question on trigonometry?
967960
What is it?  Send an im
333253
You can solve this using x, y, and r (the lengths of the 3 sides of the triangle made by the angle).  The trigonometric functions are originally defined as:\n\nsin = y/r\ncos = x/r\ntan = y/x\nsec = r/x\ncsc = r/y\ncot = x/y\n\nYou need to find x, y, and r for angles a and b.  Do this using the given sin and cot values and the equation\n\nx^2 + y^2 = r^2\n\nSo for a:\n\nsin a = 3/5\ny=3, r=5\n\nx^2 + 9 = 25\nx^2 = 16\nx=-4\n(this is negative because the angle is in the 2nd quadrant)\n\nFor b:\ncot b = -12/5\nx=-12, y=5\n(x is the negative one because this is in the 2nd quadrant)\n\n144 + 25 = r^2\n169 = r^2\nr=13\n\nNow you need to turn tan (a-b) into a form where you can plug in your x, y, and r values, and then plug and solve.  Remember that tan = y/x.\n\ntan (a-b) = (tan a + tan b) / (1 + tan a * tan b)\n=(-3/4 + -5/12) / (1 + (-3/4)(-5/12)\n=(-14/12)/(1 + 15/48)\n=(-7/6)/(63/48)\n=-8/9\n\nTherefore, your final answer is -8/9.\n\nHope this helps!
1096282
cos^2(A) / ( 1 - sinA )\n=( 1 - sin^2(A) ) / ( 1 - sinA )\n=( 1 + sinA )( 1 - sinA ) / ( 1 - sinA )\n=1 + sinA
842369
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
858889
QUESTION 1:\n\nAsk yourself: WHICH TRIG FUNCTION IS OPPOSITE OVER ADJACENT?  IT IS THE TANGENT FUNCTION, RIGHT?\n\nThe bottom of the tree is 25 feet away from a point that measures 30 degrees from ground level to the top of the tree.\n\nAnswer: tan30degrees = x/25feet...choice 1.\n\nNotice that they are NOT asking you to solve for x.  Get it?  \n\nQUESTION 2:\n\nYou need to write the given equation 3x - 2y = 12 in the y =mx + b\nform.\n\nLet's do that right now.\n\nWe have:\n\n3x - 2y = 12 \n\nI will subtract -3x from both sides of the equation.\n\nWe get:\n\n-2y = -3x + 12...See what I did?  \n\nNext:\n\nDivide BOTH sides of the equation by -2 to find the value of y.\n\nWe get this:\n\ny = (-3x + 12)/-2\n\nWe can reduce the equation to this:\n\ny = 3/2 (x) - 6\n\nSlope is 3/2 and the y-intercept is -6.\n\nFinal answer: choice 1\n\nGuido
866713
The initial poster didn't read the entire question.  This is not as simple as the pythagorean theorem idea he/she had.  You need to use vectors to solve this problem.  Assume X is east-west and Y is north-south, with positive being east and north.\n\nThe first vector is 5.0 km at 11.5 degrees north of due east.  You need to find the X and Y components of this vector.  Use SOH CAH TOA, and you have the H, and you have the angle.  So to get the value of the X component, you need to use sin:\n\ncos (11.5) = X/5.0, so the X component is 5 cos (11.5)\n\nsimilarly for the Y component, use sin:\n\nsin (11.5) = Y/5.0, the Y component is 5 sin (11.5)\n\nGreat, so the vector for the first distance is \n<5 cos(11.5), 5 sin(11.5)>\n\nThe second distance is 1.0 km due south.  Well, this is only in the Y direction, so the vector would be <0, -1>  Use the negative sign because we assigned north as positive, and south as negative.  Now we can add the components of the vectors to get a resultant vector.\n\n<5 cos(11.5), 5 sin(11.5) -1>\n\nThis says that the resultant vector has an X component of \n5 cos(11.5) and a Y component of 5 sin(11.5) -1.\nUsing the Pythagorean theorem, the magnitude of this resultant vector would be\nsqrt [ (5 cos (11.5))^2 + (5 sin(11.5)-1)^2 ]\n\nNow to get the direction, we need to use trigonometry again.  We have the opposite, and the adjacent, which is TOA.\n\ntan(theta) = Y/X = (5 sin(11.5)-1)/(5 cos(11.5))\n\nTake arctan of both sides:\ntheta = arctan[(5 sin(11.5)-1)/(5 cos(11.5))]\n\nSo the answer?\nThe resultant displacement vector is:\n\nsqrt [ (5 cos (11.5))^2 + (5 sin(11.5)-1)^2 ] at an angle of \narctan[(5 sin(11.5)-1)/(5 cos(11.5))]
873572
Hello...\n\nWell the first part is a bit tricky to explain and I'm not sure what range of x you want your answer in, because with inequalities and trig functions the solution lies in many intervals. So I'll explain the second one first\n\nIt's actually not that bad, you just need to be familiar with some trig identities\n\nIf you recall, I'm sure you learned that 2sinxcosx = sin(2x)\n\ntherefore, you can say that sin(2x)=sqrt(5)/3, and solve for x by taking the inverse sin of both sides.\n\nNow for the first problem, you can work the inequality a bit further and get cosx greater than or equal to -1/2\n\nIf you're looking for a simple interval, then x is between -2pi/3 and 2pi/3. But the actual answer also includes (4pi/3,8pi/3), (10pi/3,14pi/3)....\n\nThe best way to do the inequality is to draw the cosine function on your paper, draw y=-1/2 (horizontal line), and you can find the first intersection by taking the inverse cosine of (-1/2), which is 2pi/3. You'd have to study the symmetry of the cosine function, so since you know that it's symmetric about x=pi, you'll figure that the next point is 4pi/3 and so on and so forth and you'll see a pattern really quickly. \n\nThis may be more complicated than what you're looking for but I hope it helps you out. Good luck on your test
182808
The Sec of X is -2sqrt3/3.\n\nThe reference angle of X is 30 degrees. My trig teacher requires that we memorize the values of all 30, 45, and 60 degree angles so I knew that off the top of my head. If you don't understand why, make a simple 30, 60, 90 degree triangle with side values x, xsqrt3 and 2x. You can see the sin (opposite/hypotenuse) of the 30 degree angle is x/2x, or 1/2. The tan (opposite/adjacent) is x/xsqrt3, or 1/sqrt3, which is sqrt3/3 when rationalized.\n\nThe 30 degree reference angle is in the 2nd quadrant. \n\nIn the 1st quadrant, all sin, cos, tan, etc. values are positive. In the 2nd, only sin and it's reciprocal (csc) are positive. In the 3rd tan and cot are positive. In the 4th, cos and sec and positive. You can remember this by the sentence "All Student Take Calculus". Starting in the first quadrant and moving counterclockwise (to 2nd,3rd and 4th) you know the positive values... A,S,T,C.. (All,Sin,Tan,Cos).\n\nKnowing that, we need to use process of elimination to pick a quadrant. Sin the tangent is negative, quadrants 1 and 3 are ruled out because tan is always positive in those quadrants. This leaves 2 or 4. We know the Sin is positive, so quadrant 2 is the right answer. In quadrant 4 only cos and sec and positive.\n\nTo figure out the sec, refer back to your 30, 60, 90 triangle picture. Sec is the reciprocal of cos. Cos is adjacent/hypotenuse, so Sec must be hypotenuse/adjacent. This is 2x/xsqrt3, or 2/sqrt3. When you rationalize the denominator you get 2sqrt3/3.\n\nSince we know the angle is in the 2nd quadrant with a 30 degree reference angle, we know X is 150 degrees. However, we don't need to know the angle measure to answer the problem. All we need to remember is only sin and csc are positive in the second quadrant, so our answer of 2sqrt3/3 must be negative.\n\nTherefore, the sec x is -2sqrt3/3.\n\nSorry this was very wordy. It's difficult to explain math over the internet! I hope it kinda cleared some of your confusion.
732786
Using Trig identity 2sin(x)cos(x) = sin2x\n\n(2sin3x)(cos3x) = sin6x
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Trigonometry is hard to do in you head, so start out by making a drawing. You have two buildings A (tall) and B (short) - Draw them far apart, not like this:\n\nA         B\n|         \n|         \n|    \n|         |\n|         |\n|         |\n|         |\n\nNow draw one line from the top of A to the top of B - this is the angle of depression of the top and it is 46 degrees. That is, if you drew a straight, horizontal line across the top of building A, the angle between that line and the line joining the top of A to the top of B is 46 degress.\n\nSimilarly, draw a line from the top of A to the *bottom* of B - this is the angle of depression of the bottom and it is 77 degrees. That is, the angle between the horizontal line across the top of A, and the line between top of A and bottom of B is 77 degrees.\n\nNow can you find out the angle between the line from A-top to B-top, and the line from A-top to B-bottom? (It's very important for you to draw this!) Let's put this value aside for a minute. \n\nLooking at your drawing, can you see that you know the value for *part* of the angle between the horizontal line across A's top and length of building A? What is that part you know? Now, see how the line across A's top and length of A make a right angle? So how many degrees in a right angle? So if you know the total of that angle, and part of that angle, how do you get the other part? **Notice that this part we're trying to get is actually the angle between the line A-top to *B*-bottom, and the line A-top to *A*-bottom. \n\nYou should also notice that now we have two triangles:\nTriangle 1: A-top to B-bottom to A-bottom. Solve this triangle for line A-bottom to B-bottom for distance between the two buildings.\nTriangle 2: A-top to B-top to B-bottom. Solve this triangle for line B-top to B-bottom for height of building. \n\nDon't forget, we have two important angles that will help solve these two triangles!
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Trying for wow #2---\n\nR = .15 m\nω = 2π/T = 13.963 rad/sec\nA little vector diagram shows that tanθ = rω²m/mg = rω²/g\nbut r = Rsinθ, so tanθ = sinθ/cosθ = Rsinθω²/g  from which\n\n1/cosθ = secθ = Rω²/g = .15*13.963²/9.8 = 2.984 → θ = 70.42°\n\nYou get to do b)\nGood luck..
844759
Look at the various values of cosx when x = pi/2, pi, etc. (recall that pi/2 = 90 degrees).\n\nFrom looking at these values in my book, I see two values for x that will satisfy this equation:  x=pi/3, and x=pi\n\ncos pi/3 = 1/2\ncos 2pi/3 = -1/2\n\ncos pi = -1\ncos 2pi = 1\n\n~ ♥ ~
966822
Draw a right angled triangle.  the vertical is 35m and the sloped line, the hypotenuse, is 50m.  We have to find the angle between the string and the ground.\n\n35 is the opposite, 50 is the hyp, so it is a sine problem.  Sin x = opp/hyp = 35/50.\n\nusing the inverse sine function on the calculator, we find that the angle is 44.4 degrees\n\nhope this helps
1015200
Let a, b, and c be the lengths of the legs of a triangle opposite angles A, B, and C. Then the law of sines states that\na/(sinA)=b/(sinB)=c/(sinC)=2R,\n\nwhere R is the radius of the circumcircle.\n\nIn other words,\n\nIn any triangle, the ratios of the sides to the sines of the opposite angles are equal.
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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
510334
x = a*sin(ang)\ny = b*cos(ang)\n\nang goes from 0 to 360 degress, a and b determine how oval it is (if a=b it is a circle)
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First off......do you know how to do basic trigonometry and use Snell's Law?  If you do, then I'll walk you through it.  Write me at fortitudinousskeptic@yahoo.com\nIf you don't know trig and Snell's law I'm not going to try to explain it.  I'm not that motivated.