It's been years!  Do I use sin or cosin or something???
367069
Given: \n1) a right triangle\n2) hypotenuse = 2\n3) the other two angles are 70 and 20 degrees\n\nFind:\nLengths of the remaining sides\n\nNote: \nWe have to be more specific here.\n\nLet:\na = side opposite the 70 degree angle (side adj. 20 degrees)\nb = side opposite the 20 degree angle (side adj. 70 degrees)\nc = hypotenuse = 2\n\nSolution:\n\n1) For side a:\n\nsin 70 = opp./hyp. = a/c\nTherefore, a = csin70 = 2sin70 = 1.88 ANSWER\nOR\ncos 20 = adj./hyp. = a/c\nTherefore, a = ccos20 = 2cos20 = 1.88 ANSWER\n\n2) For side b:\n\ncos 70 = adj./hyp. = b/c\nTherefore, b = ccos70 = 2cos70 = 0.68 ANSWER\nOR\nsin 20 = opp./hyp. = b/c\nTherefore, b = csin20 = 2sin20 = 0.68 ANSWER\nOR\nPythagorean Theorem: a^2 + b^2 = c^2\nb^2 = c^2 - a^2\nb^2 = 2^2 - 1.88^2\nb^2 = 0.4656\nTherefore:\nb = 0.68 ANSWER
105824
Trigonometry.  The good old days when I still loved math... ahhh!
791300
Use the FOIL method; First, Outer, Inner, Last.\n\nFirst, n * n\n\nOuter, n * 5\n\nInner, n * 7\n\nLast 7 * 5.\n\nThen add them all together.\n\n22 NOV 06, 0420 hrs, GMT.
1147102
pi*(r^2)\n\nIt doesn't matter which unit you use, just make sure you're consistent.
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.?
215627
well from what i learn we have a formulae for that right??!! then the formulae to find theta is.... to find cosine the formulae is adjecent over hypothenus in shortform its CAH or C=A/H. and to find sine is opposite over hypothenus or SOH or S=O/H.. and to fine tangent is opposite over adjecent or TOA or T=O/A.. so to remember it you just remmber it in the short form.. as if you have a chinesse friend name SOH CAH TOA.. soh for sine n cah for cosine n toa for tangent.. i hope thats what you want for the answer.. if not.. im sorry.. sooo sorryy..
575459
Sorry but I don't quite understand your qn.\nCan you make it clearer?\n\n(to ozert:\ncos A and sin A could be negative values if tan A is in Quadrant 3)
932480
sin(A+B)sin(A-B)/(cos²Acos²B)\n= (sinAcosB + cosAsinB)(sinAcosB - cosAsinB)/(cos²Acos²B)\n= (sin²Acos²B - cos²Asin²B)/(cos²Acos²B) (since (x + y)(x - y) = x² - y²)\n= sin²A/cos²A - sin²B/cos²B\n= tan²A - tan²B ← Option c)\n\nPlease note: sin(A + B) = sinAcosB + cosAsinB\nand sin(A - B) = sinAcosB - cosAsinB
953609
2cos(x) - sin(2x) = 0 \nremembering that sin(2x) = 2sin(x)cos(x), \n2cos(x) -2sin(x)cos(x) = 0\n2cos(x) * [1 - sin(x)] = 0\nor cos(x) = 0 or 1-sin(x) = 0\ncos(x) = 0 or sin(x) = 1\n\nπ/2, 3π/2
924056
sin φ + sin φ tan φ + cos φ cot φ + cos φ = csc φ sec φ\n\nsin φ + sin φ (sin φ / cos φ) + cos φ (cos φ / sin φ) + cos φ = csc φ sec φ\nsin φ + sin²φ / cos φ + cos²φ / sin φ + cos φ = csc φ sec φ\n\ngetting the least common denominator which is sin φ cos φ\n\nthen \n\n(sin²φ cos φ + sin²φ sin φ + cos²φ cos φ + sin φ cos²φ) / sin φ cos φ = csc φ sec φ\n\ncombining similar terms\n\n(sin²φ cos φ + cos²φ cos φ) + (sin²φ sin φ + cos²φ sin φ) / sin φ cos φ = csc φ sec φ\n\nthen\n\n[cos φ(sin²φ + cos²φ) + sin φ(sin²φ + cos²φ)] / sin φ cos φ = csc φ sec φ\n\nsin²φ + cos²φ = 1\n\n(cos φ + sin φ) / sin φ cos φ = csc φ sec φ\n\n1/sin φ + 1/cos φ = csc φ sec φ\n\ncscφ + sec φ = cscφ sec φ\n\ni think you forgot to place a plus sign in between the csc φ and sec φ. i you do, then my solution is right. hope this will help.
1092612
In quadrant one, sine, cosine, and tangent are positive. In quadrant two, sine is positive, and cosine and tangent are negative. In quadrant three, tangent is positive, and sine and cosine are negative. And in quadrant four, cosine is positive, and sine and tangent are negative.\n\nQuadrant 1 – Sin(+) Cos(+) Tan(+)\nQuadrant 2 – Sin(+) Cos(–) Tan(–)\nQuadrant 3 – Sin(–) Cos(–) Tan(+)\nQuadrant 4 – Sin(–) Cos(+) Tan(–)
608164
I hope that this will help: \nsin(70) = 0.7739,sin(110) = -0.0442,sin(47) = 0.1236,sin(133) = 0.8689, sin(69) = -0.1148,\nsin(111) = -0.8645,cos(70) = 0.6333, cos(110) = -0.9990,cos(47) = -0.9923,cos(133) = 0.4948,\ncos(69) = 0.9933, cos(111) = -0.5025
94089
Open a trigonometry book!
895625
This may be unexpected, but a good site with the formulas you need.\n\nhttp://en.wikipedia.org/wiki/Trig
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
1127640
I'm not sure if you have learned the unit circle yet, but the sin and cos functions follow it perfectly...For any position coordinate on the unit circle, sin=y and cos=x... Simply look at your unit circle and draw draw out the sin and cos functions according to the values given for each one by the coordinates found on the unit circle. You should have clear intersections.
248717
Here is what I would suggest. \n\nMultiply the left fraction by sinx/sinx; leave the bottom and do out the top:\n\nsinx + sin^2x + cosxsinx\n--------------------------\n(1+sinx - cosx)(sinx)\n\n\nReplace sin^2x with 1 - cos^2x in numerator\n\nsinx + 1 - cos^2x + cosxsinx\n\nSubtract then add cosx in numerator\n\nsinx + 1 - cosx + cosx - cos^2x + cosxsinx\n\nRearrange (not really necessary but it helped me see the next step)\n\n1 + sinx - cosx + cosx + sinxcosx - cos^2x\n\nFactor cosx from the last 3 terms:\n\n(1 + sinx - cosx) + cosx(1 + sinx - cosx)\n\nFactor this and put it back over the denominator:\n\n(1 + cosx)(1 + sinx - cosx)\n----------------------------\n(1 + sinx - cosx)(sinx)\n\nCancel and get:\n\n(1 + cosx)\n-----------\n  sinx\n\n\nWhew!