Trigonometric Equation

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IstekOlympiadTeam

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IstekOlympiadTeam

#1 h Jan 2, 2016, 2:05 PM • 3 Y

Y by nicky-glass, Adventure10, Mango247

Prove that $\sin^2\alpha-\sin^2\beta=\sin(\alpha+\beta)\cdot \sin(\alpha-\beta)$

This post has been edited 1 time. Last edited by IstekOlympiadTeam, Jan 2, 2016, 4:02 PM

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#2 h Jan 2, 2016, 2:59 PM • 2 Y

Y by nicky-glass, Adventure10

Proof

Note: In general, for trig proofs, when you're asked to prove that $A = B$ :

1) If $A$ is complicated but $B$ is simple, start with the expression $A$ and do "legal moves" to it to turn it into $B$ .
2) If $B$ is complicated but $A$ is simple, start with $B$ and transform it to $A$ .
3) If they're both complicated, work with one end and simplify as much as you can to a third, simpler expression - say, $C$ . You've shown that $A=C$ . Then, take $B$ , and transform it to $C$ . Since $A=C$ and $B=C$ , you now have $A=B$ .

What you shouldn't do is start with the statement you want to prove and use a non-existent symbol like $\stackrel{?}{=}$ until you get a statement like $0=0$ that's true. That might be a way to do scratch work, but you have to write your proof going the opposite way - either starting with a true statement and getting to what you want, or starting with one expression on a side of the equation and transforming it to another expression.

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IstekOlympiadTeam

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#3 h Jan 2, 2016, 3:27 PM • 2 Y

Y by nicky-glass, Adventure10

talkinaway wrote:

Proof

Note: In general, for trig proofs, when you're asked to prove that $A = B$ :

1) If $A$ is complicated but $B$ is simple, start with the expression $A$ and do "legal moves" to it to turn it into $B$ .
2) If $B$ is complicated but $A$ is simple, start with $B$ and transform it to $A$ .
3) If they're both complicated, work with one end and simplify as much as you can to a third, simpler expression - say, $C$ . You've shown that $A=C$ . Then, take $B$ , and transform it to $C$ . Since $A=C$ and $B=C$ , you now have $A=B$ .

What you shouldn't do is start with the statement you want to prove and use a non-existent symbol like $\stackrel{?}{=}$ until you get a statement like $0=0$ that's true. That might be a way to do scratch work, but you have to write your proof going the opposite way - either starting with a true statement and getting to what you want, or starting with one expression on a side of the equation and transforming it to another expression.

Thanks for your solution. I solved it before posting .I found this when i was solving Balkan Mathematical Olympiad Geometry (P2) problem at the end i had to prove this which is easy and good problem for this forum.

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#4 h Jan 2, 2016, 4:19 PM • 3 Y

Y by nicky-glass, Adventure10, Mango247

Another Solution

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#6 h Jan 28, 2016, 6:59 PM • 2 Y

Y by Adventure10, Mango247

$\underline{\underline{2\left(\sin^2x-\sin^2y\right)}}=2\sin^2x-2\sin^2y=(1-\cos 2x)-(1-\cos 2y)=$

$\cos 2y-\cos 2x=$ $\underline{\underline{2\sin(x+y)\sin (x-y)}}\implies$ $\boxed{\sin^2x-\sin^2y=\sin(x+y)\sin (x-y)}\ .$

This post has been edited 5 times. Last edited by Virgil Nicula, Jan 28, 2016, 7:04 PM

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