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prove that 2sinxcosx-cosx/1 -sinx+sin^2x-cos^2x=cotx

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  1. trig

    express this in sinx (1/ cscx + cotx )+ (1/cscx- cotx) i got 2sinx is that right?? and B) express in cosx problem: is 1 + cotx/cscx - sin^2x i get to the step of 1 + cos-sin^2x and im stuck..help! (1/cscx + cotx) + (1/cscx - cotx) = (sinx + cosx/sinx) +
  2. Trigonometry.

    ( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! Generally these are done by changing everything to sines and cosines, unless you see some obvious identities. Also generally, it is best to start with the more complicated side and try to change it to
  3. math

    prove that 2sinxcosx-cosx/1 -sinx+sin^2x-cos^2x=cotx
  4. Trig.......

    I need to prove that the following is true. Thanks (2tanx /1-tan^x)+(1/2cos^2x-1)= (cosx+sinx)/(cosx - sinx) and thanks ........... check your typing. I tried 30º, the two sides are not equal, they differ by 1 oh , thank you Mr Reiny I'll tell my teacher
  5. Simplifying with Trigonometry Identities

    Hi, I am a senior in High School having a really difficult time with two problems. I have to prove using the trigonometric identities that they equal each other but I am having a really hard time trying to get them to equal each other. I've spent over 2
  6. Precalculus/Trig

    I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1 - cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1-cosx Simplified: cosx + sin^3x/sin^3x = cscx/1-cosx I don't know where to go from there.
  7. Trig........

    I need to prove that the following is true. Thanks (cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds of problems when I was still teaching. it requires a little "trick" L.S. =cosx/(1-sinx) multiply top and bottom by 1+sinx,
  8. Trigonometry

    Prove the following trigonometric identities. please give a detailed answer because I don't understand this at all. a. sin(x)tan(x)=cos(x)/cot^2 (x) b. (1+tanx)^2=sec^2 (x)+2tan(x) c. 1/sin(x) + 1/cos(x) = (cosx+sinx)(secx)(cscx) d. tan^2 (x)(1+1/tan^2 x)
  9. math;)

    The equation 2sinx+sqrt(3)cotx=sinx is partially solved below. 2sinx+sqrt(3)cotx=sinx sinx(2sinx+sqrt(3)cotx)=sinx(sinx) 2sin^2x+sqrt(3)cosx=sin^2x sin^2x+sqrt(3)cosx=0 Which of the following steps could be included in the completed solution? a.
  10. Math - Trig - Double Angles

    Prove: sin2x / 1 - cos2x = cotx My Attempt: LS: = 2sinxcosx / - 1 - (1 - 2sin^2x) = 2sinxcosx / - 1 + 2sin^2x = cosx / sinx - 1 = cosx / sinx - 1/1 = cosx / sinx - sinx / sinx -- Prove: 2sin(x+y)sin(x-y) = cos2y - cos2x My Attempt: RS: = 1 - 2sin^2y - 1 -
  11. Confused! Pre-Cal

    Verify that each equation is an identity.. tan A= sec a/csca I have notes (i wasn't here that day and teacher refuses to reteach) but I don't understand them here is the notes... Problem w/ same directions: Cos x= cotx/csc x = Cosx/Sin x / 1/sinx = cosx I
  12. Calculus

    Integrate 1/sinx dx using the identity sinx=2(sin(x/2)cos(x/2)). I rewrote the integral to 1/2 ∫ 1/(sin(x/2)cos(x/2))dx, but I don't know how to continue. Thanks for the help. Calculus - Steve, Tuesday, January 12, 2016 at 12:45am 1/2 ∫
  13. trigonometry

    can i use factoring to simplify this trig identity? the problem is sinx + cotx * cosx i know the answer is cscx and i know how to get it but i want to know if i can do factoring to get it bc i tried to but it wont give me the answer . this is the step i
  14. Pre-calc

    prove the identity: (cosx)(tanx + sinx cotx)=sinx+cos(squared)x i need steps to show how i got the answer generally, it is a good idea to change all trig ratios to sines and cosines, and start with the more complicated-looking side. so.... LS =
  15. Pre-Calculus

    How would you prove/Verify this Identity. x=theta (cosx+sinx-sin^3x)/(sinx)=cotx+cos^2x
  16. Trig

    prove the identity (sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2 sinX^6= sinx^2 ^3 = (1-cosX^2)^3 = (1-2CosX^2 + cos^4) (1-cosX^2) then multiply that out 1-2CosX^2 + cos^4 - cosX^2 + 2cos^4 -cos^6 add that on the left to the cos^6, and combine terms..
  17. Math Trigonometric Identities

    Is this correct? (sinx - cosx)^2 = 1 - 2sinxcosx LS = sin^2x - 2sinxcosx + cos^2x = 1 - cos^2x - 2sinxcosx + cos^2x = 1 - cos^2x + cos^2x - 2sinxcosx = 1 - 2sinxcosx LS = RS
  18. Math trigonometric Identities

    Is this correct? (sinx - cosx)^2 = 1 - 2sinxcosx LS = sin^2x - 2sinxcosx + cos^2x = 1 - cos^2x - 2sinxcosx + cos^2x = 1 - cos^2x + cos^2x - 2sinxcosx = 1 - 2sinxcosx LS = RS
  19. trig

    I need to prove equality. a) (sina + cosa)^2 -1 / ctga - sinacosa = 2tg^2a b) (sin^2x/sinx-cosx) - (sinx+cosx/tg^2x+1) = sinx + cosx c) sin^4a - sin^2a - cos^4a + cos^2a = cosπ/2 How to do these?
  20. math

    i have a few questions cosx + cosx =2secx ----- ----- 1+sinx 1-sinx cos(x-B)-cos(x-B) = 2sinxsinB csc2x= 1 secx cscx --- 2 cotx= cos5x+cos3x ----------- sin 5x-sin 3x
  21. Pre-Calc

    Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx + cosx/cosx)/(sinx/cos x - 1/cosx + cosx/cosx)= tanx + cosx (just working on the left side) ((sinx + 1 - cosx)/cosx)/((sinx - 1 +
  22. Math

    Im really struggling with these proving identities problems can somebody please show me how to do these? I'm only aloud to manipulate one side of the equation and it has to equal the other side of the equation at the end Problem 1. Sinx/(cotx+1) +
  23. Math (trigonometric identities)

    I was given 21 questions for homework and I can't get the last few no matter how hard and how many times I try. 17. Sinx-1/sinx+1 = -cos^2x/(sinx+1)^2 18. Sin^4x + 2sin^2xcos^2x + cos^4x = 1 19. 4/cos^2x - 5 = 4tan^2x - 1 20. Cosx - sinx - cos^3x/Cosx =
  24. Trig Identities

    Prove the following identities: 13. tan(x) + sec(x) = (cos(x)) / (1-sin(x)) *Sorry for any confusing parenthesis.* My work: I simplified the left side to a. ((sinx) / (cosx)) + (1 / cosx) , then b. (sinx + 1) / cosx = (cos(x)) / (1-sin(x)) I don't know how
  25. TRIG..............

    Q.1 Prove the following identities:- (i) tan^3x/1+tan^2x + cot^3x/1+cot^2 = 1-2sin^x cos^x/sinx cosx (ii) (1+cotx+tanx)(sinx-cosx)/sec^3x-cosec^3x = sin^2xcos^2x.
  26. TRIGONOMETRY *(MATHS)

    Q.1 Prove the following identities:- (i) tan^3x/1+tan^2x + cot^3x/1+cot^2 = 1-2sin^x cos^x/sinx cosx (ii) (1+cotx+tanx)(sinx-cosx)/sec^3x-cosec^3x = sin^2xcos^2x.
  27. Math - Pre- Clac

    Prove that each of these equations is an identity. A) (1 + sinx + cos x)/(1 + sinx + cosx)=(1 + cosx)/sinx B) (1 + sinx + cosx)/(1 - sinx + cosx)= (1 + sin x)/cosx Please and thankyou!
  28. Calculus 2 Trigonometric Substitution

    I'm working this problem: ∫ [1-tan^2 (x)] / [sec^2 (x)] dx ∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx) ∫cosx-sinx(sinx/cosx) ∫cosx-∫sin^2(x)/cosx sinx-∫(1-cos^2(x))/cosx sinx-∫(1/cosx)-cosx sinx-∫secx-∫cosx sinx-sinx-∫secx =-ln |secxtanx|+C Can
  29. maths

    hey, i would really appreciate some help solving for x when: sin2x=cosx Use the identity sin 2A = 2sinAcosA so: sin 2x = cos x 2sinxcosx - cosx = 0 cosx(2sinx - 1)=0 cosx = 0 or 2sinx=1, yielding sinx=1/2 from cosx=0 and by looking at the cosine graph, we
  30. drwls

    My previous question: Verify that (secx/sinx)*(cotx/cscx)=cscx is an identity. (secx/sinx)*(cotx/cscx) = (secx/cscx)(cotx/sinx) = (sinx/cosx)*cotx*(1/sinx) "The last steps should be obvious" Not to me. I can convert (sinx/cosx) to tanx if that's even what
  31. Please help

    Write in terms of cos and sin function. cotx*secx I know cotx = cosx/sinx and sec x= 1/cos x. but how do i solve it?
  32. another precal

    i have two questions for you to check please! 6cos^2x+5cosx-4=0 where 0degrees < or equal to 0 which is < or equal to 360. I got that this factors to (3x+4)(2x-1) so the answer is x= -4/3 and x= 1/2 ?? right? and the second question: determine sin2theta,
  33. Trigonometry Check

    Simplify #3: [cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] = [cosx-((1)cosx-(0)sinx)sinx]/[cosx-((-1)cosx+(0)sinx)tanx] = [cosx-cosxsinx]/[cosx+cosxtanx] = [cosx(1-sinx]/[cosx(1+tanx] =
  34. Math help again

    cos(3π/4+x) + sin (3π/4 -x) = 0 = cos(3π/4)cosx + sin(3π/4)sinx + sin(3π/4)cosx - cos(3π/4)sinx = -1/sqrt2cosx + 1/sqrt2sinx + 1/sqrt2cosx - (-1/sqrt2sinx) I canceled out -1/sqrt2cosx and 1/sqrt2cosx Now I have 1/sqrt sinx + 1/sqrt2 sinx And that
  35. math

    Write in terms of cos and sin function. cotx*secx Show work. I know cotx = cosx/sinx and secx = 1/cosx, would that just be the answer or can i solve it?
  36. Calculus

    Q: If y=sinx/(1+tanx), find value of x not greater than pi, corresponding to maxima or minima value of y. I have proceeded thus- Equating dy/dx=0 we get{ (1+tanx)cosx-sinx.sec^2 x}/(1+tanx)^2=0……..(A) Or cosx+sinx=sinx.sec^2 x or cosx= sinx.sec^2-sinx
  37. Math

    1) evaluate without a calculator: a)sin(3.14/4) b) cos(-3(3.14)/4) c) tan(4(3.14)/3) d) arccos(- square root of three/2) e) csctheata=2 2) verify the following identities: a) cotxcosx+sinx=cscx b)[(1+sinx)/ cosx] + [cosx/ (1+sinx)]= 2secx c) (sin^2x +
  38. Precalculus/Trig

    Prove the following identity: 1/tanx + tanx = 1/sinxcosx I can't seem to prove it. This is my work, I must've made a mistake somewhere: Converted 1/tanx: 1/sinx/cosx + sinx/cosx = 1/sinxcosx Simplified 1/sinx/cosx: cosx/sinx + sinx/cosx = 1/sinxcosx Found
  39. Math

    Verify the identity . (cscX-cotX)^2=1-cosX/1+cosX _______ sorry i cant help you (cscX-cotX)=1/sinX - cosX/sinX = (1-cosX)/sinX If you square this you have (1-cosX)^2/(sinX)^2 Now use (sinX)^2 = 1 - (cosX)^2 to get (1-cosX)^2 / 1 - (cosX)^2 =
  40. Trigonometry

    Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx? Simplify #2: sin2x/1+cos2X = ??? I'm stuck on this one. I don't know what I should do. Simplify #3:
  41. Math 12

    Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx? Simplify #2: sin2x/1+cos2X = ??? I'm stuck on this one. I don't know what I should do. Simplify #3:
  42. trigonometric identities

    Prove the following trigonometric identities 1. tanx = sinx + sin^2x/cosx(1+sinx) 2. cos^3+(cosx)(sin^2x) = 1/secx
  43. Trigonometry

    How do you simplify: (1/(sin^2x-cos^2x))-(2/cosx-sinx)? I tried factoring and creating a LCD of (sinx+cosx)(sinx cosx) (cosx-sinx), but cannot come up with the right answer. The answer is (1+2sinx+2cosx)/sin^2x+cos^2x,but I don't know how the book arrived
  44. pre-calc

    Solve: cos(2x-180) - sin(x-90)=0 my work: cos2xcos180 + sin2xsin180= sinxcos90 - sin90cosx -cos2x - sin2x= cosx -cos^2x + sin^2x -2sinxcosx=cosx I'm stuck here. I tried subtracting cosx from both sides and making sin^2x into 1- cos^2x, but I still can't
  45. maths - trigonometry

    I've asked about this same question before, and someone gave me the way to finish, which I understand to some extent. I need help figuring out what they did in the second step though. How they got to the third step from the second. [sinx + tanx]/[cosx + 1]
  46. Trigonometic

    Solving Trigonometic Equations solve for x and give the answers as a equations : ( by radian) 1)cos(sinx)=1 We know sin 2x = 2(sinx)(cosx) so (sinx)(cos)=1/2(sin 2x) So we can change your equation from (sinx)(cosx)=1 to 1/2(sin 2x) = 1 (sin 2x) = 2 But the
  47. Trig

    Prove: (cotx sinx)(secx-cosx)=sin^2(X)
  48. Math

    I'm only aloud to manipulate one side of the problem and the end result has to match the other side of the equation Problem 1. sinx + cosx + sinx + tanx + cosxcotx = secx + cscx Problem 2. ((sinx + cosx)/(1 + tanx))^2 + ((sinx - cos^2x)/(1 - cotx))^2 = 1
  49. trig

    Verify that each equation is an identity. 16. 1+tanx/sinx+cosx =secx ok i have a clue on how to do it. i multiplyed the denominator by sinx-cosx and i also did the top but when i do i get this weird fraction with all these cos and sin and then i get
  50. Mathematics - Trigonometric Identities - Reiny

    Mathematics - Trigonometric Identities - Reiny, Friday, November 9, 2007 at 10:30pm (sinx - 1 -cos^2x) (sinx + 1 - cos^2x) should have been (sinx - 1 + cos^2x) (sinx + 1 - cos^2x) and then the next line should be sin^2x + sinx - cos^2xsinx - sinx - 1 +
  51. trig

    Solve for [0, 360) 2sinxcosx + cosx =0 2sinxcosx = -cosx 2sinx = -cosx/cosx sinx = -1/2 {210, 330) Is this correct?
  52. Trig (Last URGENT)

    sin2x+cosx=0 , [-180,180) = 2sinxcosx+cosx=0 = cosx(2sinx+1)=0 cosx=0 x1=cos^-1(0) x1=90 x2=360-90 x2=270 270 doesn't fit in [-180,180) what do I do? Or maybe I did something wrong. sinx=1/2 x1=sin^-1(1/2) x1=30 x2=180-30 x2=150 is this correct? Please and
  53. Pre-calculus

    Prove the following identities. 1. 1+cosx/1-cosx = secx + 1/secx -1 2. (tanx + cotx)^2=sec^2x csc^2x 3. cos(x+y) cos(x-y)= cos^2x - sin^2y
  54. math

    1. Prove the identity:cosx/1-tanx+sinx/1-cotx=sinx+cosx 2.Prove the identity: cosx/1+sinx+1+sinx/cosx=2secx
  55. trig

    (cotxsec2x - cotx)/ (sinxtanx + cosx) = sinx prove. i keep getting (if i start from the left side) = cosx instead of sinx please help!
  56. trigonometry (please double check this)

    Solve the following trig equations. give all the positive values of the angle between 0 degrees and 360 degrees that will satisfy each. give any approximate value to the nearest minute only. 1. sin2Į = (sqrt 3)/2 2. sin^2Į = cos^2Į + 1/2 3. sin 2x
  57. math

    this is for proving identies and its fustrating i can't solve this one question! lol x=feta (btw the first part is supposed to be divided by the bottom) 1 + 2sinxcosx + sinxcosx sinx + cosx (1 + 2sinxcosx / sinx + cosx) + sinxcosx
  58. Calculus

    Find the dervatives: 1. f(x)=(3x+1)e^x^2 2. y = e^(sin x) ln(x) 3. f(x)=(x^(2)+x)^23 4. f(x)=sin(2x)/cosx 5. square root of x/(3x+1) 6. f(x)=sin^4(3x=1)-sin(3x+1) 7. x+y=cos(xy) The answers I got: 1. 6x^(2)e^(x^2)+2xe^(x^2)+3e^(x^2) 2. e^sinx/x +
  59. Mathematics-Integration

    Question: For any positive integer n,show that integrate.[(sin x)^2n dx ] from o - π/2 = [(2n)!*π]/[(2)^(2n+1)*(n!)^2 ] What I thought: Let I =int.[(sinx)^2n dx] And again I= int.[ (sin(π/2-x))^2n dx] = int.[ (cos)^2n dx] 2I= int.[(sin x)^2n + (cos
  60. pre-cal

    Simplify the given expression........? (2sin2x)(cos6x) sin 2x and cos 6x can be expressed as a series of terms that involve sin x or cos x only, but the end result is not a simplification. sin 2x = 2 sinx cosx cos 6x = 32 cos^6 x -48 cos^4 x + 18 cos^2 x -
  61. Trig

    Prove the following functions: (sinx+sin2x)/(1+cosx+cos2x)=tan x (cos3x/sinx)+(sin3x/cosx)=2cot2x tan2x=(2/cotx-tanx) theses are due in the am. Fastness would be great?
  62. trig

    prove that (sin^4x-cos^4x)/(sinx-cosx) = sinx+cosx
  63. calculus

    Could someone check my reasoning? thanx Find the derivative of the function. sin(sin[sinx]) I need to use the chain rule to solve. So I take the derivative sin(sin[sinx) first. Then multiply that by the inside which is the derivative of sin[sinx]. And
  64. Trig Help

    Prove the following: [1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)] =[sinx+sin^2x]/[sinx+1] =[sinx+(1-cos^2x)]/[sinx+1] =??? This is where I'm stuck. Can someone help me. Please check what I got is
  65. Pre-Calculus

    How can this identity be proved/verified? cscx-sinx=1/(secxtanx) I have tried starting on both sides of the equation to get to the other. By starting on one side of the equation and manipulating it to make it look like the other side, the identity will be
  66. calc

    Where do I start to prove this identity: sinx/cosx= 1-cos2x/sin2x please help!! Hint: Fractions are evil. Get rid of them. Well, cos2x = cos2x - sin2x, so 1-coscx = 1 - cos2x - sin2x = 1 - cos2x + sin2x You should be able to simplify this to 2*something
  67. Pre-Calc : Verify the Identity

    Verify the Identity: sin(x+π)/cos(x+3π/2) =tan^2x-sec^2x I've done: sinxcosπ+cosxsinπ / cosxcos(3π/2) - sinxsin(3π/2) sinx(-1) + cosx(0) / cosx(0)- sinx(-1) -sinx/sinx What do I do from here? Or what did I do wrong?
  68. Trig

    #1 (1/sinx)(sin^2x + cos^2x(sinx/cosx) )/(sinx + cosx) = (sinx + cosx)/(sinx+cosx) = 1 Reiny helped with this question but I don't how Reiny did it. Please explain
  69. Calculus

    If y=3/(sinx+cosx) , find dy/dx A. 3sinx-3cosx B. 3/(sinx+cosx)^2 C. -3/(sinx+cosx)^2 D. 3(cosx-sinx)/(sinx+cosx)^2 E. 3(sinx-cosx)/(1+2sinxcosx)
  70. Math(Please help)

    1)tan Q = -3/4 Find cosQ -3^2 + 4^2 = x^2 9+16 = sqrt 25 = 5 cos = ad/hy = -4/5 Am I correct? 2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos
  71. Math(Please help)

    2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos pi/4 = sqr2/2 2sin^x(sqrt2/2) = -1 sin x = -sqrt2 x = 7pi/4 and 5pi/4 Am I correct?
  72. math, calculus

    Prove that if y = cotx then dy/dx = - (cscx)^2. Hint: cotx = cosx/sinx
  73. Pre-Calc

    How do I solve this? My work has led me to a dead end. tan(45-x) + cot(45-x) =4 my work: (tan45 - tanx)/(1+ tan45tanx) + (cot45 - cotx)/(1 + cot45cotx) = 4 (1-tanx)/(1+tanx) + (1-cotx)/(1+cotx) = 4 Then I found a common denominator, giving me this:
  74. Math - help really needed

    Prove each idenity. 1+1/tan^2x=1/sin^2x 1/cosx-cosx=sinxtanx 1/sin^2x+1/cos^2x=1/sin^2xcos^2x 1/1-cos^2x+/1+cosx=2/sin^2x and (1-cos^2x)(1+1/tan^2x)= 1 I haven't even gotten 'round to sny of the quedtions because the first one is just so hard. I'm not
  75. Maths

    Solve this equation fo rx in the interval 0
  76. maths

    if sinX+cosX=m prove that sin^6X+cos^6X=4-3(m^2-1)^2/4 where m^2=
  77. Math - help really needed

    I'm sorry to double post; I don't want to seem impatient, but I really need help with this. Prove each idenity. 1+1/tan^2x=1/sin^2x 1/cosx-cosx=sinxtanx 1/sin^2x+1/cos^2x=1/sin^2xcos^2x 1/1-cos^2x+/1+cosx=2/sin^2x and (1-cos^2x)(1+1/tan^2x)= 1 I haven't
  78. Math

    Prove each of the following identities. a)tan^2x = 1 – cos^2x/cos^2x b)sinx/1 – sin^2x = tanx/cosx
  79. math, calculus

    Calculate the derivatives of the following: (also please could you show the steps, i'm so confused) a) y = sinx - cosx b) f(x) = x^2 + cosx c) y = sin(x^2) d) f(x) = x^5 sin (2x^3) e) y = cos (sinx) f) f(x) = sin^2 x
  80. Pre-Calculus

    Hi, I need a bit of help with verifying identities. The problems are as follows: sinx+cosx/cotx+1=sinx sin2x/sin - cos2x/cosx=secx 1-tan^2θ/1+tan^2θ=cos2θ
  81. Mathematics-Integration

    Question: Prove that [integrate {x*sin2x*sin[π/2*cos x]} dx] /(2x-π) } from (0-π) = [ integrate {sin x*cos x*sin[π/2*cos x} dx ] from (0-π). My thoughts on the question: We know that integrate f(x) dx from (0-a) = integrate f(a-x) dx from (0-a) From
  82. Mathematics - Trigonometric Identities

    Prove: sin^2x - sin^4x = cos^2x - cos^4x What I have, LS = (sinx - sin^2x) (sinx + sin^2x) = (sinx - 1 -cos^2x) (sinx + 1 - cos^2x) = sin^2x + sinx - sinx - cos^2xsinx - cos^2xsinx - 1 - 1 + cos^4x = sin^2x - 2cos^2xsinx - 2 + cos^4x Where did I go wrong?
  83. Math, please help

    Which of the following are trigonometric identities? (Can be more then one answer) tanx cosx cscx = 1 secx-cosx/secs=sin^2x 1-tanxtany=cos(x+y)/cosxcosy 4cosx sinx = 2cosx + 1 - 2sinx Find all solutions to the equation cosx cos(3x) - sinx sin(3x) = 0 on
  84. Calc II

    For the following question, we need to find the length of the polar curve: r= 2/(1-cosx) from pi/2
  85. Math

    Choose the three true statements from the following. Options A) The period of y =cosx is 90◦. B) The period of y =sinx is ð radians. C) The period of y =sin(2x) is four times the period of y =sinx. D) The period of y =sin(2x) is half of the period of y
  86. Math

    Choose the three true statements from the following. Options A) The period of y =cosx is 90◦. B )The period of y =sinx is ð radians. C) The period of y =sin(2x) is four times the period of y =sinx. D) The period of y =sin(2x) is half of the period of y
  87. math trigonometry

    Prove the identity Sin^3X sinx*cos^2X=tanx/cosX
  88. TC Maths

    Prove sin^2(x)-cos^2(y)/Sinx cosx-Sinycosy=TAN(x+y)
  89. math trigonometry

    Prove the identity Sin^3X+sinx*cos^2X=tanx/cosX
  90. Trig

    Simplify sin x cos^2x-sinx Here's my book's explanation which I don't totally follow sin x cos^2x-sinx=sinx(cos^2x-1) =-sinx(1-cos^2x) =-sinx(sin^2x) (Where does sine come from and what happend to cosine?) =-sin^3x
  91. Calculus

    Can someone help me find the derivatives of these questions? For these can you only need to tell me the answer so i can see if i have the same thing. y=cos(-4x) y = sinx^2 y = xcosx I need help with these 6 please. y = xcos(1/x) y = (1+sinx)/(1-sin^(2)x) y
  92. Math

    How do I solve this? tan^2x= 2tanxsinx My work so far: tan^2x - 2tanxsinx=0 tanx(tanx - 2sinx)=0 Then the solutions are: TanX=0 and sinX/cosX = 2 sin X Divide through by sinX: we have to check this later to see if allowed (ie sinX is not zero) cosX=1/2
  93. mathematics

    Knowing that derivative of sinx is cos⁡x and derivative of cosx is -senx, prove by induction, that the 2n derivative of sinx = ((- 1) ^ n ) * sin⁡x⁡ and the 2n derivative of cos⁡x = ((- 1) ^ n)* cos⁡x.
  94. Maths

    Prove that 1+cosx+sinx/1-cosx+sinx=cotx
  95. Math

    integrate (sinx)^(5)(cosx)^(12)from 0 to (pi/2) I'm having problems just even integrating it... I tried breaking it up (sin^(2)x)^(2)(cos^(2)x)^(6)(sinx)dx then doing u-sub with u=cosx and du= -sinx but then I get stuck... the answer is 8/3315 just don't
  96. Trig math

    PLEASEE HELPPMEEE!!!! cot2x+sec2x=tan2x+csc2x and 1+2sinxcosx/sinx+cosx=sinx + cosx
  97. calculus

    express in sinx 1 1 ---------- + -------- cscx + cotx cscx - cotx and express in cosx 1 + cot x ------- - sin^2x cscx = 1/sinx so what do i do w. that extra one on the top!?? im so confused cscx Your fractions aren't lining up. This question is also not
  98. Math (Calculus AB)

    If y=cos^2x-sin^2x, then y'= a) -1 b) 0 c) -2(cosx+sinx) d) 2(cosx+sinx) e) -4(cosx)(sinx) I thought the answer was C but the answer key says it is E. Please help. Thanks in advance.
  99. math (trig)

    i have some problems doing trig the first one is: Show that cos(x/2) sin(3x/2) = ½(sinx + sin2x) i know that you are supposed to substitute all those trig function things in it but i kind of forgot how to the only that i can see substituting in is the
  100. Mathematics - Trigonometric Identities

    Prove: (tanx)(sinx) / (tanx) + (sinx) = (tanx) - (sinx) / (tanx)(sinx) What I have so far: L.S. = (sinx / cosx) sinx / (sinx / cosx) + sinx = (sin^2x / cosx) / (sinx + (sinx) (cosx) / cosx) = (sin^2x / cosx) / (cosx / sinx + sinxcosx)

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