Double Angle Trig Proof/Identity?
952626
Rewrite cos 4A as cos 2(2A)\n\ncos 2(2a) is 2 cos^2 2A - 1\n\nso you have cos^2 2A = (1 + 2 cos^2 2A - 1)/2\n= (2 cos^2 2A) /2\n= cos^2 2A
161223
law of cosines : a^2 = b^2 + c^2 -2bc * cos(A)\nwhere a, b,c are side lengths and A is the angle oppisite side a.
960686
The equation is correct but only if A is 0, 180 or integer multiples of 180 degrees.\nsecA/(1+sinA) - cosA/(1+sinA) = tanA\n(secA - cosA)/(1+sinA) = tanA\ncosA(secA - cosA)/(1+sinA) = tanA * cosA\n(1-cos^2A)/(1+sinA) = sinA\nsin^2A/(1+sinA) = sinA\nsin^2A = sinA + sin^2A\nsinA = 0  which occurs when A=0,180,360,etc.
610758
sin(2x) = 2sinxcosx is a trigonometric identity called the double angle identity. It was derived from another identity:\n\n   sin(x + y) = sinxcosy + cosxsiny\n \nhere x and y are two different angles but what if x = y\nreplace y with x to make all the equations in terms of x, thus the equation becomes:\n\n   sin(x + x) = sinxcosx + cosxsinx\n   \nnote that sinxcosx is also equal to cosxsinx, so simplifying the equation it becomes:\n\n  sin(2x) = 2sinxcosx or sin(2x) = 2cosxsinx
1159205
pair up the  opposite sines, for example sin(10) and sin(350) and add them:\nsin(10) + sin(350) = 0\nsin(20) + sin(340) = 0\n...\nuntil you only have sin(180) and sin(360) which both equal 0\n\nPS. I assumed you were using degrees (sexagesimal). Won't work in radians (RAD) or metric (decimal) angles.
60232
Well, Vanessa almost got it....the side opposite the 90 degree angle is the hypotenuse "C", the adjacent sides to the angle are "A" & "B". The correct formula is A squared plus B squared = C squared.\n\nor\n\nA^2 + B^2 = C^2\n\npi is an infinately long number about equal to  3.14
713469
Can't u prove it ?\nuse simple geometrical angle laws.\n\nif we have a triangle : \nupside-- the inclined plane\ndownside--- the horizontal\nright side--- the gravitational force (vertical)\n\nDRAW IT........ as u know \n\nand then draw the perpendicual component on the plane     \n\nuse this: sum of the two angles inside the triangle    =180-90=90\n\nsum of angles at the gravitional point= 180-90=90 also\n\nu will prove the equality
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.
748572
sin(x)tan(x)=(cos(x)csc^2(x))/(cot^2(x)+cot^4(x))\nStep 1. Starting with the right side (the obviously more complicated one) factor out a cotangent squared out of the denominator.\n\n(cos x csc^2 x)/ cot^2 x (1+cot^2 x)\n\nStep 2. Use the identity csc^2 x= cot^2 x+1 in the numerator  and separate the cosine x in the top and the cotangent squared x in the bottom\n\n(cosx/cot^2 x) * (cot^2 x+1/ cot^2 x +1)\n\nStep 3. This leaves you with a cosine x divided by cotangent squared x times 1.  Use the identity that says cot x = cos x/ sin x\n\n(cos x) / (cos^2 x/ sin^2 x)\n\nStep 4. Divide the double fraction and get\n\n(sin^2 x cosx)/ (cos^2 x)\n\nStep 5. Separate your multiple factors from the squares and separate the fraction into three parts, one will be the sine x, the other will be the tangent x (because tan x = sin x/ cos x), and the cosine x divided by cosine x will go to 1.\n\nsin x * (sin x/ cos x) *(cos x/ cos x)\n\nStep 6.  Write final answer\n\nsin x tan x = sin x tan x
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
528735
tan by definition is equal to sine/cosine.\n\nProof:\n\ntan = opp/adj\nsin = opp/hyp\ncos = adj/hyp\n\nsin/cos = (opp/hyp)/(adj/hyp) \n= opp/hyp*hyp/adj \n= opp/adj\n= tan\n\nSorry about that, knew that was too simple :P!\n\nNEW ANSWER:   You nedd to draw out the triangle and solve from there.  Let there be a triangle with adjacent side a,  and opp side b.  \n\nThen if tanbeta = 2, opp= 2, adj = 1 and hyp = sqrt(5)   from the pythagorean theorem.\n\nThus cos = 1/sqrt(5) and sin = 2/sqrt(5)\n\nThe quadrant only tells you the sign of cos and sin or how the angle is measured.  Since you're in QI, both sin and cos are positive so no modification to the above answers are needed.\n\nNow you might wonder why I said opp was 2 and adj was 1.  It doesn't matter (can't remember the proof for that right now).  As long as you pick a solution that works, say 4 and 1 or 3 and 1.5, it will ALWAYS simplify to the answer I just gave you.
905611
If u r saying that u have three sides of a triangle as a,b,c\nAnd u want to determine angle b/w them, u can use the formula\ncosA=(b^2+c^2-a^2)/2bc where A is the angle formed by the sides b and c.similarly if B is the angle formed by a and c, it can be calculated as\ncosB=(a^2+c^2-b^2)/2ac\nonce u have calculated A and B, subtract their sum from 180, u will get third angle.\nRemember u will have to use inverse table for cos to find A and B
624769
You can use sin and cosine.
844799
Wallstream is correct -- start with a right triangle.\n\nLabel the interior angles a, b and c, with c being the 90-degree angle.\n\nNow notice that a and b are complementary angles that add up to 90-degrees.\n\nLet x = sina.  Since the angles are complementary, we also have that x = cosb.\n\nSo we can rewrite as follows:\n\nIf x = sina, then a = arcsinx\nIf x = cosb, then b = arccosx\n\nBut a+b = 90-degrees (aka pi/2).\n\nSo arcsinx + arccosx = pi/2\n\nHope that helped!
484958
sin 2x=2 sinx . cos x\ncos 2x=(cos x)^2-(sin x)^2\ncos 2x=2 (cos x)^2 -1\ncos 2x=1- 2 (sin x)^2\ntan 2x=(sin 2x)/(cos 2x)\n\nor you can \nsin 2x=1\nlet 2x=y\nsin y=1\ny=90 degree + 360 degree\n2x=90 degree + 360 degree\nx=45 degree + 180 degree\nso x={45,225,....}degree\n\ntan x+cot x=2/sin 2x:\ntan x+cot x\n=(sin x/cos x )+(cos x/sin x)\n=((sin x)^2+(cos x)^2)/(sin x . cos x)\n=1/(sin x . cos x)\n=2/2(sin x . cos x)\n=2/sin 2x\nso it is an identity
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^_^
567245
simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard