Trigonometry and inverses?
955832
The problem is that thw word 'inverse' is used in two different ways here. It is perhaps easier to think of sec x as the *reciprocal* of cos x. Thus, if, say x=pi/3, cos x=1/2 and sec x=2.\n\nOn the other hand, arccos x is the function that *undoes* cos x. Thus, since cos(pi/3)=1/2, arccos(1/2)=pi/3.
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!
885716
cot(-x)cos(-x) + sin(-x) = -csc x\n\nLHS = cot(-x)cos(-x) + sin(-x)\n= cos(-x)/sin(-x) * cos(-x) + sin (-x)\n= cos²(-x)/sin(-x) + sin²(-x)/sin(-x)\n= (cos²(-x) + sin²(-x))/sin(-x)\n= 1/sin(-x)\n= 1/-sin(x)\n= -csc(x)\n= RHS
349408
The answer is B) 1.122\n\nFor the different trigo. functions:\n\nLet:\na = side opposite angle n\nb = side adjacent angle n\nc = hypotenuse of the right triangle\n\nTherefore:\n\ncos n = adj./hyp. = b/c\nsin n = opp./hyp. = a/c\ntan n = opp./adj. = a/b\nsec n = hyp./adj. = c/b\ncsc n = hyp./opp. = c/a\ncot n = adj./opp. = b/a\n\nYou will notice the following:\n\n1) the secant is the reciprocal of cosine.\n2) the cosecant is the reciprocal of sine.\n3) the cotangent is the reciprocal of tangent.\n\nTherefore: \n\ncsc n = 1/sin n = 1/0.891 = 1.122\n\nFor the remaining trigo. functions, solve for n first:\n\nn = arcsin 0.891 = 62.99917663˚ or 62˚59'57.04"\n\nTherefore:\n\n1) cos 62˚59'57.04" = 0.454\n2) tan 62˚59'57.04" = 1.963\n3) sec 62˚59'57.04" = 1/cos 62˚59'57.04" = 2.203\n4) cot 62˚59'57.04" = 1/tan 62˚59'57.04" = 0.510\n\nI hope that the solution was clear enough...
843808
If sin A = x and sinB = -x  then \nA = .... , -4π - B, -3π + B, -2π - B, -π + B,.-B, π + B, 2π - B, 3π + B, 4π - B, .... \n= nπ + (-1)^nB (n is any integer)\n\nSo arcsin(x) - arcsin(-x) = nπ + (-1)^nB - B\n= nπ + ((-1)^(n) -1)B\n\nNo this looks too clumsy ....new strategy!!\n\nsinA + sinB = 0\nSo 2 sin(½(A + B))cos(½(A - B)) = 0\n\nThus sin(½(A + B)) = 0 or cos(½(A - B)) = 0\nie ½(A + B) = 0, ±π, ±2π, ±3π, .... , ±nπ, \nie A + B = ±2nπ\nie arcsinx + arcsin (-x) = ±2nπ\n\nOr ½(A - B) =  ±π/2, ±3π/2, ±5π/2, ... , ±(2n - 1)π/2\nSo A - B = ±(2n - 1)π\nSo arcsinx - arcsin(-x) = ±(2n - 1)π
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
895625
This may be unexpected, but a good site with the formulas you need.\n\nhttp://en.wikipedia.org/wiki/Trig
433978
Simply equate the functions as it asks you to;\nso you get 1-sinx = 2sin2x\nThen use the trig identity sin2x=2sinxcosx\nso you get 1-sinx=4sinxcosx\n\nCan you solve it from there?\nI'd next use cosx = sqrt (1-(sinx)^2)\nThen square the equation and solve it for x as you've probably done a million times before.\n\nYou'll probably get a quadratic equation in sinx. Use the quadratic formula to get some values out, and remember that you have to give answers between -360 and 360 degrees.
1159612
1) You meant to have a minus sign not "over". I've used x in place of "theta"\n    secx = 1/cosx and tanx = sinx/cosx \nSo 1 becomes (1/cosx)*sinx - tanx = 0 \nor tanx - tanx = 0, which is true.\n\n2) tanx = 1/cotx, so tanx*cotx = 1 \nSo 2 becomes 1- tanx*cosx = 1-sinx\nBut tanx = sinx/cosx so we have:\n1- (sinx/cosx)*cosx = 1-sinx, so\n1-sinx =1-sinx, which is true.\n\nI'll do one more:\n\n3) tan^2 x - 3sinx*tanx*secx = - 2tan^2 x\nAgain secx = 1/cosx and tanx = sinx/cosx\nSo 3. becomes:\ntan^2 x - 3(sinx/cosx)*tanx = - 2tan^2 x, so\ntan^2 x - 3tan^2 x = -2tan^2 x, so\ntan^2 x = tan^2 x, which is true.\n\nYou need to learn to do these yourself: Here is a link which gives you all the trig identities to solve the rest. They are not difficult.\n \nhttp://www.sosmath.com/trig/Trig5/trig5/trig5.html
880627
Do you mean verifying trig identities?\n\nTry: ask dr math\n\nIf you have a specific trig question, write back.\n\nGuido
684975
cosx=4/5\nor, x=cos inverse (.8)\nor, x= 36.87\n\nuse scientific calculator.
962689
To solve for #1, you need to use trigonometric substitution.  When doing trig substitution, you have to note the form of it.  They can be in one of these forms:\na) 1 - [junk]^2\nb) [junk]^2 -1\nc) 1 + [junk]^2 or [junk]^2 + 1\n\nThe above cases mirror the following trig identities.\na) 1 - (cos(x))^2 = (sin(x))^2\nb) (sec(x))^2 - 1 = (tan(x))^2\nc) 1 + (tan(x))^2 = (sec(x))^2\n\nWhen doing trig substitution, you have to keep this in mind.\n\nThe appropriate trig substitution in #1 is c).  You'll want to let x=1/(sqrt(5)tany) and work from there.  The answer is much too huge to put on here, and my goal was to get you started anyway.\n\nTo solve #2, Note that sqrt(2/x) is the same as sqrt(2)/sqrt(x).\n\nIntegral (sqrt(2/x))dx = Integral(sqrt(2)/sqrt(x))dx\n= sqrt(2) Integral (1/sqrt(x))dx (I just pulled out the constant).\n\nAnd then from here, all you have to do is note that 1/sqrt(x) is the same as 1/(x^(1/2)) which is the same as x^(-1/2).  Now it's the matter of using the reverse power rule.\n\nsqrt(2) * Integral (x^(-1/2)) = sqrt(2) * 2 * x^(1/2) + C
699547
sin[x-(pi/2)]=kcosx\n-sin[(pi/2)-x]=kcosx\n-cosx=kcosx\nk=-1\ncos[x+{pi/2)]=ksinx\n=cos[(pi/2)+x]=ksinx\n=-sinx=ksinx\nk=-1
756840
cosx=tanx\ncosx=sinx/cosx\ncos^2(x)=sinx\n1-sin^2(x)=sinx    {pythagoras}\nsin^2(x)+sinx-1=0\nuse quadratic formula\nsinx=(-1+or-sqrt(5))/2\n=0.6180339888\nor -1.6180339888\nsine has to lie between +/-1\nsinx= 0.6180339888\nx=arcsin(0.6180339888)\n  =38.17270763\n+ve value for sine lies in 1st and 2nd\nquadrants \ntherefore,x=38.17270763 degrees or\n141.8272924 degrees\n\nthese two values were checked in the \noriginal equation and were found \nto be correct\n\ni hope that this helps
282780
1.sin(2x)=2sin(x)cos(x)\n2.tan(x)=sin(x)/cos(x)\n3.cos(x)=(1-sin^{2}(x))^(1/2)\n\nwe can use (3) in (2) and (1) to eliminate cos(x). Then substitute sin(x)=y:\n\n1.sin(2x)=2y.(1-y^{2})^(1/2)\n2.tan(x)= y/(1-y^{2})^(1/2)\n\ntherefore,\n\nf(y)=2y(1-y^{2})^{1/2}\n     +y/(1-y^{2})^{1/2}\n\nI hope that's what you wanted.
1094719
It's quite dificult to explain trigonometry this way, so it;s better for you to have an electronic teacher, this teacher is at your service day or night, sun or rain and  you could find it\n\nhttp://aleph0.clarku.edu/~djoyce/java/trig/\n\nhttp://www.sosmath.com/trig/trig.html\n\nhttp://fergusmurray.members.beeb.net/trig.htm\n\nhttp://mathworld.wolfram.com/Trigonometry.html\n\ngood luck
1088145
The answer is no, it isn't necessarily true, and the angle could be 150 degrees.\n\nSine is a repeating function there are an infinite number of values x which would yield the value 1/2 from the function y = sin x.\n\nBecause the sine function repeats, the inverse of the sine function is not itself a function, but it is common in mathmatics to, by limiting the range of the inverse, to make it a function.\n\nSo when using the inverse sin FUNCTION (y = Arcsin x), which has a range from negative 90 degrees to positive 90 degrees, inclusive, the only value in the range which will yield an answer of 1/2 is, indeed, 30 degrees.\n\nhopefully that made sense.