trigonometry. thank u for ur help
121046
Factor the left side: 2cosx(tanx+1)=(tanx + 1)\n\nThis is true whenever tanx+1 = 0, or 2cosx = 1\n\ntanx = -1 in degrees, -135 or 315\ncosx=1/2 in degrees, 60 or 300\n\n(or multiples of 360 plus the above)
373284
whats the equation or formula u need help with?
346551
tan(a+b)=sin(a+b)/cos(a+b)=\n(sinacosb+sinbcosa)/\n(cosacosb-sinasinb)\ndivide top and bottom by cosacosb\n(tana+tanb)/(1-tanatanb)
211121
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!
552872
Depends\n\nlim θ→π/2- (ie from below) tanθ = +∞\n\nlim θ→π/2+ (ie from above) tanθ = -∞ \n\nThis is because in the 1st quadrant tanθ >0 and in the second quadrant tanθ <0
437258
To understand you should read over this section on Trigonometry  identities on the site www.wikipedia.org and form here you will be able to reagrane the equation since somethings equal others
202300
Knowing trig makes learning calculus easier.
207416
Go to http://mathworld.wolfram.com/Cosine.html
547300
sorry, i don't have a calculator w/ me right now but i'l try\n1. use the law of cosines\na squared = b squared + c squared -2bc*cosA\n2. in a 30-60-90 triangle the sides are x- the side across from the 30; x*square root 3- the side across from the 60; 2x-the side across from the 90\n3. i would try the law of sines\na/sinA = b/sinB = c/sinC\nremember that you know two angles, the 90 and the 61\nso the 61 is the degree the ladder and the ground forms.  set up the problem like this: x/sin61 = 18/sin90\nit should work\n\nto find secant, type into the calculator 1/cos(36) and check what mode you're in (degree or radians)\ni'm pretty sure cot60 = 1/the square root of 3\nsec30=2/the square root of three
672883
When I was first trying to do trigonometry problems like this, I found it helpful to draw a picture.\n\nDraw a horizontal line to represent the ground, and then put a vertical line to represent the radio tower.  We want to know the height of the tower, so label it "x".  The angle between the tower and the ground is 90 degrees, so label it too.\n\nNow draw a dot (or a little stick figure with a protracter if you feel like it) to the side of the tower, on the ground.  Label the line between the tower base and the dot "50", as it's 50 meters from the tower.\n\nNow draw a line between the dot and the top of the tower, representing the person sighting to the top and measuring the angle.  Lable the angle between this sight line and the tower top "67 degrees".\n\nYou now have a drawing of the triangle with two angles and one adjacent side known.  This should be enough to get you started picking the right trig function to apply here.\n\nIf that's not enough to get going, here's another hint: Remember the unit circle with a triangle and the sides labeled with "sin(theta)", "tan(theta)" etc?  Are there any similarities to your tower drawing?  I always found being able to draw that unit circle on my scratch paper at the beginning of a problem set or test helped immensely.\n\nGood luck.
517195
= 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)
447009
See the sources.\n\nYou start with\n\nds = a y ''\n\nds = √[1 +( y ')² ] dx
150155
Well let's look at the dates first\nA. 1538\nB. 1636\nC. 1802\nD. April 23, 1564\nE. January 4, 1643\nSources below in order of parts (so a is first)\nNow, the word trigonometry was first used in 1595 (last source), so the answer is a and d
150389
The answer is B) 1.122\n\nFor the different trigo. functions:\n\nLet:\na = side opposite angle n\nb = side adjacent angle n\nc = hypotenuse of the right triangle\n\nTherefore:\n\ncos n = adj./hyp. = b/c\nsin n = opp./hyp. = a/c\ntan n = opp./adj. = a/b\nsec n = hyp./adj. = c/b\ncsc n = hyp./opp. = c/a\ncot n = adj./opp. = b/a\n\nYou will notice the following:\n\n1) the secant is the reciprocal of cosine.\n2) the cosecant is the reciprocal of sine.\n3) the cotangent is the reciprocal of tangent.\n\nTherefore: \n\ncsc n = 1/sin n = 1/0.891 = 1.122\n\nFor the remaining trigo. functions, solve for n first:\n\nn = arcsin 0.891 = 62.99917663˚ or 62˚59'57.04"\n\nTherefore:\n\n1) cos 62˚59'57.04" = 0.454\n2) tan 62˚59'57.04" = 1.963\n3) sec 62˚59'57.04" = 1/cos 62˚59'57.04" = 2.203\n4) cot 62˚59'57.04" = 1/tan 62˚59'57.04" = 0.510\n\nI hope that the solution was clear enough...
76829
First of all, the best way to learn trigonometry is to learn it from the beginning, which means take your first textbook which has trigonometry and study from the beginning.\nhttp://mathworld.wolfram.com/Trigonometry.html\nhttp://www.du.edu/~jcalvert/math/trig.htm#Trig\nhttp://www.syvum.com/math/trigonometry.html
641664
First off......do you know how to do basic trigonometry and use Snell's Law?  If you do, then I'll walk you through it.  Write me at fortitudinousskeptic@yahoo.com\nIf you don't know trig and Snell's law I'm not going to try to explain it.  I'm not that motivated.
345463
Cool geometry im learning this in class 2bad i cant help u cuz im faling -.-