Results 1 to 7 of 7
Like Tree4Thanks
  • 1 Post By Prove It
  • 1 Post By Esteban
  • 1 Post By Esteban
  • 1 Post By Prove It

Math Help - Quick one about trig identities

  1. #1
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Quick one about trig identities

    Hey there MHF.

    I got a quick question with hopefully a quick answer.

    How does \frac{sin^2x}{cos^2x}+1=sec^2x ?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,408
    Thanks
    1294

    Re: Quick one about trig identities

    Quote Originally Posted by Paze View Post
    Hey there MHF.

    I got a quick question with hopefully a quick answer.

    How does \frac{sin^2x}{cos^2x}+1=sec^2x ?

    Thanks!
    \displaystyle \begin{align*} \sin^2{(x)} + \cos^2{(x)} &= 1 \\ \frac{\sin^2{(x)} + \cos^2{(x)}}{\cos^2{(x)}} &= \frac{1}{\cos^2{(x)}} \\ \frac{\sin^2{(x)}}{\cos^2{(x)}} + 1 &= \sec^2{(x)} \end{align*}
    Thanks from Paze
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie Esteban's Avatar
    Joined
    Feb 2013
    From
    Santiago, Chile
    Posts
    16
    Thanks
    3

    Re: Quick one about trig identities

    it's simple, look:

    to get the proof of this identity you have to choose a side of equation to start and get the other side. Using the left side we have:

    \dfrac{\sin^2 x}{\cos^2 x}+1=\dfrac{\sin^2 x +\cos^2 x}{\cos^2 x}=\dfrac{1}{\cos^2 x}=\sec^2 x

    I hope you understood, greetings
    Thanks from Paze
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Quick one about trig identities

    Thanks guys. I did not realize that sin(x)+cos(x)=1. Is there a quick proof for that one or? Is it only when they are squared maybe..?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie Esteban's Avatar
    Joined
    Feb 2013
    From
    Santiago, Chile
    Posts
    16
    Thanks
    3

    Re: Quick one about trig identities

    No! be carefully it's cos^2 x + \sin^2 x =1 pay attention with the squares! that's called "fundamental trigonometric's identity" and it's valid for all x \in \mathbb{R}

    there's a proof I think you can find it in wiki or any precalculus book
    Thanks from Paze
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,408
    Thanks
    1294

    Re: Quick one about trig identities

    Think of the unit circle. If you draw in a radius, it makes an angle \displaystyle \begin{align*} \theta \end{align*} with the positive x-axis. Then draw a segment from the end of the radius to the x-axis, you will have a right-angle triangle. The horizontal length is \displaystyle \begin{align*} \cos{(\theta)} \end{align*} and the vertical length is \displaystyle \begin{align*} \sin{(\theta)} \end{align*}. Since it's a right-angle-triangle, Pythagoras' Theorem holds, and so \displaystyle \begin{align*} \left[ \sin{(\theta)} \right] ^2 + \left[ \cos^2{(\theta)} \right]^2 = 1^2 \end{align*}, or \displaystyle \begin{align*} \sin^2{(\theta)} + \cos^2{(\theta)} = 1 \end{align*}. This is true no matter the value of \displaystyle \begin{align*} \theta \end{align*}.
    Thanks from Paze
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Quick one about trig identities

    Quote Originally Posted by Prove It View Post
    Think of the unit circle. If you draw in a radius, it makes an angle \displaystyle \begin{align*} \theta \end{align*} with the positive x-axis. Then draw a segment from the end of the radius to the x-axis, you will have a right-angle triangle. The horizontal length is \displaystyle \begin{align*} \cos{(\theta)} \end{align*} and the vertical length is \displaystyle \begin{align*} \sin{(\theta)} \end{align*}. Since it's a right-angle-triangle, Pythagoras' Theorem holds, and so \displaystyle \begin{align*} \left[ \sin{(\theta)} \right] ^2 + \left[ \cos^2{(\theta)} \right]^2 = 1^2 \end{align*}, or \displaystyle \begin{align*} \sin^2{(\theta)} + \cos^2{(\theta)} = 1 \end{align*}. This is true no matter the value of \displaystyle \begin{align*} \theta \end{align*}.
    Spot on as always! Thank you, Prove It!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving Trig Identities with Fundamental Identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 11th 2013, 02:41 AM
  2. Replies: 2
    Last Post: November 26th 2012, 03:57 PM
  3. trig identities II
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: November 17th 2009, 05:09 PM
  4. 3 quick question's on Identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 01:29 PM
  5. Quick and easy question trig identities
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: January 12th 2008, 03:37 PM

Search Tags


/mathhelpforum @mathhelpforum