when dealing with trigonometric functions. How would you do this with out usind a calculator? Step by Step
129107
x can be any value.\n\ntan x = sin x / cos x\n\ncot x = cos x / sin x\n\nso tan x * cot x = sin x/cos x * cos x/sin x\n\nsin x's cross out each other\n\ncos x's cross out each other\n\nso you're left with 1.
444374
(secx - tanx)^2 = (1/cosx - sinx/cosx)^2 = ((1-sinx)/cosx)^2\n\n(1-sinx)/(1+sinx) \n= [(1-sinx)(1-sinx)/(1+sinx)(1-sinx)]\n\n= [(1-sinx)^2/(1-(sinx)^2)] \n= [(1-sinx)^2/(cosx)^2] \n= ((1-sinx)/cosx)^2
576555
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.
468828
(sin x+ cos x)^2 = sin^2x + 2sin x cos x + cos^2 x using (a+b)^2\n= sin^2 x + cos ^2 x + 2 sinx cos x\n= 1 + sin 2x\nbecuase sin^2x+ cos^2x =1 \nand sin 2x = sin(x+x) = sin x cos x + cosx sin x = 2sin xcos x\nusing the formula sin(A+B) = sin A cos B + cos A sin B
390092
Finding max's and mins is done thru the first derivitive and setting it to 0, so\n\nf(x)= sinx - cosx\nf1(x)= cosx + sinx\n0=cosx + sinx\n-cosx = sin x\nthis occurs when x= (3pi)/4\nmax is root 2, as was stated.\n\nThe answer that used to be directly above is wrong.
28448
I forgot how to do those, but it's some sort of trig identity isn't it?  Like\n\nsqrt (1-tan [x])\n\nProbably not though.
591362
I think...\n\n# means theta\n\nsin(90-#)*cos(90-#)=tan#/sec^2#\ncos#sin#                =(sin#/cos#)/(1/cos^2#)\ncos#sin#                =(sin#cos^2#)/cos#\ncos#sin#                =cos#sin#\n\nUsing: sin(90-#)=cos#\n           cos(90-#)=sin#\n           tan#= (sin#/cos#)\n           sec^2#=(1/cos^2#)\n\nThat work?
362067
Use sin(arcsinx) = x...\n\ncos(arcsinB) = sqrt(1-sin^2(arcsinB))\n= sqrt(1-B^2)\n\nSimilarly, sin(arccosA) = sqrt(1-A^2).\n\nUse the cos(x+y) equality to compute the last one...
425755
As I mentioned in another post here the other day, there is one technique that *always* works, even if it isn't the fastest.\n\nChange everything to sin and cos. Use sin^2 x + cos^2 x = 1. Simplify. Thats it.\n\nSo, the LHS is \n1/tan x - tan x\n= cos x / sin x  - sin x / cos x\n= (cos^2 x - sin^2 x) / (sin x cos x).\nNow, we have the right denominator, but the numerator isn't right yet. What else can we simplify? We know how to change cos2x into something just involving cos x or sin x:\nYou may already know that cos 2x = cos^2 x - sin^2 x. If not, maybe you know this one:\ncos 2x = 2 cos^2 x - 1\n= 2 cos^2 x - (sin^2 x + cos^2 x)\n= cos^2 x - sin^2 x.\nAnd thus the two sides are equivalent.
414846
Have you learned trig identities yet?  I'd check to see if there was a trig identity you could substitute here.\n\nBTW, I am assuming you meant cos²x rather than cosx².\n\nTry this:  cos²x + sin²x = 1, so substitute 1-sin²x for cos²x:\n\n6cos²x + sinx - 4 = 0\n\n6(1-sin²x) + sinx - 4 = 0\n\n-sin²x + sinx + 2 = 0\n\n(multiply by -1 to make it a little easier)\n\nsin²x - sinx - 2 = 0\n\n(sinx-2)(sinx+1) = 0\n\nSo, sinx = 2 or sinx = -1\n\nSolving for x, this means x = sin^-1(2) or sin^-1(-1)
115364
So you know the ones for sin(2a) and cos(2a). Cool.\n\nNow, tan(2a) is the same as sin(2a)/cos(2a), since tan = sin/cos (definition of tangent).\n\nSub in the formulae for sin(2a) and cos(2a) :\n\n2sin(a)cos(a)\n----------------\ncos(a)^2-sin(a)^2\n\nDivide both numerator and denominator by (cos(a))^2\n\n2tan(a)\n----------\n1-tan(a)^2
142019
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!
440237
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.?
475814
2 - csc²z =\n2 - (1/sin²z) = \n(2sin²z - 1) / sin²z = \n(sin²z + sin²z - 1 ) / sin²z =\n(sin²z - cos²z) / sin²z =  {using trig id sin²+cos²=1}\n1 - (cos²z / sin²z) =\n1 - cot²z\n\n~ ~ ~ ~\n\ncot³t / csct =\n(cos³t/sin³t) / (1/sint) =\n(cos³t/sin³t) * sint =\ncos³t/sin²t =\ncost (cos²t/sin²t) =\ncost (cot²t) = \ncost (csc² -1)  {using trig id 1+cot²=csc²}\n\nHope that helped!\n\n~ ♥ ~
117274
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.
645632
u can factor it out to be (2sinx+1)(cosx)=0.. then find x for which either sinx=-1/2 or cosx=0\n\n(sinx+1)cosx=0 same method applies
364846
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck