Proving LS=RS in Trigonometry?

[tahayassen]

Homework Statement

http://img829.imageshack.us/img829/3413/daumequation13237287425.png

Prove that LS=RS.

Homework Equations

There are no relevant equations.

The Attempt at a Solution

http://img829.imageshack.us/img829/3413/daumequation13237287425.png

 

[Simon Bridge]

if LHS=RHS, then:

$$(\cos x-\sin y)(\cos x+\sin y)=(\cos y+\sin x)(\cos y-\sin x)$$

... right?
So why not expand it out and look for the usual identities?

 

[tahayassen]

if LHS=RHS, then:

$$(\cos x-\sin y)(\cos x+\sin y)=(\cos y+\sin x)(\cos x-\sin y)$$

... right?
So why not expand it out and look for the usual identities?

According to my teacher/school (I'm from Ontario), you're not allowed to work on both sides like that. You have to work on each side independent from the other. Why? I have no idea why... :(

edit: My teacher says you can't do that, because we don't know if LS=RS. We're trying to see if LS=RS by proving it.

 

[tahayassen]

http://img9.imageshack.us/img9/4701/daumequation13237312394.png

That's another identity I'm having trouble with. I have no idea how to change the exponents from 4 to 2. I'm totally lost with this one. I can't even think of a first step.

 

[eumyang]

Regarding your first post in this thread: do NOT multiply out the numerators. Leave them in factored form. So you have this:
$$\frac{(\cos x - \sin y)(\cos y + \sin x)}{cos^2 y - sin^2 x}$$
Now use the Pythagorean identities to rewrite BOTH terms in the denominator.

If you figure this out, you'll probably ask, how did I know NOT to multiply out the numerator? I noticed that when you multiplied top and bottom by (cos y + sin x), that is also the numerator of the RHS. You'll see what happens by leaving the numerator in factored form.

 

[eumyang]

http://img9.imageshack.us/img9/4701/daumequation13237312394.png

That's another identity I'm having trouble with. I have no idea how to change the exponents from 4 to 2. I'm totally lost with this one. I can't even think of a first step.

Note that on the LHS,
sin² x + cos⁴ x = sin² x + (cos² x)²
Use the Pythagorean identity for what's inside the parentheses and see what happens.

 

[tahayassen]

Regarding your first post in this thread: do NOT multiply out the numerators. Leave them in factored form. So you have this:
$$\frac{(\cos x - \sin y)(\cos y + \sin x)}{cos^2 y - sin^2 x}$$
Now use the Pythagorean identities to rewrite BOTH terms in the denominator.

If you figure this out, you'll probably ask, how did I know NOT to multiply out the numerator? I noticed that when you multiplied top and bottom by (cos y + sin x), that is also the numerator of the RHS. You'll see what happens by leaving the numerator in factored form.

Amazing! I feel so happy solving these. After that, I noticed that the denominator was a difference of square, and I canceled out the factors.

 

[tahayassen]

Note that on the LHS,
sin² x + cos⁴ x = sin² x + (cos² x)²
Use the Pythagorean identity for what's inside the parentheses and see what happens.

Haha. I got this identity too! Thanks!

The interesting part about this one is that when I wrote it down on paper, I couldn't solve it. It wasn't until I was inputting into my equation editor to generate the images did I realize what I missed earlier.

 

[Simon Bridge]

If LHS=RHS then I can propose (what I did) - and attempt to disprove it. If the proposition is false, then surely the equation will yield an inconsistency like 2=3? Of course, not all will reduce easily.

You could then start by putting both sides over a common denominator ... then you only have to prove the numerators are the same. This will be mathematically equivalent to the above approach.

On the other one - without looking: you can reduce the power by using the half-angles.

[ah - beat me to it]