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 + cos^2 x + cos^2xsinx - cos^4x
= cos^2 x - cos^4 x
= RS

an easier way would have been

LS = sin^2x(1-sin^2x) by common factor
=sin^2x(cos^2x)

RS = cos^2x(1-cos^2x) also by common factor
=cos^2x(sin^2x)
= LS


--------------------


What happened to the negative in the first line?

You wrote:


an easier way would have been

LS = sin^2x(1-sin^2x) by common factor
=sin^2x(cos^2x)

--

Shouldn't it be LS = sin^2x (-sin^2x) ?




And for my way ...


You wrote:

(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 + cos^2 x + cos^2xsinx - cos^4x
= cos^2 x - cos^4 x
= RS


--

and I'm asking why is the "cos^2x" a negative from "sinx - 1 + cos^2x"?

This line, "sin^2x + sinx - cos^2xsinx - sinx - 1 + cos^2 x + cos^2xsinx - cos^4x" ----> I get "sin^2x + sinx - cos^2xsinx - sinx - 1 + cos^2x + cos^2xsinx + cos^2x - cos^4x" ----> I get an extra "cos^2x" ...

  1. 👍 0
  2. 👎 0
  3. 👁 113
  1. sin^2x - sin^4x = cos^2x - cos^4x

    1. 👍 0
    2. 👎 0
  2. Ok, let's start from the beginning

    sin^2x - sin^4x = cos^2x - cos^4x

    LS = (sinx - sin^2 x)(sinx + sin^2 x)

    then you had:
    (sinx - 1 -cos^2x) (sinx + 1 - cos^2x)

    I will put in the in-between step

    =(sinx - (1 -cos^2x)) (sinx + 1 - cos^2x)

    = (sinx - 1 + cos^2x) (sinx + 1 - cos^2x)
    = sin^2x + sinx - cos^2xsinx - sinx - 1 + cos^2 x + cos^2xsinx + cos^2 x - cos^4x
    (and yes, you are right, I left out a +cos^2 x , so.......

    = sin^2x+cos^2x + sinx-sinx - 1 + cos^2xsinx-cos^2xsinx + cos^2 x - cos^4 x
    = 1 + 0 - 1 + 0 + cos^2 x - cos^4 x
    = cos^2 x - cos^4 x
    = RS

    For my second solution I worked on both the LS and the RS, reaching the same expression at both ends.

    Surely your teacher must have shown you that as an acceptable method.

    Here is what you wrote:

    "hat happened to the negative in the first line?"

    regarding:

    LS = sin^2x(1-sin^2x) by common factor
    =sin^2x(cos^2x)

    didn't you recognize the replacement of
    1 - sin^2 x with cos^2 x ???

    1. 👍 0
    2. 👎 0
    posted by Reiny
  3. I see it now, thanks

    1. 👍 0
    2. 👎 0

Respond to this Question

First Name

Your Response

Similar Questions

  1. 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

    asked by amanda on March 7, 2007
  2. 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)

    asked by Jessy on November 24, 2013
  3. 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 -

    asked by Dave on January 2, 2007
  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 30ยบ, the two sides are not equal, they differ by 1 oh , thank you Mr

    asked by abdo on April 18, 2007
  5. 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

    asked by Anonymous on February 21, 2012
  6. 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/

    asked by Emily on August 22, 2013
  7. trigonometric identities

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

    asked by noel on June 22, 2012
  8. Math (trigonometry)

    Trigonometry identities are so hard... I need some help proving these identities: *Oh, and I'm only in grade 11, so the identities we use are quotient identity and Pythagorean identity. sinx/(sinx + cosx) = tanx/(1 + tanx) cos^2x

    asked by Lucy on April 13, 2008
  9. 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

    asked by mo on April 18, 2007
  10. 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

    asked by Jill on January 14, 2014

More Similar Questions