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I'm trying to do my PreCalculus homework but I have forgotten how to find the conjugate. The problem is $(\sin C) / (1+\cos X)$. The problem says to multiply the numerator and denominator by the conjugate of the denominator. Any help?

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2 Answers 2

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The conjugate of $a+b$ is $a-b$. So, you have $\dfrac{\sin(C)}{1+\cos(X)}\cdot\dfrac{1-\cos(X)}{1-\cos(X)}=\dfrac{\sin(C)(1-\cos(X))}{1-\cos^2(X)}=\dfrac{\sin(C)(1-\cos(X))}{\sin^2(X)}$.

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  • $\begingroup$ To add to what Sanath said, we multiply by the conjugate because $(a+b)(a-b) = a^2-b^2$, and this will lead to an interesting property involving trigonometric functions. $\endgroup$
    – 2012ssohn
    Commented Mar 11, 2014 at 1:52
  • $\begingroup$ What about Sin C multiplied by 1-cosX $\endgroup$
    – Trevor
    Commented Mar 11, 2014 at 1:53
  • $\begingroup$ I need to multiply the numerator also. $\endgroup$
    – Trevor
    Commented Mar 11, 2014 at 1:54
  • $\begingroup$ @Trevor See my edited answer. $\endgroup$
    – user122283
    Commented Mar 11, 2014 at 2:00
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by multiplying both sides by the conjugate ($1 - \cos x$) we have $\frac{\sin x (1 - cos x)}{1 - \cos^2 x}$. By using the pythagorean identity, we have $\frac{\sin C (1 - cos x)}{\sin^2x}$. Sanath, you wrote the problem wrong; 1 + cos x, not 1 - cos x.

For a better definition of what a conjugate is, see this link: http://www.mathsisfun.com/algebra/conjugate.html

It defines the conjugate as "where you change the sign in the middle of two terms, like

3x+1 to 3x-1". This is pretty legit.

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  • $\begingroup$ Thanks. In your answer, you didn't really explain the what the conjugate of $1+\cos(x)$ is. $\endgroup$
    – user122283
    Commented Mar 11, 2014 at 1:59

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