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I'm doing a refresher on high school math and wanted to prove the $\sin(\theta + \phi) = \sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi)$ identity but I can't get it right. I drew two right triangles $ACD$ and $ABC$ on on top of another like can be seen in this picture: setup Let's assume that $\overline{AD}=1$ for simplicity. Also $DF$ is perpendicular to $AB$. Then $\sin(\theta+\phi)=\overline{DF} = \overline{FE}+\overline{DE}$. Notice that $\angle{DEC}=\angle{AEF}= \pi/2 - \theta$. Thus $\overline{DE}=\overline{DC}/\sin(\angle{DEC})=\sin(\phi)/\cos(\theta)$.

Note that $\cos(\phi)=\overline{AC}$ and that $\overline{EC}=\overline{DE}\cos(\angle{DEC})=\overline{DE}\sin(\theta)$. From this we get $\overline{AE}=\overline{AC}-\overline{EC}=\cos(\phi) - \overline{DE}\sin(\theta) = \cos(\phi)-\sin(\phi)\tan(\theta)$. From this $\overline{FE}$ can be calculated as $\overline{AE}\sin(\theta)=\cos(\phi)\sin(\theta)-\sin(\phi)\tan(\theta)\sin(\theta)$. So finally, $$\sin(\theta+\phi)=\overline{DF}=\overline{FE}+\overline{DE}=\cos(\phi)\sin(\theta)-\sin(\phi)\tan(\theta)\sin(\theta)+\sin(\phi)/\cos(\theta),$$ which is not the correct relation.

Even though I double checked the steps, I can't find an error in my work. I'd be thankful for pointing out what went wrong here.

AJB
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2 Answers2

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While it's true that $DF = DE + EF,$ that fact is not really helpful because neither $DE$ nor $EF$ is equal to either term in the sum-of-angles formula.

Rather than arbitrarily dividing $\overline{DF}$ in this way just because $E$ is the first intersection point you notice while making a diagram, try looking for the terms in the diagram instead. You could start with $\cos(\phi)\sin(\theta).$

You already have found that $AC = \cos(\phi)$; now observe that $\overline{AC}$ is the hypotenuse of a triangle with angle $\theta.$ Using that triangle, you can show that $BC = \cos(\phi)\sin(\theta).$

How you just need to find where $\sin(\phi)\cos(\theta)$ is hiding in the figure and find a way to show that the combined lengths of segments $\overline{AC}$ and the segment of length $\sin(\phi)\cos(\theta)$ are the length of $\overline{AD}.$

For example, you could cut $\overline{DF}$ at $G$ (in the figure below) instead of at $E.$ Just let $\overline{CG}$ be parallel to $\overline{BF}.$

enter image description here

There is a beautiful graphical proof of both the sine and cosine angle-sum formulas in this answer using your triangles $\triangle ABC$ and $\triangle ACD.$ It works by circumscribing a rectangle around your figure rather than cluttering the interior of the figure with lines that cross each other.

In my opinion, the advantage of an improved diagram over algebraic recombination of the trig terms (as you could do in order to salvage the formulas you got from $\overline{DE}$ and $\overline{EF}$) is that in a well-constructed and well-labeled figure you can practically just read the desired final formula directly off the figure.

David K
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  • I guess you're right. Using your diagram with the extra GC segment, we can produce a proof without referring to any algebraic manipulations, which qualifies as a "proof without words" I think – AJB Jun 19 '24 at 06:07
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Simplifying the expression gives the desired result: $$ \cos(\phi)\sin(\theta)+\frac{\sin(\phi)}{\cos(\theta)}(-\sin^2(\theta)+1) = \cos(\phi)\sin(\theta)+\sin(\phi)\cos(\theta). $$ [using the fact that $\sin^2(\theta)+\cos^2(\theta)=1$]

AJB
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