Help on Trigonometry, please?
466401
Trigonometry is hard to do in you head, so start out by making a drawing. You have two buildings A (tall) and B (short) - Draw them far apart, not like this:\n\nA         B\n|         \n|         \n|    \n|         |\n|         |\n|         |\n|         |\n\nNow draw one line from the top of A to the top of B - this is the angle of depression of the top and it is 46 degrees. That is, if you drew a straight, horizontal line across the top of building A, the angle between that line and the line joining the top of A to the top of B is 46 degress.\n\nSimilarly, draw a line from the top of A to the *bottom* of B - this is the angle of depression of the bottom and it is 77 degrees. That is, the angle between the horizontal line across the top of A, and the line between top of A and bottom of B is 77 degrees.\n\nNow can you find out the angle between the line from A-top to B-top, and the line from A-top to B-bottom? (It's very important for you to draw this!) Let's put this value aside for a minute. \n\nLooking at your drawing, can you see that you know the value for *part* of the angle between the horizontal line across A's top and length of building A? What is that part you know? Now, see how the line across A's top and length of A make a right angle? So how many degrees in a right angle? So if you know the total of that angle, and part of that angle, how do you get the other part? **Notice that this part we're trying to get is actually the angle between the line A-top to *B*-bottom, and the line A-top to *A*-bottom. \n\nYou should also notice that now we have two triangles:\nTriangle 1: A-top to B-bottom to A-bottom. Solve this triangle for line A-bottom to B-bottom for distance between the two buildings.\nTriangle 2: A-top to B-top to B-bottom. Solve this triangle for line B-top to B-bottom for height of building. \n\nDon't forget, we have two important angles that will help solve these two triangles!
284409
Ex:\nA man stands 60 feet away from a building and sees the top of the building at a 60 degree angle. How tall is the bulding?\n \n1)Choosing an equation: (sin, cosine or tan)\nGoing to those three equations which one do we use?\nWell, we want to find the hight of the bulding (opposite b/c it is opposite the angle), so lets use tan.\n \n2)Setting the equation:\n \ntan 60 (the measure of the angle is θ) = (x) (opposite) / 60 (side adjacent to angle is adjacent)\n \n3)Press on Calculator: tan 60\n \nU should get: 1.732050...\n \n4) Multiply by 60. (b/c ur solving for x , see original equation),  \n5)Answer:  about 103.9230485\n \n6)Using 30-60-90 Triangle tricks we check our answer and find the height is 60 X (sqrt 3) which is aproximately 103.9230485 so we are right.
472052
ok so i assume you know the basic rules of trigonomerty for a triangle.\ni.e sinR = opposite/hypotenuse and cosR= adjacent/hypotenuse.\n\ndraw the triangle on a piece of paper.\nYou will see that sin R = ST/SR\n\n(where ST is the edge running from S to T and similarly SR is the edge running from Sto R)\n\nalso cos R= RT/SR\n\nso sinR^2 + cosR^2 = (ST)^2/(SR^2) + (RT)^2/(SR^2)\n\n= [(ST)^2 + (RT)^2]/(SR^2)\n\nNow we use pythagoras' theorem. I assume u know this? The sum of the squares of the two shorter sides of the triangle equals the square of the longer side (the hypotenuse)\n\nso (ST^2) + (RT)^2 = SR^2\n\nSO WE NOW HAVE\nsinR^2 + cosR^2 = SR^2/SR^2 = 1\n\nthere you go\n\n\nCANT E-MAIL YOU BACK WESTILE BECAUSE IT SAYS YOUR E-MAIL IS NOT VERIFIED
109151
It looks like your equation got cut off, but I think you were writing\n\nR = (V^2/2g)(sin (2 theta) +sin (2 theta))\n\nwhich can be immediately simplified, by adding the sin (2 theta) terms, as \n\nR= (V^2/2g )(2)(sin(2 theta) = (V^2/g )sin (2 theta)\n\nwhich is the formula that the other correspondent already gave.  To solve it, simply multiply both sides by g, divide by V^2, and you get \n\n(R g)/v^2 = sin (2 theta)\n\n[Arcsin (Rg/V^2) ]/2 = theta
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.
329665
Well I'm not much help, but to solve trig. identies you basically just use the given identies and equal them to each other. You substitue whutever equation is possible to simply the identity down to it's simplest form. \n\nLook at the bottom of this website for more: http://library.thinkquest.org/20991/alg2/trigi.html\n\nExcept that website might actually be more focused on trig. Equations not specifically Identities... lol So try using this website instead =) :\nhttp://www.intmath.com/Add_Trig/1_TrId.php\n\nHave fun.
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!
779945
1. cosθ=A*B / (|A|*|B|)=(1*5+0*2+3*6) / 5√ 26\n            =81.171\n\n2. assume the angle of vectors of a and b is "m"  \n   so cos m= a*b/(|a|*|b|)=0   because cos90=0\n   m=90...so it is a right triangle...\n   |a| and |b| are representing the right-angled sides of the triangle.\n   |a|=√17; |b|=√26\n   S=|a|*|b| / 2 =10.5\n\n3. P(1,3,5) to cylindrical coordinates.......\n                         r=√1+9 =√10....\n                         θ= tan^(-1)3=71.6\n                         z= 5 \n       In terms  of x , y , and z\n                         x= r.cosθ =√10*71.6 =226\n                         y= r.sinθ = 0.95*√10= 3\n                         z=5\n                        so its (226,3,5)\n            ditto, T 's doesn't have cylindrical coordinates \n                     S 's cylindrical coordinates is (0.6, -0.8, -10)\n  ......................(that's all i know..)
129693
1) square both sides\n\n2) then make trig substitutions to get same 'variable'. Get sin^2[x] throughout equation. \n\ncos^2[x] + sin^2[x] = 1, so\n\n1 - cos^2[x] = sin^2[x].\n\n3) sin[x] cos[x] = ...\n\n4) Set equal to zero\n\n5) Solve.
830578
Adding these fraction expressions (common denominator  etc) gives\n(sin3θsinθ + cos3θcosθ)/(cosθsinθ)\n\nThe top is the expansion of\ncos(3θ - θ) which is cos2θ\n\nThe bottom is half of \nsin2θ = 2sinθcosθ,\nso multiply top and bottom by 2 and you have the result, since\ncos2θ/sin2θ = cot2θ
1073595
(cosx)^2 - 2sinx = 2\ncosx^2 = 1 - sinx^2\n(1 - sinx^2) - 2sinx = 2\n1 - sinx^2 - 2sinx = 2\n-sinx^2 - 2sinx + 1 = 2\n-sinx^2 - 2sinx - 1 = 0\n-(sinx^2 + 2sinx + 1) = 0\n-(sinx + 1)(sinx + 1)\n\n-sinx - 1 = 0\n-sinx = 1\nsinx = -1\nx = sin^-1(-1)\nx = 270°\n\n--------------------------------------\n\n1.5^2 = 1.7^2 + 1.8^2 - 2(1.7)(1.8)cosA\n2.25 = 2.89 + 3.24 - 6.12cosA\n2.25 = 6.13 - 6.12cosA\n-6.12cosA = -3.88\ncosA = (388/612)\ncosA = (97/153)\nA = 50.6551127\n\n1.7^2 = 1.5^2 + 1.8^2 - 2(1.5)(1.8)cosB\n2.89 = 2.25 + 3.24 - 5.4cosB\n2.89 = 5.49 - 5.4cosB\n-2.6 = -5.4cosB\ncosB = (13/27)\nB = 61.2177953\n\n1.8^2 = 1.5^2 + 1.7^2 - 2(1.5)(1.7)cos(C)\n3.24 =  2.25 + 2.89 - 5.1cosC\n3.24 = 5.14 - 5.1cosC\n-1.9 = -5.1cosC\ncosC = (19/51)\nC = 68.1270919\n\nAs you can see, the angle opposite of side A, is the smallest angle.\n\nANS : about 51°\n\n----------------------------------------------\n\ncos(A) = 4/7\nx = 4\nr = 7\n\nr = sqrt(x^2 + y^2)\n7 = sqrt(4^2 + y^2)\n49 = 16 + y^2\ny^2 = 33\ny = sqrt(33)\n\nsin(A) = y/r\nsinA = (-sqrt(33))/7\n\ntanA = y/x\ntanA = (-sqrt(33))/4\n\ncscA = r/y\ncscA = -7/(sqrt(33))\ncscA = (-7sqrt(33))/33\n\nsecA = r/x\nsecA = 7/4\n\ncotA = x/y\ncotA = -4/(sqrt(33))\ncotA = (-4sqrt(33))/33
1041436
I got  the first part. just get the cosx by itself. then you wind up with the cosx = -(the square root of 2/2)\nwhich is 225+360n and 315 +360n(in degrees)\nSorry i can't help you on the other one. Good luck, just keep trying new things with it, would be my advice.
1159612
1) You meant to have a minus sign not "over". I've used x in place of "theta"\n    secx = 1/cosx and tanx = sinx/cosx \nSo 1 becomes (1/cosx)*sinx - tanx = 0 \nor tanx - tanx = 0, which is true.\n\n2) tanx = 1/cotx, so tanx*cotx = 1 \nSo 2 becomes 1- tanx*cosx = 1-sinx\nBut tanx = sinx/cosx so we have:\n1- (sinx/cosx)*cosx = 1-sinx, so\n1-sinx =1-sinx, which is true.\n\nI'll do one more:\n\n3) tan^2 x - 3sinx*tanx*secx = - 2tan^2 x\nAgain secx = 1/cosx and tanx = sinx/cosx\nSo 3. becomes:\ntan^2 x - 3(sinx/cosx)*tanx = - 2tan^2 x, so\ntan^2 x - 3tan^2 x = -2tan^2 x, so\ntan^2 x = tan^2 x, which is true.\n\nYou need to learn to do these yourself: Here is a link which gives you all the trig identities to solve the rest. They are not difficult.\n \nhttp://www.sosmath.com/trig/Trig5/trig5/trig5.html
730760
θ = α + β  where α = tan^-1 (1/3) and β = tan^-1 (1/2)\nie tanα = 1/3 and tanβ = 1/2\n\ntanθ = tan(α + β)\n= (tanα + tanβ)/(1 - tanαtanβ)\n= (1/3 + 1/2)/(1 - 1/3 * 1/2)\n= (5/6)/(5/6)\n= 1\nSo θ = ....., π/4, 5π/4, .....\nSo cosθ = ±1/√2 = ±√2/2
844759
Look at the various values of cosx when x = pi/2, pi, etc. (recall that pi/2 = 90 degrees).\n\nFrom looking at these values in my book, I see two values for x that will satisfy this equation:  x=pi/3, and x=pi\n\ncos pi/3 = 1/2\ncos 2pi/3 = -1/2\n\ncos pi = -1\ncos 2pi = 1\n\n~ ♥ ~
610143
I think ur right cos^2(theta) should be cos(theta). then the answer would be perfect.
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.