How can the calculator calculate the sinus and the cosine of a number, without considering any triangle?
369033
Most handheld calculators just use Taylor series expansions to approximate the value.  For example, if the angle x is in radians, then\n\ncos(x) = 1 - (1/2) x^2 + (1/24) x^4 - (1/720) x^6 + ...\nsin(x) = x - (1/6) x^3 + (1/120) x^5 - (1/5040) x^7 + ...\ntan(x) = x + (1/3) x^3 + (2/15) x^5 + (17/315) x^7 + ...\n\nYour calculator has a polynomial (maybe of degree 50?) written into its memory which can calculate sine and cosine to a remarkable accuracy -- but it's only an approximation in the end.\n\nOther more powerful programs, such as Mathematica, MATLAB, or even Maple use slightly more sophisticated ideas.  They have a built in database which contain the *exact* value of the trigonometric functions for certain angles x.  For instance, they would know that cos(pi/3) = 1/2 -- something which is not obvious from the Taylor series expansion above!
459338
Go to http://mathworld.wolfram.com/Cosine.html
854998
No sinx/cosx = tanx not tan2x
741746
They don't have values, they are operations, which means each one tells you to do something different (and shame on your teacher for not explaining this to the class BEFORE having you do them on your calculator). \n\nYou always have to have right triangles when dealing with these three trig functions (that's what sine, cosine and tangent are, trig functions). They are each ratios, which means a comparison of two numbers. Sine means to take the length of the opposite side (the side opposite the angle you're measuring) and put that number in a fraction as the numerator, then in the denominator put the length of the hypotenuse of the triangle. So for short, sine means "opposite over hypotenuse." Similarly, cosine means "adjacent over hypotenuse" and tangent means "opposite over adjacent." The adjacent side is the side forming the angle that is NOT the hypotenuse (this is really difficult to explain without being able to draw you a picture). My students called it SOHCAHTOA for short to remember them all.\n\nI'll try and walk you through an example as best I can...Draw an isoceles right triangle (which means the triangle has one right angle and two 45 degree angles). Since it's isoceles, the two sides that are NOT the hypotenuse must be the same length. So let's say those two sides each have a length of 1. If you do the pythagorean theorem (a squared plus b squared equals c squared) you'll get that the length of the hypotenuse is equal to the square root of 2, or approximately 1.414. Did you draw the pictures and label all the parts? \n\nNow pick one of those 45 degree angles and circle it. Let's calculate the sine of 45 degrees, WITHOUT using a calculator. Using that angle you chose, find the length of the opposite side (the side of the triangle that isn't touching the angle at all). It's 1 right? So put that in a fraction as the numerator, the number on top. Sine means opposite over hypotenuse, so in the denominator we need the length of the hypotenuse, which is the square root of 2 or 1.414. So your equation now says....\n\nsin 45=1/1.414\n\nOn your calculator, punch in 1 divided by 1.414 and you'll get the number 0.70721....\n\nMake sure your calculator is in degree mode, if it's not you'll get the wrong answer, it's a very common mistake!\n\nNow punch in sin 45 and hit enter. What number do you get? 0.70710....\n\nThe reason the two numbers are different is because we rounded the length of the hypotenuse. If we'd actually used the square root function on our calculators and typed in the square root of 2 instead of the decimal, they would have been exactly the same number. \n\nAnd it doesn't matter what the lengths of the sides of the triangles are, as long as they're right triangles. IF YOU DON'T HAVE RIGHT TRIANGLES THOUGH, YOU CAN'T USE THESE FUNCTIONS. So be careful. But that's all the sine, cosine and tangent represent. They are the ratios of the side lengths of right triangles. \n\nHere's a final example. The tangent of 51 degrees (tan 51) equals 1.23489..... That means that in a right triangle with a 51 degree angle, if you take the length of the side opposite the 51 degree angle and divide it by the length of the side adjacent to that angle (remember, tangent means "opposite over adjacent"), then you'd get 1.23489....\n\nI hope that helped. Good luck! I'm sorry I couldn't draw you a picture.
1015891
Isn't it sinx divided by 6?\nSimple just cancel 'n' from the numerator and the denominator.\nU will be left with six=6
879956
2 - csc²z =\n2 - (1/sin²z) = \n(2sin²z - 1) / sin²z = \n(sin²z + sin²z - 1 ) / sin²z =\n(sin²z - cos²z) / sin²z =  {using trig id sin²+cos²=1}\n1 - (cos²z / sin²z) =\n1 - cot²z\n\n~ ~ ~ ~\n\ncot³t / csct =\n(cos³t/sin³t) / (1/sint) =\n(cos³t/sin³t) * sint =\ncos³t/sin²t =\ncost (cos²t/sin²t) =\ncost (cot²t) = \ncost (csc² -1)  {using trig id 1+cot²=csc²}\n\nHope that helped!\n\n~ ♥ ~
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.
730722
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck
1023518
Use the Law of Sines.\n\nsin A/a = sin B/b\n(sin 36)/24 = sin B/47\nsin B = (47 sin 36)/24\nsin B = 1.151079452\nAt this point you have a problem--it is impossible for the sine of an angle to have a value greater than 1.\n\nSo either you copied the problem wrong, or you are being asked to recognize that a triangle cannot be formed with these dimensions.
985667
They taught us SOHCAHTOA back in OL trig, but I never really felt the need for it, I mean just picture a right angled triangle and the longest side is the hypotenuse,h. Take an angle and the side opposite it is h.sin x , where x is the angle. The other side is h.cos x
880627
Do you mean verifying trig identities?\n\nTry: ask dr math\n\nIf you have a specific trig question, write back.\n\nGuido
835259
I'll give you a hint and let you try to figure out the rest yourself (unless someone else tells you the exact answer). =)\n\nUse your trig identities:\n\ntan A = sinA/cosA\n\ncot A = 1/tanA or cot A = cosA/sinA
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Take a unit circle and a point p on the circle. Place the circle so that it is centered at the origin in a Cartesian xy-plane. The cosine of p, cos(p), is defined to be the x-coordinate, or abscissa, of p. The sine of p is the y-coordinate. These functions are the just the projection functions for the unit circle by the standard embedding in the Cartesian plane as described above.\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 corresponds to a unique point on the unit circle.\nThis relationship can be seen by drawing a radius from the origin to mark the angle from the positive x-axis and assigning that angle to 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.\nIt is now however easy to see why all the basic trigonometric functions have a neighbor with prefix co-. The cosine is short for complementary sine. This is because the cosine of a point is obviously the same as the sine of the complementary angle.\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.\nThe arc trigonometric functions are so named because they measure arc length; they give you back the radian measure of the sine or cosine when you restrict these functions to one hemicircle (otherwise you get two possible answers).\nThe secant function does not give you an arclength; it is another function of points on the unit circle (and thus by our unit circle coordinate system, it is a function of arc length). In the right triangle formed by the tangent to the circle, the secant is the length of the hypotenuse. The arcsecant function returns the arc length on the unit circle that the secant intersects. See the sources below for illustrations.
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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
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
824139
All you need to do is enter these on a scientific or graphing calculator.  Make sure it is in degree mode.  For the second part use the 2nd or shift button to get sin^-1 etc
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Switch your calculator from degrees (DEG) to radians (RAD). That should do the trick. \nHere are some quick trig rules you need to know:\nIn radians just remember that pi is 180 degrees. Pi/3 is 60 degrees, pi/2 is 90, pi/4 is 45 degrees and so on. \nIn a unit circle, the hypotenuse is always 1.\nSin is always the far side away from the angle (the vertical side). We call that the opposite side.\nCos is always the adjacent side (or horizontal side).\nIn a right triangle with angles 60 and 30, the shortest side is always 1/2 the hypotenuse, so in a unit triangle, it is just 1/2. You can use the pathagorean theorem to find that the longer side is the square root of (3/4), or (sqrt(3))/2.\nIn a right triangle with angles 45 and 45, the opposite and adjacent sides are equal, and we can use the pathagorean theorem to find that the sides are the square root of (1/2), or 1/(sqrt(2)).\n\n1. Sin 1/3 pi. This is the same as sin pi/3 which is a 60 degree angle. If you draw the triangle on a unit circle you would see that the opposite side is the longer side of a 90/60/30 triangle. We have seen above that the length is (sqrt(3))/2.\n\n2. Cos pi. Easy. pi is 180 degrees, so draw the unit circle. You see that it is just a striaght horizontal line going the other way so the answer is -1.\n\n3. cos pi/4 is a 45 degree angle. We know from above that both sides have to be 1/(sqrt(2)).\n\n4. sin pi/4. Same as #3.\n\n27. What is 315 degrees? Well draw your unit circle. Go, 90, then to 180, then to 270, then go 45 more degrees and you will end up at 315 degrees. Draw your triangle. We know that the hypotenuse is 1, so the coordinates will simply be, (cos 315, sin 315). Cos 315 is just a 90/45/45 triangle in the bottom right quadrant so the adjacent side is positive. The opposite is negative, so you end up with coordinates (1/(sqrt(2)), -1/(sqrt(2))). Do the same to solve the last question youeself.