If sinA=12/13 and A is in the first quadrant, find the value of sinA/2 and cosA/2?
435670
lets assume b = a/2\ntherefore 2b = a\n\nnow sin 2b = 12/13\n\nsin 2b is also equal to 2sinbcosb\n\nso 2sinbcosb = 12/13\n\nsinb * cosb = 6/13\n\nnow u know that cosb = sqrt ( 1 - sin^2 b)\n\nso sinb * sqrt(1-sin^2 b) = 6/13\n\nor sin^2 b * (1 - sin^2 b) = 36/169\n\nsubstituting sin ^ b = x\n\nyou get x - x^2 - 36/169 = 0\n\nThis is a simple quadratic equation\n\nsolving for x you get\n\nx = 234/338 or x = 104/338\n\nso you get sin b is sqrt (234/338) or sin b is sqrt(104/338)\n\nsince a is in the first quadrant, the value of sin a/2 will be less than that of cos a/2 since a/2 is less than 45 degrees.\n\nSo, sin a/2 = sqrt of (104/338)= 0.554\n\ncos a/2 = sqrt of (234/338) = 0.832
411302
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=1
1135417
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
960686
The equation is correct but only if A is 0, 180 or integer multiples of 180 degrees.\nsecA/(1+sinA) - cosA/(1+sinA) = tanA\n(secA - cosA)/(1+sinA) = tanA\ncosA(secA - cosA)/(1+sinA) = tanA * cosA\n(1-cos^2A)/(1+sinA) = sinA\nsin^2A/(1+sinA) = sinA\nsin^2A = sinA + sin^2A\nsinA = 0  which occurs when A=0,180,360,etc.
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
743368
For cos t to be in quadrant 3, it would look like this:\n(-cos t,-sin t)\n\nSince b=-3/7, you do this:\na^2+(-3/7)^2=1^2\na^2+9/49=1\na^2=49/49-9/49\na^2=40/49\n√a^2=-√40/49\na=-√40/7\n\nCheck:\n(-√40/7)^2+(-3/7)^2=1^2\n40/49+9/49=1\n49/49=1\n1=1\n\nI hope this helps!
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.
575459
Sorry but I don't quite understand your qn.\nCan you make it clearer?\n\n(to ozert:\ncos A and sin A could be negative values if tan A is in Quadrant 3)
982428
LHS\n= (sin^2B-tan^2B)/(1-sec^2B)\n= (sin^2B-tan^2B)/(-tan^2B)\n= (sin^2B/-tan^2B) + (-tan^2B/-tan^2B)\n[splitting the numerator up]\n= (-cos^2B) + 1\n[cosx= sinx/tanx]\n= 1 - cos^2B\n= sin^2B\n= RHS
788868
cosθ = 2cos²θ/2 - 1\nSo cos²θ/2 = ½(1 + cosθ)\nSo cos²(π/12) = ½(1 + cos(π/6))\n= ½ + √3/4\n= ¼(2 + √3)    \nSo cos(π/12) = + √{¼(2 + √3)}\n= ½√(2 + √3)\n\nLet √(2 + √3) = a + b√3\nSquare both sides\n2 + √3 = a² + 3b² +2ab√3\n\nSo a² + 3b² = 2 and 2ab = 1\na = 1/√2 and b = 1/√2 \nSo √(2 + √3) = ½(√2 + √6)\n\nSo cos π/12 = ¼(√2 + √6)
429515
The site below, around one-third of the way down, shows the geometry needed to answer your question.\n\nOnce you have the formula for cos(A+B), namely cosAcosB - sinAsinB, the proof is easy. Just substitute -B for B, and you get\n\ncos(A-B) = cosAcos(-B) - sinAsin(-B)\n\nNow, cos(-B) = cosB and sin(-B) = -sin(B). Thus\n\ncos(A-B) = cosAcosB + sinAsinB
1182723
sinx+cosecx=2\nsin+(1/sinx)=2\nfrom here try to cross multiply!
381123
Yes. I assume to want to find sin 1/2x, cos 1/2x, and so on.\nIt's quite simple:\n1. Write the expanded form of cos 2x\ncos^2x - sin^2x\n2. Write that twice, in once in a column (remember it equals cos 2x)\ncos^2x - sin^2x  = cos 2x =  cos^2x - sin^2x\n3. Rewrite cos^2x in the first column as 1-sin^2x, and in the second column rewrite sin^2x as 1-cos^2x\n(1-sin^2x) - sin^2x = cos 2x = cos^2x - (1 - cos^2x)\n4. Combine like terms\n1-2sin^2x   and -1+2cos^2x\n5. In order to find cosx and sinx, we must isolate each:\n-2sin^2x = cos 2x -1  and 2cos^2x = cos 2x + 1\n6. Divide by the coefficent of each\nsin^2x = 1/2 - 1/2(cos 2x) and cos^2x = 1/2 + 1/2(cos 2x)\n7. Find the square root of both sides\nsin x = sqrt(1/2 - 1/2(cos 2x)) and \ncos x = sqrt(1/2 + 1/2(cos 2x))\n8. We found what sin x and cos x equal, we want what 1/2x equals. Let x = 1/2x\nsin 1/2x = sqrt(1/2 - 1/2(cos 2(1/2x)) and\ncos 1/2x = sqrt(1/2 + 1/2(cos 2(1/2x))\n9. Simplify\nsin 1/2x = sqrt(1/2 - 1/2(cos x)) and\ncos 1/2x = sqrt(1/2 + 1/2(cos x))\n10. You're done. You can find tan 1/2x, csc 1/2x, sec 1/2x, and cot 1/2x by using the trigometric identities with sin 1/2x and cos 1/2x:\n\ntan 1/2x = (sin 1/2x)/(cos 1/2x)\ncsc 1/2x = 1/(sin 1/2x)\nsec 1/2x = 1/(cos 1/2x)\ncot 1/2x = (cos 1/2x)/(sin 1/2x)\n\nObviously there are other expansions for the half angle formulae. These are just the easiest.
258058
you know that sin^2(t)+cos^2(t)=1. so cos^2(t) is 144/169. but because it is on the positive part of the axis, cos(t)=12/13.
847590
sin (2x ) = 2.sinx.cosx so;\n6- 6(2.sinx.cosx) = 6-12sinxcosx;\n6(1 - 2sin2x) is the simplified form.
1081842
1)x is an angle in quadrant III and sin x = -1 / 3. Find cos x.\n2)x is an angle in quadrant IV and tan x = -5. Find sin x.\n3)Simplify the trigonometric expression\n(sin x + cos x)2 + (sin x - cos x)2\n4)Find the exact value of sin(15o)\n5)Simplify cos(x - pi/2)\n6)Find the inverse of the linear function f.\nf(x) = 3x - 2\n7)Find the inverse of the quadratic function f.\nf(x) = x 2 - 2x , x >= 1\n\nFor the answers to the above questions log on to www.google.co.in where u can find many such problems.
851033
(cosx-sinx)^2=cos^2x+sin^2x-2sinxcosx so\nit can be 1-2sinxcosx+cosx+sinx