Trigonometric Ratios of Any Angle?
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SOH-CAH-TOA.\n\nRight triangle:\nOpposite side = 15\nAdjacent side = 8\nHypotenuse = 17 (sqrt(8² + 15²) = 17)\n\nSo\nsin(angle) = O/H = 15/17\ncos(angle) = A/H = 8/17\ntan(angle) = O/A = 15/8\n\nI agree with you and think the book has it wrong...\n\nAnother confirmation:\ntan(angle) should equal sin(angle) / cos(angle).\ntan(angle) = 15/17 divided by 8/17\ntan(angle) = 15/8\n\nSo cos(angle) should definitely be 8/17... must be a typo.\n\nThe only explanation is that perhaps it was the cotangent?  cot(angle) = 1/tan(angle) = 8/15...
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sin=opposite/hypotenuse\ncos=adjacent/hypotenuse\ntan=opposite/adjacent
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With these data a lot of angles fulfill that requirement; If  you have alpha first look at cos alpha, after divide by1.33  and look for theta.\n\nDid you not make a mistake. I suppose you mean sin and not cos\n\nSnell law uses sinus
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The first quadrant angle x whose sine is 1/2 is 60 degrees.\n\nThis angle has cosine of sqrt(3)/2.\n\nAnother way that we can do this is to remember that:\n[sin(x)]^2 + [cos(x)]^2 = 1 for all x.\n\n(1/2)^2 + [cos(x)]^2 = 1\n[cos(x)]^2 = 3/4\n[cos(x)] = sqrt(3/4) = sqrt(3)/2
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You can use sin and cosine.
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You're looking for the angle that has a sine of 1/2.\nDrawing a triangle representing this is very helpful.  You will see that it is a 30-60-90 triangle.  \nIf you can find it in degrees, then you can convert it to radians by multiplying it by pi/180.\nsine(30) = 1/2\n30 degrees = 30(pi/180) = pi/6
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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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The half-angle formula for tangent is:\n\ntan(x/2) = sin(x)/(1+cos(x))\n\npi/8 is not a very easy angle to work with, but pi/4 is. So let x = pi/4 in that half-angle formula it'll give you tan(pi/8).
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Here are some real tough ones.(Looked tough to me)\n\nProve that cot(90 - theta)sin(90 - theta) = sin theta\n\nFind Cos(3x)\n\nFind Sin(3x)
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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.
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Job one, convert everything to trig functions of a single angle, namely x.\n\nsin2x=2sinxcosx\ncos2x=cos^2(x)-sin^2(x)=2cos^2(x)-1=1-2sin^2(x)\n\nThen substitute in and simplify:\n2sinxcosx+2(2cos^2(x)-1)=1+sinx-4cosx\n2sinxcosx+4cos^2(x)-2-1-sinx+4cosx=0\n4cos^2(x)+2sinxcosx-sinx+4cosx-3=0\n\nFactor this into:\n(2cosx-1)(2cosx+sinx+3)=0\n\nset each parantheses equal to zero\n2cosx-1=0   \ncosx=1/2\nx=arccos(1/2)\nx={pi/3, 5pi/3} + 2pi*n  where n=0,1,2....\n\nAnd\n\n2cosx+sinx+3=0\n\nSolved numerically gives imaginary solutions of\n{-i*ln((-2/5)-(i/5),-i*ln(-2-i)}\n\nSolved....
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Go to http://mathworld.wolfram.com/Cosine.html
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cot = o/a\ncsc = r/o\n\no = opposite side\na = adjacent side\nr = radius\n\ncalculate csc(2x) in terms of o and r.\nSubstitute o/a for cot(x) and substitue the value you determine for csc(2x) \nmultiply these, then convert to a trig function in terms of x.\n\nIf you know trig identities better than I do, you can reduce this expression using identities, without resorting to o, a, and r.  However, either route will work.
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Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
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The ratios of the indices of refraction will be the same as the ratio of the sin of the angles.\n\nsin(32)/sin(27) = x/1.61
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Here are a couple i learned for double angle formulas.\n\nFor sin(a+b)= sin(a)cos(b)+cos(a)sin(b)\nSay out loud: "sine cosine cosine sine"\n\nFor cos(a+b)= cos(a)cos(b)-sin(a)sin(b)\nSay out loud: "cosine cosine SIGN sine sine"\nNotice the word "SIGN". The word SIGN is there to remind you that the SIGN of the rightmost term has the opposite SIGN of the input to the cosine on the left hand side of the equation (if its a+b, then subtract. if its a-b, then add).\n\nJust say it out loud like 3 times. It has a really nice rythm:\n\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n\nSee? Easy.
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Since there are four quadrant that make up the 360degrees. Draw the two axis x and y. Assume the top right part as quadrant 1, top left part as quadrant 2, bottom left as 3 and bottom right as 4. Now the calculation of angle starts from quadrant 1 in anticlockwise direction to make an angle of 360degrees. Now use "ASTC" rule i.e. assign 'A' to quadrant 1, 'S' to quadrant 2, 'T' to 3 and 'C to 4. \n\n'A' states that in this quadrant all i.e. sin, cos and tan have positive values.\n\n'S' states that in this quadrant sin have positive value and cos and tan have negative values.\n\n'T' states that in this quadrant tan have positive value and cos and sin have negative values.\n\n'C' states that in this quadrant cos have positive value and sin and tan have negative values.\n\nthat's it simple enough.