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March 1st, 2012, 03:51 PM   #1
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Trig Identities

I am so lost.

How do I solve these three problems, please use descriptive explanation. I missed the day we did these in class.

1. sin x + cos x cot x = csc x


2. sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1 (I realize this is some kind of perfect square, but where do I began with this? I'm confused)

3. csc^4 x - cot^4 x = cot^2 x + csc^2 x

Thank you!
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March 1st, 2012, 04:05 PM   #2
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Re: Trig Identities

All three of them are based on using sin^2 x + cos^2 x = 1. For example - the second problem is simply squaring this expression. The others will become clearer if you convert everything to sin or cos, using cot = cos/sin and csc = 1/sin.

Note that problem 3 can be simplified by observing that the right side is a factor in the left side.
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March 1st, 2012, 04:30 PM   #3
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Re: Trig Identities

Quote:
Originally Posted by mathman
All three of them are based on using sin^2 x + cos^2 x = 1. For example - the second problem is simply squaring this expression. The others will become clearer if you convert everything to sin or cos, using cot = cos/sin and csc = 1/sin.

Note that problem 3 can be simplified by observing that the right side is a factor in the left side.
Ok so for 1. are you telling me that it should look like this:

sin x + cos x cos/sin = 1/sin ? (I changed cot= cos/sin and csc x = 1/sin) Did I do this correctly?
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March 1st, 2012, 04:36 PM   #4
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Re: Trig Identities

Yes (although don't forget the arguments for the trig functions...cot(x) = cos(x)/sin(x) etc.), now combine the terms with a common denominator, and the resulting numerator will be a familiar expression.
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March 1st, 2012, 04:49 PM   #5
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Re: Trig Identities

Oh my gosh, I'm getting ready to bang my head against my keyboard. So this is what I got. What am I doing wrong?

Original problem: sin x + cos x cot x = csc x

I changed the problem to: sin x + cos x cosx/sin x = 1/sinx

Then I got a common denominator (which was obviously sin x) : sinx(sinx)/sinx + cosx(sinx)/sinx cosx/sinx = 1/sinx

Then I got: sin^2 x/sinx + cosx sin x cos x / sinx = 1/sinx

How do I get rid of the mess in the middle? am I changing cosx to 1/sec? I am so confused.
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March 1st, 2012, 04:56 PM   #6
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Re: Trig Identities

Quote:
Originally Posted by swish
...
Then I got a common denominator ( which was obviously sin x) : sinx(sinx)/sinx +[color=#BF0000] cosx(sinx)/sinx cosx/sinx[/color] = 1/sinx
...
No.

The second term already had sin(x) in the denominator.

You should have sin^2(x)/sin(x) + cos^2(x)/sin(x)
on the left hand side.

i.e.

[sin^2(x) + cos^2(x)]/sin(x)

or
1/sin(x)
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March 1st, 2012, 06:18 PM   #7
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Re: Trig Identities

1.) Another approach:



2.)

Another approach:





3.)

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