Sine and Cosine Ratios (can you help with this?)?
488283
I think that what you really want here is the tangent ratio. You know the slope of the line, right (hint, it is right there in the equation)? You know that the tangent ratio is a over b (where a and b are the non-hypotenuse sides of the triangle), right? You should be able to use that to figure out the tangent of the angle made by the line. Now all you have to do is figure out the angle that has that number as a tangent. That would be the "tan -1" (inverse tangent) function on your calculator.\n\nIf you are working from trig tables rather than a calculator, find the number in the second column of the tangent table that is closest to your tangent number then look in the first column for the angle.\n\nIf you are required to use sin and cosine remember your Pythagorean theorem and use it to find the hypotenuse.\n\nASIDE TO Drakestone: Your solution only finds the sine of the angle, not the angle itself. You need to apply "sin -1" to your answer to get the final answer.\n\n\nASIDE TO eternal...: You are wrong too. graph y=5x+3 and you will see that it is way too steep to be an 11 degree angle. You mixed up your opposite and adjacent values in calculating your tangent.
438454
Change cscx to 1/sinx and sin(-x) = -sinx\ncscx(cscx + sin-x)\n= (1/sinx)(1/sinx - sinx)\nGet a common denominator\n= (1/sinx)([1-(sinx)^2]/sinx)\n1-(sinx)^2 = (cosx)^2\nSo\n(1/sinx)([1-(sinx)^2]/sinx)\n= (1/sinx)((cosx)^2/sinx)\n= (cosx)^2/(sinx)^2\n= (cosx/sinx)^2\n= (cotx)^2
994111
First, factor the left hand side of the equation:\n\n2sin(x)*cos(x) + cos(x) = 0\n\ncos(x)*[2sin(x) + 1] = 0\n\nNow, this equation is satisfied whenever either cos(x) = 0 or (2sin(x)+1) = 0\n\ncos(x) = 0 at x = pi/2 and 3*pi/2  (You should memorize when the cosine and sine functions are equal to zero!)\n\nThese are the first two solutions to the original problem.\nNow deal with the factor containing the sine term.  We need to find the values of x when:\n\n2sin(x) + 1 = 0\n\nRearrange this equation to read:\n\n2sin(x) = -1\n\nthen\n\nsin(x) = -1/2\n\nx = 11pi/6 and 7pi/6 .  These are the other two solutions to the original problem.\n\nYou should also memorize when the sine and cosine functions are equal to 1/2.  For the sine function, this happens at x = pi/6 = 60 degrees and 5pi/6 = 150 degrees.  If the sine or cosine function has a certain value at x, then it will have -1 times that value at x + 180 degrees (x + pi radians), so in this case if sin(pi/6) = 1/2, then sin(pi/6 + pi) = sin(7pi/6) = -1/2.  Similarly, if sin(5pi/6) = 1/2, then sin(5pi/6 + pi) = sin(pi/6+pi) = sin(7pi/6) = -1/2.\n\nNote that despite what the previous answerer said, the sin(2pi/3) and the sin(4pi/3) do not equal -1/2.  [It's the cosine that's equal to -1/2 at these points.)
1186234
Actually, when the answers come out as negative, it's calculating in radians. So the sine of 6 is negative, as it is actually calculating about 324 degrees.\n\n2nd+0 resets your calculator, but i don't recommend it.
921194
sin(90)=1\ncos(90)=0\ncsc(90)=1\nCot(90)=0
969737
sine 30° 15'\n\nCos ( 90 - A)\n\n89 60\n30 15\n- - - - -\n\n59 45\n\nThe answer is cos 59° 45'\n\n- - - - - - - -s-
329959
1-CosA = 2 sin^2 A/2     -----1\n\nFirst whe have to find the value of the cos \n By Cos A = Root(1-Sin^2A)\nCos A = 5/13\nAnd \nThe apply the formaula of the 1\n1- 5/13 = 2Sin^2 A/2\nSinA/2 = Root(8/26) it gives the answer\n And the value of the cos A/2\nIs get by equation 1+ cos A = 2 Cos ^2 A/2
676590
Sorry, I can't see the photo...but:\n\nSOH - sin = opp/hyp\nCAH - cos = adj/hyp\nTOA - tan = opp/adj
732786
Using Trig identity 2sin(x)cos(x) = sin2x\n\n(2sin3x)(cos3x) = sin6x
1021739
Since tanx = sinx/cosx,\n\ntan(A-B) = sin(A-B)/cos(A-B)\n\nNow, write out the expression for each term on the right side of this equation.\n\nWhen you finish that, divide each term by cosAcosB and you will get the result you are seeking.
289664
This ratio is called the cosine.
158498
sin(5)=0.08715574.....
1176396
sin=opposite/hypotenuse\ncos=adjacent/hypotenuse\ntan=opposite/adjacent
1150095
its sooo easy  \nsinA-sinB = 2 sin[(A-B)/2] * cos[(A-B)/2]\ncosA - cosB = 2 cos[(A-B)/2] * cos[(A-B)/2]\nthus LHS = \n(2 sin[(A-B)/2] * cos[(A-B)/2]) / (2 cos[(A-B)/2] * cos[(A-B)/2])\n= sin[(A-B)/2] / cos[(A-B)/2]\n= tan[(A-B)/2 \n= tan (A/2-B/2)\n=RHS \n\nhence proved
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.
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
656507
Use the law of sines\n\nn/sin(N)=m/sin(M)\n\nso \n\nm = n*sin(M)/sin(N)\n\nPlug in the values and use your calculator.