Using a half angle identity, help me to find the exact value of sin 105degrees?
359761
When you need to use the HALF ANGLE formula, think of the given angle has HALF the ANGLE you need to use.  So:\n\nsin(105) = sin(210/2)\n\nThen use the formula using the angle 210.  And, since you know other ways to get the answer, you can check that you did it right!  (That's probably why you were given that angle.)
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.
546529
1. Make sure the angle of interest is in radians.  Radians are the "natural" (non)-units of angles, because they relate the angle measurement to the length of arc the angle cuts off on a unit circle.  So 2*Pi radians are equivalent to 360 degrees.  Basically, if you have that sucker in degrees, multiply it by 2 * Pi / 360.  Unit conversion is done. \n \n2. Make sure the angle is between -Pi and Pi.  Because sine is a periodic function, with period 2*Pi, just subtract or add 2*Pi enough times to get the angle between those two points.\n\nnow for the meat of the problem:\n\nsin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - ... + \n+ (-1)^(n-1)*x^(2n-1)/(2n-1)! + ...\n\nExample: sin(15)\n\nStep 1: 15 degrees = 15*(2*Pi/360) radians = Pi/12 radians \n \nStep 2: I need Pi/12 in decimal (let's see, Pi=3.14159 (memorize that much, it's pretty easy... "Pi when calculating sine, 3 point 1 4 1 5 9!")) \nPi/12 approx = .261799 (we only had 6 sig figs, so we can't add any more than that) \n \nNow to the series: \n         first term:  x=.261799 \n         1st estimate: .261799 \n         second term: -x^3/3! = -.0029905  \n         2nd estimate: .258809 \n         third term: x^5/5! = .0000102     \n         3rd estimate: .258819 \n         fourth term: -x^7/7! = -.167*10^-7  \n         stop here - past precision\n\nThe answer is positive \nsin(15) = 0.258819\nsin(18) = 0.309017
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^_^
59453
Using your unit circle, you can see that:\nsin 30 = sin 150 = -sin 210 = -sin 330 and\ncos 10 = cos 350 = -cos 170 = -cos 190\n\nThis illustrates some basic identities which you should memorize, namely that\n\nsin (180 - x) = sin x and sin (-x)= - sin x\nand\ncos (180 - x) = -cos x and cos (-x) = cos x\n\nThese are probably in the front or back cover of your textbook.  :)
757793
1. Using exact values, verify that Cos 60°= 1-2 sin² 30°\nRHS=1-sin^2 30*\n=1-(1/2)^2\n=1-1/4\n=3/4\nLHS=[(rt3)/2]^2\n=3/4\nLHS=RHS\n\n2. Using exact values, find the value of:\n\na. 12 sin 30° -6 tan 45° + sec45°/52 \n12(1/2)-6(1)+(rt2)/52\n\nb. sin 2A + tan 3A/2 – cos A + sec (A+15°), When A= 30°\n=sin60*+tan45*-cos30*+sec45*\n=rt3/2+1-rt3/2+rt2\n=1+rt2\n\n3. Write the other functions of x if cos x= 3/5..(show work)\nthe right triangle will bea 3,4,5 triangle using the pythagorean triplet\nsinx=4/5\ntanx=4/4\ncotx=3/4\nsecx=5/3\ncosecx=5/4 \n.\n4. In a right triangle sin A =3/4 and c =17. Find a and b. Be sure to draw a right triangle and letter it properly.\nc=17\nletthe oppositeside be 3x and the hypotenuse be 4x\nadjacent side=rt(16x^2-9x^2)\n=(rt7)x\n4x=17\nx=17/4\nrt7 x=17/4\nsquaring\n7x^2=289/16\nx^2=(289/16)*(1/7)\nx=rt[9289/16)(1/7)\n5. The vertex angle of an isosceles triangle is 57° 24’ and each of its equal sides is 375.5 \nft. long.Find the altitude of the triangle.\naltitude/side=sin of base angle\nbase angle=(180-57*24')/2\naltitude=375.5 times sin[(180*-57*24')/2]\n6. Solve the following right triangles:\na. a=117 ft.--B=16.35ft.\ncapital letters are used only for angles\nso B=16.35' is wrong\nb. B=8°29, --a=32.8 ft. \nare you taking <c as the right angle\nif so <a=(90*-8*29')\nuse a/sinA=b/sinB=c/sinC\n\n7. A man standing on top of a building sees an automobile on a street below him.The angle of depression is 36 degrees 22. If eye is 550 ft. above the level of the automobile ,how far is the automobile from him?\nlet the horizontal distance be x\nx/550=cot36*22'\nx=550cot 36*22'\n=>550tan(90-36*22')\n\n8. When the suns angle of elevation is 30 degrees. The shadow of a post is 6 ft.longer than when the angle is 45 degrees . Find the height of the post.(show work)\nlet the height of the post be x\nthe length of the shadow at 30*=xtan30*\nthe length of the shadow at 45*=x tan 45*\ngiven the difference x(tan45*-tan30*)=6\nsolve\n\n9. In a right triangle, given a= 32.8 ft. and A=28° 30’ 15’’,solve for B and C\ntaking <C=90* by convention\n<B=90*-28*30'15" \nnow use a/sinA=b/sinB=c/sinC
509207
Yes. You can use sine or cosine to determine the angle.\n\nSine x = Opposite / Hypotonuse\nSin x=5/13\nSin x=.38\nx=22.6*\n\nCosine x = Adjacent / Hypotonuse\nCos x=12/13\nCos x=.92\nx=22.6*
1050391
EDIT: I was looking at the answer above mine, and the cos(0) and sin(0) values are reversed, so it is not x=0.\n\nOK!! No problem. First of all, let me just say I teach math, but there is no way I would want to memorize all of the trig identities, so open up your precalc book, and if it is a good book, the major identities will be located on the front or back cover. If not, then go to the applicable page.\n\nStart with what is given:\n(sq root:3)sinx + cosx = 1\n\nAll I did was go through the identity lists and see what seemed to match closest to what you have there, and I noticed that:\n\ntan(x/2) = (1 - cosx) / sinx\n\nThen I visualized doing a little bit of algebra with your problem, and saw that if you subtracted cosx from both sides you would have:\n\nsqrt(3) sinx = 1 - cosx\n\nWhich is kind of similar to tan(x/2), except we need a sinx in the denominator. And if you divide both sides by sinx, that is what you get! So now we have:\n\nsqrt(3) = (1 - cosx) / sinx\n\nBut we already know that this is the same as tan(x/2)\n\nSo, we have:\n\nsqrt(3) = tan(x/2)\n\nFrom here, since you want to find the actual value for x which will be an angle measure, you go to inverse tangent:\n\ntan(-1) sqrt(3) = (x/2)\n\nI also have a handy table of all the major trig values open (it is in the back cover of my precalc book) and notice that tan(60) = sqrt(3)\n\nWe are almost done:\n\n60 degrees = x / 2\n\nMultiply by 2,\n\nx = 120 degrees.\n\nOf course, you should check that answer and make sure it is correct!!
497748
In rt triangle ABC with B= 90degrees (draw figure)      \nIf C is theta, cosC=7/8(ignoring sign)\ni.e., BC = 7, CA = 8 (in simplest terms)\nusing pythagoras theorem, AB= sqrt(15)\n \nsince C=theta is in the second quadrant tan C will be negative\n\nso tan C = AB / BC \n\n  tan C= - { sqrt(15) } / 7\n\ntan(theta) = -0.5533
809910
cosine 75*=cos(90-15*)=sin15*
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!
461039
sin2A = 2sinA.cosA\nSo,\nsin2A = 2*12/13*5*13\n      = 120/169\n\ngood luck!!
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
86738
The angles all add up to 180 degrees.\n\nIf you know the lengths of all the sides you can use the "Law of Tangents" to find all the angles.  This is advanced geometry and Trigonometry, so rather than type, I am including a link to a pretty good explanation.\n\nhttp://answers.google.com/answers/threadview?id=38131
718931
The only possible answer for your fixed number would be zero.  So you just have to do some inverse trig functions to get the value for x.  \n\nSolved:\nLet x=1.0038848219 [radians]\nTan(1.0038848219)=1.570796327=pi/2\nCos(pi/2)=0\nSin(0)=0
166573
:)\nNice trick question.\nYou should mention that the whole 'equations' reads:\n2 times (pi) radians = 360 degrees (measuring angles).\ntherefore (pi) radians equales 180 degrees (half circle)\nBut the number (pi) is same old 3.14......
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