There seems to be a relation between trigonometric functions (like cosine, sine, and tangent)?
445038
Take a unit circle and a point p on the circle. Embed the circle to be centered at the origin on a Cartesian xy-plane. The cosine of p, cos(p), is the x-coordinate, or abscissa, of p. The sine of p is the y-coordinate. These functions are the projection functions for the unit circle. \nIt is common to give the points p on the 1-dimensional circle the coordinate referring to the arclength between p and the positive x-axis (as measured clockwise). Thus, the point (1, 0) on the unit circle is given the native coordinate (0), or 0 for short. Inversely, the point 0 in the unit circle has Cartesian coordinates (cos(0), sin(0)). Since the circumference of a circle is 2Pi, a point that is a quarter of the way around the circle from the positive x-axis is given the coordinate 2Pi/4 = Pi/2 and so forth. This system of coordinates is called radians and is the native way of measuring angles, as each angle has a unique radian measure. \nThis relationship can be seen by drawing a radius from the origin to mark the angle and assigning it the radian measure of wherever it intersects the unit circle. There are more analytical reasons as well, as you go on to study calculus.\nThe radius drawn above also gives the relationship between these circular projection functions, and the way one first encounters trigonometric functions, as ratios of the sides of right triangles. Now you should be able to visualize the triangles inside the unit circle (or any larger circle, dependent on the triangle's hypotenuse), and see why the trigonometric functions are cyclic in 2Pi, and derive some properties of the trigonometric functions that are not easily apparent from just looking at right triangles.\nThe tangent function is so named because it is the length of a certain tangent to the circle; a tangent drawn from the positive x-axis to where it meets an extended radius of the circle. You may be able to work out the equation tan(p) = sin(p)/cos(p) from the geometry involved. If you're familiar with the secant function, (1/cos(p)), you may want to know that it is a geometric secant as well. See the sources below for pictures.
226513
csc(x) is the same as 1/sin(x) and sec(x) is the same as 1/cos(x) so the left side of the equation can be simplified to \n\nsin(x)sin(x) + cos(x)cos(x) \nor sin^2(x) + cos^2(x)    {"sin squared x plus cos squared x"}\n\nUsing the same steps, the right side of the equation becomes sin(x)/sin(x) or 1.\n\nSo the identity has been reduced to  sin^2(x) + cos^2(x) = 1 which is another common trigonometry identity!
830426
The crux of this problem is for you to realize that the given angles are "complementary pairs", which add up to 90 degrees.\n\nYou are expected to know that the "co-function" of angle A gives the same value as the "function" of angle A's complement.\n\nSpecifically:\n\n cos(A) = sin(90-A)\n\nA graphical way of looking at this statement is that "cosine and sine are the same function, but offset by 90 degrees!"  We call that "out of phase by pi/2 or 90 degrees."\n\nThe same works for tangent and its co-function:\n\n tan(A) = cot(90-A)\n\nSo the answer to the second problem is 0.\n\nIf one of the fuctions in the first problem had been sin, the quotient would have been 1.  However, that's not the case, and I leave the upper problem as an exercise for another Answerer.\n\n Hope this helps!
908924
Because of (one of) the definition(s) of sin and cos.\nWhich is: take a unit circle, and draw in any radius at some angle theta. Then sin theta = the y value, and cos theta = the x value. Those are obviously less than 1.\n\nOr you could just take a typical right angled triangle - the length of the opposite/adjacent sides are always less than the length of the hypotenuse.
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!!
463801
The cosine laws apply to triangles which are NOT right angled. I think you have maybe read that wrong:\n\nhttp://www.themathpage.com/aTrig/law-of-sines.htm\n\ngives you a full guide to using the cosine laws to solve problems like these.
420145
The antideriative is the same as an integral.  For this, we can use the fact that sin^2(t) + cos^2(t) = 1.  Then,\n\n[int] cos^2(t)/sin(t) dt = [int] (1 - sin^2(t))/sin(t) dt\n= [int] (1/sin(t)) - sin(t) dt\n= [int] csc(t) - sin(t) dt\n= ln(tan(t/2)) + cos(t) + C
885632
Yes you are correct.\n\nx=tan t = (sin t)/(cos t)\nso cos t = (sin t)/x\nSo y^2 = (sin^2 t)/x^2 = (1 - cos^2 t)/x^2 = (1 - y)/x^2\n\nRearranging this you get y = 1/(x^2 +1)
680870
sounds like you need to use some basic trig functions.\n\nthe button "tan" on the calculator stands for tangent. the tangent of an angle of a right triangle (other than the right angle) is = to the ratio of the side opposite the angle to the side adjacent to the angle.\n\nthe button "sin" on the calculator stands for sine.  the sine of the angle of a right triangle is = to the ratio of the side opposite the angle to the hypotenuse.\n\nthe button "cos" on the calculator stands for cosine. the cosine of an angle is = to the ratio of the side adjacent to the angle to the hypotenuse.\n\nbasically, you decide which function to use that best fits the information given.  you enter the given angle in the calculater and press the appropriate function.  then you set that value (which you round off to the nearest ten-thousandth) equal to the appropriate ratio and then you solve.
105824
Trigonometry.  The good old days when I still loved math... ahhh!
200647
sin^2(x)=0.5*(1-cos(2x)).this can be integrated
831941
You're supposed to print a table, right? The question is poorly written, so it's not clear, but I'll assume that's the goal.\n\nWell, it sounds like you can write your loop, so shouldn't you just print out the sine, cosine, and tangent of the angle in the body of the loop?  It's not clear what the stumbling block is.  You realize that there are built in functions that compute sine, cosine, etc.?
349408
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...
619642
Think about it. tangent is an odd function such that\nf(-x) = -f(x)\ntan(-3pi/4) = -tan(3pi/4)\n\nObviously we have a measure in radians, we want degrees.\nYou should know that:\n1 radian = 180/pi degrees\n1 degree = pi/180 radians\n\nConvert to radians\n3pi/4 radians\n3pi/4 (180/pi degrees)\n45 x 3\n135\n\nThus 3pi/4 radians = 135 degrees\n\nPlug back in \n-tan (135)\n\nWe have a reference angle of 45 degrees with the x axis. Tangent in the second quadrant is negative. What is the tangent of 45? It's 1, but because we're in the second quadrant, tan (135) = -1.\n\nHowever, don't forget we have that -tan. So, -1 x -1 = 1.\n\nAs a result, tan(-3pi/4) = 1.
109287
The trig functions are not one-to-one hence they don't have an inverse.  The only way for them to have an inverse is to make them 1-1.  You can only do this by restricting the domain.  In the case of the tangent, this is from -pi/2 <x<pi/2.  The inverse trig functions are all positive in the first quadrant only.  Look at the graphs of the trig functions to find the restrictions on the domain.  As for your problem, draw a triangle in QI.  arctan1/2 means "the angle whose tangent is 1/2."  Draw a triangle with tan = 1/2.  Since tan = y/x, y = 1 and x = 2.  Solve for r and you get r = sqrt5.  Since sin = y/r, you get 1/sqrt5 or sqrt5/5.  Since you have all three sides, you can define the other 5 functions of arctan1/2.
545074
The word "Trigonometry" is derived from two Greek words meaning measurement or solution of triangles. Trigonometry is a branch of mathematics that deals with the ratio between the sides of a right triangle and its angles. Trigonometry is used in surveying to determine heights and distances, in navigation to determine location and distances, and in fields like nondestructive testing for determining things such as the angle for reflection or refraction of an ultrasound wave.\n\nThere are three principle functions in trigonometry: sine A, cosine A, and tangent A where A is an angle. These are typically abbreviated for use in algebra to: sinA, cosA, and tanA. These terms are defined in terms of a right triangle. SinA is equal to the side opposite of the angle A (side a) divided by the hypotenuse of the triangle (side c). From the right triangle below, it can be seen that the value of angle A is directly linked to the ratio of side a and side c. In other words, if the length of side a is changed (and side c is not changed by the same amount), then angle A must change. When the division of the length of side a by side c is performed, the resulting value is directly related to angle A. The relationship between the decimal value of the ratio of the side and the angular value of angle A can be looked up in trigonometry tables or, as is more common these days, programmed into a scientific calculator. The relationship between angle A and the ratio of any two of the three sides can be determined by using either the Sin, Cos or Tan functions. CosA is equal to the side adjacent to the angle A divided by the hypotenuse and tanA is the sine divided by the cosine and is therefore the side opposite the angle divided by the side adjacent.
388496
There are six ratios of the three sides of a right triangle. These ratios have been identified by name throughout the centuries. In the USA they are known as sine, cosine, tangent, cotangent, secant, and cosecant. Your calculator has buttons for three of these: [SIN], [COS], and [TAN]. Remember them with the non-word SohCahToa. \n\nThe tangent ratio of an angle is simply the ratio of the opposite side to the adjacent side of a right triangle. \n\nIf you know the angle, type \n\n[TAN] > angle > [ENTER].\n \nThe answer will be the tangent (ratio of the opposite side to the adjacent side).\n\nIf you know the opposite and adjacent sides and need to find the angle, type \n\n[2nd] > [TAN] > opposite > [/] > adjacent > [ENTER]. \n\nThe answer given will be the desired angle.\n\nCaution: \n(1) Different calculators require these buttons to be pushed in a different order.\n(2) If your angle is in degrees, then make sure your calculator's mode is in degrees.\n(3) Don't get the numbers backwards or you'll find the cotangent of the angle and not the tangent.