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Here is the question: simplify the expression

$$\frac{\sin(f+g)+\sin(f-g)}{\cos(f+g)+\cos(f-g)}.$$

For this questions, are all of the addition and subtraction identities of sin and cos required? I am not certain how to approach this question. If someone could help it would be very appreciated!

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  • $\begingroup$ Yes, or an equivalent. $\endgroup$
    – Chappers
    Apr 5, 2015 at 23:50
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    $\begingroup$ Has the sum only one sine [cosine] in the numerator [denominator] or two? $\endgroup$
    – Karanko
    Apr 5, 2015 at 23:52
  • $\begingroup$ @Karanko Are you referring to the question itself or the answer? $\endgroup$
    – Ash
    Apr 5, 2015 at 23:55
  • $\begingroup$ @Elaqqad, that was what I was going for, I just an still trying to figure out the math notations on this site. I know how to make the expressions more easily detected, but I am not so good with making the fractions $\endgroup$
    – Ash
    Apr 5, 2015 at 23:58

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Okay

$\sin(f+g)=\sin f \cos g + \sin g \cos f $

$\sin(f-g)= \sin f \cos g - \sin g \cos f$

$\cos(f+g)= \cos f \cos g - \sin f \sin g$

$\cos (f-g)= \cos f \cos g + \sin f \sin g$

$$\sin(f+g)+\sin(f-g)=\sin f \cos g + \sin g \cos f+ \sin f \cos g - \sin g \cos f=2\sin f \cos g$$

$$\cos(f+g)+\cos(f-g)= \cos f \cos g - \sin g \sin f + \cos f \cos g + \sin f \sin g=2\cos f \cos g$$

So dividing gives $$\frac{2\sin f \cos g}{2 \cos f \cos g}=\frac{\sin f \cos g}{ \cos f \cos g}$$

$$=\frac{\sin f}{\cos f}= \tan f$$

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  • $\begingroup$ I had thought that this was going to be how to do this question. Pulling out each identity individually aand then combining them together. Thanks for the help, this really clarified my work. $\endgroup$
    – Ash
    Apr 6, 2015 at 0:05

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