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Trig question

  1. Apr 8, 2007 #1
    How can you solve for X, an angle, if you get an equation like this?

    5=2sinX + cosX

    I couldn't think of any trig identities that can solve this, even though this may be extremely easy.

    Can someone show me? Thanks
  2. jcsd
  3. Apr 8, 2007 #2
    Well, what you cay say immediately is that the solution is going to be complex. What are the maximum values of sin and cos on the reals? :wink:
  4. Apr 9, 2007 #3
    srry about that...i just randomly put a few numbers down...

    If the equation is like:: 2=3sinX+cosX

    How would you solve that?
  5. Apr 9, 2007 #4
  6. Apr 10, 2007 #5


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    Staff Emeritus
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    That's a bit complicated but you could do this: Let y= cos(X). Since [itex]sin(x)= \sqrt{1- cos^2(x)}[/itex] we have [itex]2= 3\sqrt{1- y^2}+ y[/itex]. Rewrite that as [itex]3\sqrt{1- y^2}= 2- y[/itex] and square both sides: [itex]9(1- y^2)= 4- 4y+ y^2[/itex] or [itex]9- 9y^2= 4- 4y+ y^2[/itex] so [itex]10y^2- 4y- 5= 0[/itex]. Solve that using the quadratic formula to find y= cos(X) and take the arcsine to find X.
  7. Apr 10, 2007 #6
    Substitute [tex]sin(x)= \sqrt{1- cos^2(x)}[/tex] and solve the quadratic for [tex]cos(x)[/tex]. Then take the arcsine of you two solutions. Basically what HallsofIvy said.
  8. Apr 10, 2007 #7
    or try to substitute :

    sin(x) = 2tan(x/2)/1+tan^2(x/2)
    cos(x)= (1-tan^2(x/2))/1+tan^2(x/2)

    it might solve the problem !
  9. Apr 10, 2007 #8
    Why not get sin on one side and cos on the other side. Square both sides (possibly introducing extraneous roots) and substituting for either sin^2 or cos^2 (using sin^2+cos^2=1). This would result in a quadratic equation. Check the solutions.
  10. Apr 10, 2007 #9


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    Gold Member

    You could write it in the form of [tex]2=Acos(x-\theta)[/tex]
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