solving Natural Logs for x and a Trig identity question.

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March 18th 2009, 01:19 AM
squrrilslayer

solving Natural Logs for x and a Trig identity question.

Hey.

Code:

2Ln2 - Ln(x-1) = ln(2x)

That is what im having trouble sorting out. I know its something to do with the quadratic formula but its just not clicking in my head.

also, i cant seem to finish this off:

Code:

Prove that (1/cot^2 θ) - (1/csc^2 θ) = (sin^4 θ)/(cos^2 θ)

This is what i have so far:

Code:

-> (1/(1/tan^2 θ)) - (1/(1/sin^2 θ)) -> 1/(tan^2 θ - sin^2 θ) -> (1/((cos^2 θ)/(sin^2 θ))) - (1/sin^2 θ) -> ((sin^2 θ)/(cos^2 θ)) + ((sin^2 θ)/ 1)

i am stuck there... unless of course the actually DO add together to make

Code:

(sin^4 θ)/(cos^2 θ)

Any help would be greatly appreciated!
March 18th 2009, 01:43 AM
ADARSH

Quote:

Originally Posted by squrrilslayer

Hey.

Code:

2Ln2 - Ln(x-1) = ln(2x)

That is what im having trouble sorting out. I know its something to do with the quadratic formula but its just not clicking in my head.

b ln(a) = ln(a^b)

Thus

ln(4) - ln(x-1) = ln(2x)

But
ln(a) - ln(b) =ln(a/b)

Thus

ln(4/(x-1)) = ln(2x)

Thus

4/x-1 = 2x

Hence
2x^2 - 2x = 4

x^2 - x - 2 = 0

x^2 - 2x + x -2 = 0

x(x-2) + (x-2) = 0

(x+1)(x-2)=0

x = -1 or x= 2

But since ln(-ve) is not defined thus x = 2
March 18th 2009, 01:49 AM
ADARSH

Quote:

Originally Posted by squrrilslayer

Code:

Prove that (1/cot^2 θ) - (1/csc^2 θ) = (sin^4 θ)/(cos^2 θ)

This is what i have so far:

Code:

-> (1/(1/tan^2 θ)) - (1/(1/sin^2 θ)) -> tan^2θ - sin^2 θ -> sin^2 θ /cos^2 θ - sin^2 θ -> sin^2θ( 1/cos^2 θ -1) -> sin^2 θ ( sec^2 θ -1) -> sin^2θ (tan^2 θ) -> sin^2 θ x sin^2 θ/cos^2 θ -> sin^4 θ/cos^2 θ

I have corrected in code
March 18th 2009, 02:08 AM
squrrilslayer

wow thanks that was quick. I see now on the first one how you did that :D
as for the second, you got the answer so im reading through step by step analyzing what you did.

Thanks again!
[resolved]
March 18th 2009, 03:46 AM
squrrilslayer

sorry re on trig idents

i still dont understand what's been done:

-> (1/(1/tan^2 θ)) - (1/(1/sin^2 θ))
-> tan^2θ - sin^2 θ
-> sin^2 θ /cos^2 θ - sin^2 θ <<here
-> sin^2θ( 1/cos^2 θ -1) <<here
-> sin^2 θ ( sec^2 θ -1) <<here
-> sin^2θ (tan^2 θ) <<here
-> sin^2 θ x sin^2 θ/cos^2 θ
-> sin^4 θ/cos^2 θ

like how did

sin^2 θ /cos^2 θ - sin^2 θ

become:

sin^2θ( 1/cos^2 θ -1) ? (Surprised)

and so on so forth?

Thanks if someone could clear this up!
March 18th 2009, 03:55 AM
stapel

Line (3) is (effectively) "x/y - x".
Line (4) is (effectively) "x(1/y - 1)".
The author factored the "x" out front.

To go from Line (5) to Line (6), apply one of the trig identities they gave you. (Wink)
March 18th 2009, 04:10 AM
ADARSH

$tan^2(\theta) - sin^2(\theta) = \frac{sin^2(\theta)}{cos^2(\theta)} - sin^2 (\theta)$

$\frac{sin^2(\theta)}{cos^2(\theta)} - sin^2(\theta)$

$= sin^2(\theta)\times \frac{1}{cos^2(\theta)} - sin^2(\theta) \times 1$

Take $sin^2(\theta)$ common

$= sin^2(\theta) (\frac{1}{cos^2(\theta)} -1)$ -------(1)

Using the rule that

sec(x) = 1/cos(x)

Replace $1/cos^2(\theta)$ by $sec^2(\theta)$

= sin^2 θ ( sec^2 θ -1)......................(2)

Using the property that

sec^2 (x) = 1 + tan^2 (x)

Replace $sec^2(\theta)$ by $tan^2(\theta) - 1$

which gives

$= sin^2(\theta)\times (tan^2(\theta))$

$\frac{sin^2 \theta \times sin^2\theta}{cos^2 \theta}$

-----------------------------

Tell the step you fail to understand now