1] / (2cos^2x - 1)\n                            = 1/cos2x    (because 2cos^2x - 1 = cos2x)\n\n     Then 1/1+tanxtan2x = cos2x
282780
1.sin(2x)=2sin(x)cos(x)\n2.tan(x)=sin(x)/cos(x)\n3.cos(x)=(1-sin^{2}(x))^(1/2)\n\nwe can use (3) in (2) and (1) to eliminate cos(x). Then substitute sin(x)=y:\n\n1.sin(2x)=2y.(1-y^{2})^(1/2)\n2.tan(x)= y/(1-y^{2})^(1/2)\n\ntherefore,\n\nf(y)=2y(1-y^{2})^{1/2}\n     +y/(1-y^{2})^{1/2}\n\nI hope that's what you wanted.