As the title says, I'm having trouble with understanding the proof for the law of tangents, provided in wikipedia here is the: proof and here is picture of the line where I'm lost I can't understand how this: $$ \frac{2*\sin(\frac{\alpha-\beta}{2})*\cos(\frac{\alpha+\beta}{2})} { 2*\sin(\frac{\alpha+\beta}{2})*\cos(\frac{\alpha-\beta}{2})} $$ got to this: $$ \frac{\sin(\frac{\alpha-\beta}{2})}{ \cos(\frac{\alpha-\beta}{2})} \div \frac{\sin(\frac{\alpha+\beta}{2})}{\cos(\frac{\alpha+\beta}{2})} $$
Asked
Active
Viewed 88 times
-1
-
3Possible duplicate of Proof of dividing fractional expressions – Martin R Dec 20 '15 at 20:07
-
1This is because $;\dfrac{AB}{CD}=\dfrac{\cfrac AC}{\cfrac DB}$. – Bernard Dec 20 '15 at 20:09
2 Answers
0
Notice in the final statement, the first fraction be thought of as the numerator, and the second fraction can be thought of as the denominator. You could then multiply both the numerator and denominator by $\cos(\frac{\alpha+\beta}{2})$ to be left with $$\frac{\sin(\frac{\alpha-\beta}{2})\cos(\frac{\alpha+\beta}{2})}{\cos(\frac{\alpha-\beta}{2})} \div \sin(\frac{\alpha+\beta}{2})$$ Then we just multiply the numerator and denominator by $$\frac{1}{\sin(\frac{\alpha+\beta}{2})}$$ and we end up with what we started with. (Of course you could just add back in the twos that cancel out to be left with exactly what it was).
Will Fisher
- 5,112
0
From:
$$
\frac{2*\sin(\frac{\alpha-\beta}{2})*\cos(\frac{\alpha+\beta}{2})} { 2*\sin(\frac{\alpha+\beta}{2})*\cos(\frac{\alpha-\beta}{2})}
$$
i:
$$
\frac{\sin(\frac{\alpha-\beta}{2})*\cos(\frac{\alpha+\beta}{2})} {\sin(\frac{\alpha+\beta}{2})*\cos(\frac{\alpha-\beta}{2})}
$$
ii:
$$
\frac{\sin(\frac{\alpha-\beta}{2})}{\cos(\frac{\alpha-\beta}{2})}*\frac{\cos(\frac{\alpha+\beta}{2})}{\sin(\frac{\alpha+\beta}{2})}
$$
Finish:
$$
\frac{\sin(\frac{\alpha-\beta}{2})}{ \cos(\frac{\alpha-\beta}{2})} \div \frac{\sin(\frac{\alpha+\beta}{2})}{\cos(\frac{\alpha+\beta}{2})}
$$
Drunk Cynic
- 145