Here are some more problems that need checking from my other thread I made.

1)

2)

)

For this problem, does it matter if the right side's answer is the same value on the left side but flipped?

3)

Doing the right side:

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- March 24th 2011, 01:52 PMMajorJohnsonProving Trig identities Problems that need correction
Here are some more problems that need checking from my other thread I made.

1)

2)

)

For this problem, does it matter if the right side's answer is the same value on the left side but flipped?

3)

Doing the right side:

- March 24th 2011, 02:06 PMdwsmith
- March 24th 2011, 02:07 PMpickslides
What are you trying to do with these?

for the first one, I get,

- March 26th 2011, 06:42 AMMajorJohnson
My objective is to find the other side.

I also want to know if my method for this problem is legal:

Cross multipling ( sin x / cos x) * (cos x / sin x) would give me one.

Here i canceled the ones out and got ' cot ^ 2 x + tan ^ 2 x '.

Canceling the ones again would give me csc^ 2 x + sec^2x. - March 26th 2011, 08:22 AMQuacky
Edit: I misunderstood your approach, I think it works, other than the fact that you cancel the ones at the end. When you reach the stage , you can just use identities to show that it is the same as . I'll leave my method here, but after rereading, I think your proof is otherwise fine.

Generally, you'd start with the side which is easiest to simplify. Here, that's the right hand side. Start by putting everything over a common denominator:

which is the left hand side.

You also haven't finished 2) - you can simplify the fraction further using the identity - March 27th 2011, 09:04 AMMajorJohnson
Thank you. In solving these problems in general, can there be more than one possible answer?

- March 27th 2011, 01:07 PMQuacky
- March 27th 2011, 01:20 PMpickslides
- March 27th 2011, 04:16 PMMajorJohnson
Okay, but even if i have to find the other side of an identity, such as with some of the other ones i had posted in my other thread, there can only be one answer right?

- March 31st 2011, 02:59 PMMajorJohnson
How would i go about solving this problem?

So far I have:

cosPI/2 - cos x 1/ cos x - March 31st 2011, 03:14 PMtopsquark
- April 1st 2011, 03:52 AMMajorJohnson
Okay that helped me a little bit, I also have another question unrelated to this topic but with dealing with Solving Right triangles. When going about solving the outer sides, how do I know whether or not I should take the given outside side and times it by whatever degree given inside the triangle, as opposed to dividing the outside side by the inside? (Example

- April 1st 2011, 04:20 AMmasters