What is is cos 67 and a half degrees? Use a half-angle identity to find each value?
368801
67.5 times 2 is 150.\nThe formula is\ncos(B/2) = ± sqrt([1 + cos B] / 2)\nB = 150\nThe answer will be positive because 67.5 is in the first quadrant and all trig functions are positive.\ncos(150) = -cos(30) = -sqrt(3)/2\nPlug this into the formula\nsqrt([1-sqrt3)/2]/2)\n= .258819045103
1030265
a=c*sin(30)=.5c since it is apposite the 30 degree angle\nb=sqrt(c^2-a^2)= .866c\nor b=c*cos(30)=.866c\n\na=b=c*sin(45)=.707c=
809910
cosine 75*=cos(90-15*)=sin15*
846581
Expand the exponent: (cos (x/2) - sin(x/2))^2 = (cos x/2)^2 - 2*cos(x/2)*sin(x/2) + (sin x/2)^2 = 1 - 2*sin(x/2)*cos(x/2) = 1-sin(x)\n\nusing the fact (cos a)^2 + (sin a)^2 = 1 and 2*sin(a)*cos(a) = sin(2a)\n\nSteve
938253
2cos²(45º-0.5x) – 1 = sinx\n\nUsing: cos 2a = 2cos² a -1\nSo: 2cos²(45º-0.5x) – 1 = cos 2(45º-0.5x) = cos (90º - x)  = sinx
932480
sin(A+B)sin(A-B)/(cos²Acos²B)\n= (sinAcosB + cosAsinB)(sinAcosB - cosAsinB)/(cos²Acos²B)\n= (sin²Acos²B - cos²Asin²B)/(cos²Acos²B) (since (x + y)(x - y) = x² - y²)\n= sin²A/cos²A - sin²B/cos²B\n= tan²A - tan²B ← Option c)\n\nPlease note: sin(A + B) = sinAcosB + cosAsinB\nand sin(A - B) = sinAcosB - cosAsinB
873429
cos5x = sinx\ncos5x = cos(pi/2 - x)\n5x = 2npi +- (pi/2 - x)\n6x = (4n + 1)pi/2   ,   4x= (4n-1)pi/2\nx= (4n+1)pi/12    ,  x = (4n-1)pi/8
463511
Use the double angle identities.
761231
[cos(x)]^2 - 1.2cos(x) - 0.28 = 0\n\nMultiply through by 100 to get rid of decimals.\n100[cos(x)]^2 - 120cos(x) - 28 = 0\n\nDivide by the common factor = 4.\n25[cos(x)]^2 - 30cos(x) - 7 = 0\n\nThis expression factors, so we don't need the quadratic formula.\n[ 5cos(x) - 7 ] [ 5cos(x) + 1 ] = 0\n\nEquating each factor to zero gives :\ncos(x) = 7/5 = 1.4, but cos(x) is always <= 1,\nso this is not a solution.\n\nor\n\ncos(x) = -1/5 = - 0.2\nThis is a solution because -1 <= cos(x) <= 1.\n\nTaking the arccos by calculator gives :\nx = 101.54º\n\nThis angle is in the second quadrant.\n\nFor cos, there must also be an angle in the third quadrant,\nbecause the cos of all angles there, are also negative.\n\nThis angle is greater than 180º by the same\namount that the other angle is less than 180º.\n\nSo we find that 180º - 101.54º = 78.46º\nThus, 101.54º is less than 180º by 78.46º.\n\nTherefore, the other angle must be 78.46º more than 180º.\nSo the answer to this is 180º + 78.46º = 258.46º.\n\nTherefore, x = 101.54º or 258.46º.\n\nI'll let you calculate those values in radians.\nI still find it easier to work with degrees!!!
319082
RHS cot a - tan b \n= cos a/sina - sinb/cosb \n= cosacosb - sinasinb/sinacosb\n= cos(a+b)/sinacosb\n= LHS (proven)\n\nThe idea is to prove that both sides are the same.
610143
I think ur right cos^2(theta) should be cos(theta). then the answer would be perfect.
1096282
cos^2(A) / ( 1 - sinA )\n=( 1 - sin^2(A) ) / ( 1 - sinA )\n=( 1 + sinA )( 1 - sinA ) / ( 1 - sinA )\n=1 + sinA
733318
I think your identity should be :\ncsc^2(x) / [1 + tan^2(x)] = cot^2(x)\nas you forgot to square the tan(x).\n\nWorking on the 1 + tan^2(x) gives:\n1 + sin^2(x) / cos^2(x)\n= [sin^2(x) + cos^2(x)] / cos^2(x)\n= 1 / cos^2(x)\n\nNow we have : csc^2(x) / [1 / cos^2(x)]\n= [1 / sin^2(x)] / [1 / cos^2(x)]\n= cos^2(x) / sin^2(x)\n= cot^2(x), which was to be shown.
98672
use the tangent ratio ( trig function ) or inverse tangent of (6/ (43/8)) OR iNV TANGENT (1.116) = 48.14 degrees and to get the other angle, just subtract it from 90 degrees or 41.86 degrees. hope it helps how I got it. go over the trig function or cohsahtoa rule
324333
Look at this link.\n\nUse the first two "sum to product" identities.\n\nOn the right you will have:\n\n= (2 / 2) (sin a / cos a) (cos b / sin b)\n\n= 1 (tan a) (cot b)\n\na = (x + y) / 2,  b = (x - y) / 2\n\nso\n\n= (tan (x+y)/2) (cot (x-y)/2)\n\nOh, darn!  You're right, the signs are reversed from this.  I wonder if the problem was given to you incorrectly.  I don't see how you could go from our result to the problem the way it was given.  Ugh!\n\nI'm sorry, but I can't go any further with this.  Good luck, and let us know if the problem was stated wrong, please.
477890
Consider a triangle with sides a, b, c and angle A oppose to the side a.  Side a is the diagonal that you are going to find in each case.\n\nBy knowing 2 sides (b, c) and the angle (A) in between, then use Cosine rule:\n\na^2 = b^2 + c^2 - 2bc cos A\n\na^2 = 60^2 + 80^2 - 2 * 60 * 80 cos(56)\n\na = 68.06 ft  (shorter diagonal)\n\nConsider when A = 180 - 56 = 124, use Cosine rule again:\n\na^2 = 60^2 + 80^2 - 2 * 60 * 80 cos(124)\n\na = 123.97 ft (longer diagonal)
756724
The first quadrant angle x whose sine is 1/2 is 60 degrees.\n\nThis angle has cosine of sqrt(3)/2.\n\nAnother way that we can do this is to remember that:\n[sin(x)]^2 + [cos(x)]^2 = 1 for all x.\n\n(1/2)^2 + [cos(x)]^2 = 1\n[cos(x)]^2 = 3/4\n[cos(x)] = sqrt(3/4) = sqrt(3)/2