Results 1 to 2 of 2

Math Help - Equations and results involving double angle formulae

  1. #1
    RAz
    RAz is offline
    Junior Member RAz's Avatar
    Joined
    May 2008
    From
    Canada
    Posts
    54

    Equations and results involving double angle formulae

    1. I would like to know if this is correct:

    5sinx-2secx=0 where x is an element of 0 and 360 (2pi in radians)

    5sinx-\frac {2}{cosx}=0

    5sinxcosx-2=0

    sin5x-2=0

    \therefore sinx=\frac {2}{3}, \frac {2}{3} + \pi,  \frac {2}{3} + 2\pi

    2. I would like to know where to continue in this question:

    sin3xcosx-cos3xsinx=\frac {\sqrt{3}}{2}

    3sinx-4cos^3xcosx-4cos^3x-3cosx=

    3sinxcosx-4cos^4x-4sincos^3x-3cos^2xsinx=

    ...I don't really know where to go from here and how to find angles for x.

    3. I am being introduced to product and sum formulas and I don't know how to tackle the questions.

    eg1. sinx=\frac{3}{2}, 0<x<\frac{\pi}{2}

    Find the exact value of sin2x, cos2x.

    The extra formulas given for sinx and cosx are:

    sinx=\frac{2sin\frac{x}{2}}{1+tan^2 \frac{x}{2}}

    and

    cosx=\frac {1-tan^2 \frac{x}{2}}{1+tan^2\frac{x}{2}}

    I know all the normal ones like sin2x=2sinxcosx etc., just not how to apply them.

    If you could help me out here that would be great
    Last edited by RAz; March 1st 2010 at 12:09 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Hello RAz

    See my comments in
    red.
    Quote Originally Posted by RAz View Post
    1. I would like to know if this is correct:

    5sinx-2secx=0 where x is an element of 0 and 360 (2pi in radians)

    5sinx-\frac {2}{cosx}=0

    5sinxcosx-2=0

    sin5x-2=0 No. 5 sin x cos x = 2.5 sin 2x.

    \therefore sinx=\frac {2}{3}, \frac {2}{3} + \pi,  \frac {2}{3} + 2\pi

    2. I would like to know where to continue in this question:

    sin3xcosx-cos3xsinx=\frac {\sqrt{3}}{2}

    3sinx-4cos^3xcosx-4cos^3x-3cosx= Don't do it like this. Use sin 3x cos x - cos 3x sin x = sin (3x-x) = sin 2x

    3sinxcosx-4cos^4x-4sincos^3x-3cos^2xsinx=

    ...I don't really know where to go from here and how to find angles for x.

    3. I am being introduced to product and sum formulas and I don't know how to tackle the questions.

    eg1. sinx=\frac{3}{2}, 0<x<\frac{\pi}{2} This doesn't make sense. Do you mean tan x = 3/2?

    Find the exact value of sin2x, cos2x.

    The extra formulas given for sinx and cosx are:

    sinx=\frac{2\color{red}\tan\color{black}\frac{x}{2  }}{1+tan^2 \frac{x}{2}}

    and

    cosx=\frac {1-tan^2 \frac{x}{2}}{1+tan^2\frac{x}{2}}

    I know all the normal ones like sin2x=2sinxcosx etc., just not how to apply them. Don't use them here. Evaluate sin 2x and cos 2x, using the formula above, but substituting 2x in place of x.

    If you could help me out here that would be great
    Can you try these again now?

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Double angle formulae
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: October 5th 2011, 01:41 PM
  2. Double Angle Formulae
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: May 1st 2011, 03:03 AM
  3. Replies: 4
    Last Post: July 5th 2010, 06:20 AM
  4. Double angle formulae
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 18th 2010, 06:40 AM
  5. R formulae (double angle formulae) URGENT
    Posted in the Algebra Forum
    Replies: 8
    Last Post: April 10th 2008, 11:30 AM

Search Tags


/mathhelpforum @mathhelpforum