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

  1. 👍
  2. 👎
  3. 👁
  1. for the first one:

    LS = cosx/(1+sinx) + cosx/(1-sinx)

    getting a common denominator of (1+sinx)(1-sinx)
    = [cosx(1-sinx) + cosx(1+sinx)]/(1+sinx)(1-sinx)
    = 2cosx/(1-sin^2)
    = 2cosx/cos^2x
    = 2/cosx
    = 2secx
    = RS

    for cos(x-B)-cos(x-B) = 2sinxsinB

    use the formula cos(A-B) = cosAcosB + sinAsinB on the Left Side, it comes apart very nicely after that

    For the last two, try changing all trig ratios into sines and cosines.
    Show me what you get

    1. 👍
    2. 👎
  2. for cos(x-B)-cos(x-B) = 2sinxsinB
    you must have a typo, the LS is zero the way it stands
    I am sure you meant

    cos(x-B)-cos(x+B) = 2sinxsinB

    1. 👍
    2. 👎
  3. i don't understand how to use the formula in the second problem. what would i plug into A and B?

    1. 👍
    2. 👎
  4. cos(x-B)-cos(x+B) = 2sinxsinB

    LS
    = cosxcosB + sinxsinB - (cosxcosB - sinxsinB)
    =2sinxsinB
    = RS

    1. 👍
    2. 👎

Respond to this Question

First Name

Your Response

Similar Questions

  1. trigonometry

    how do i simplify (secx - cosx) / sinx? i tried splitting the numerator up so that i had (secx / sinx) - (cosx / sinx) and then i changed sec x to 1/ cosx so that i had ((1/cosx)/ sinx) - (cos x / sinx) after that i get stuck

  2. 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

  3. maths

    (Sin^3x-cos^3x)/(sinx-cosx) – cosx/sqrt(1+cot^2x)-2tanxcotx=-1 where x∈(0,2pi) general value of x.

  4. 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

  1. 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

  2. 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/

  3. Trigonometry

    4. Find the exact value for sin(x+y) if sinx=-4/5 and cos y = 15/17. Angles x and y are in the fourth quadrant. 5. Find the exact value for cos 165degrees using the half-angle identity. 1. Solve: 2 cos^2x - 3 cosx + 1 = 0 for 0

  4. Math

    (sinx - cosx)(sinx + cosx) = 2sin^2x -1 I need some tips on trigonometric identities. Why shouldn't I just turn (sinx + cosx) into 1 and would it still have the same identity?

  1. Calculus

    determine the absolute extreme values of the function f(x)=sinx-cosx+6 on the interval 0

  2. 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.

  3. 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)

  4. 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 +

You can view more similar questions or ask a new question.