Results 1 to 6 of 6

Math Help - Trig Identities Prob

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    4

    Trig Identities Prob

    'Solve sin x + 4 cos x = 3'
    Can anyone help me with the above problem? Trigonometric identities and formulae confuse me.

    ------------------------------------
    I have tried doing this:
    sin x = 3 - 4 cos x

    divide all by cos x
    then tan x = 3/cos x - 4
    rearrange so 4 = 3/cos x - tan x
    which is same as 4 = 3 sec x - tan x

    square both sides
    so 16 = 9 sec^2 x - tan^2 x
    this is 16 = 9(tan^2x + 1) - tan^2 x
    and 16 = 9tan^2x - tan^2x + 9

    rearrange so 7 = 8 tan^2 x
    7/8 = tan^2 x
    so tan x= sqrt(7/8)
    and x = arctan sqrt(7/8)

    Is this correct?
    If anyone could provide any help or solution to this problem, I'll be so grateful.

    Thank you!
    Herman
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    My favorite:

    \sin(x) + 4\cos(x) = \sqrt{17}\sin(x+atan(4))

    It's a lot easier with only one function.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by TKHunny View Post
    My favorite:

    \sin(x) + 4\cos(x) = \sqrt{17}\sin(x+atan(4))

    It's a lot easier with only one function.
    why would you even know such an identity?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2008
    Posts
    4
    so... is my answer right or wrong?
    if wrong, what's right? lol
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,864
    Thanks
    744
    Hello, Herman!

    You started off well, but you squared incorrectly . . .


    Solve: . \sin x + 4\cos x \:= \:3

    I have tried doing this: . \sin x \:= \:3 - 4\cos x

    Divide all by \cos x\!:\;\;\;\tan x \:= \:3\sec x - 4 \quad\Rightarrow\quad 4 \:=\:3\sec x - \tan x

    Square both sides: . 16 \:= \:9\sec^2\!x - \tan^2\!x . . . . no
    You should have: . 16 \;=\;9\sec^2\!x - 6\sec x\tan x + \tan^2\!x

    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    Quote Originally Posted by Jhevon View Post
    why would you even know such an identity?
    It's a philosophy, "Derive, Don't Memorize."
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig prob
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: April 27th 2009, 11:55 AM
  2. Replies: 1
    Last Post: April 26th 2009, 01:15 PM
  3. Trig prob
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 21st 2009, 01:15 PM
  4. Trig prob
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: March 25th 2009, 07:50 AM
  5. Triangle trig prob.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 11th 2008, 10:24 PM

Search Tags


/mathhelpforum @mathhelpforum