Simpiifying trigonometric functions...please help

**aikenfan** · November 24th 2007, 06:34 AM

I have attempted these problems, but I am stuck on them...Could anyone please assist me?....

**topsquark** · November 24th 2007, 07:12 AM

Originally Posted by aikenfan

I have attempted these problems, but I am stuck on them...Could anyone please assist me?....

Please be more specific in the future. You have already solved 5 -7.

$cos^2(x) - sin^2(x) = 2cos^2(x) - 1$

This is practically already an identity. Both of these expressions are equal to $cos(2x)$ . However, if you please:
$cos^2(x) - sin^2(x)$

$= cos^2(x) - (1 - cos^2(x))$

$= cos^2(x) - 1 + cos^2(x))$

$= 2cos^2(x) - 1$

-Dan

**aikenfan** · November 24th 2007, 05:52 PM

I'm sorry, I should have said...I did numbers 5,6, and 7 the best I could, but i needed someone to look over them...I do not think they are right because number 5 is not done...I was stuck on number 5....It is supposed to equal 1 - sin^2x but i only get 1 and I don't know why...

**Macleef** · November 24th 2007, 06:01 PM

Originally Posted by aikenfan

I'm sorry, I should have said...I did numbers 5,6, and 7 the best I could, but i needed someone to look over them...I do not think they are right because number 5 is not done...I was stuck on number 5....It is supposed to equal 1 - sin^2x but i only get 1 and I don't know why...

LS:
$= cos^2x - sin^2x$
$= 1 - sin^2x - sin^2x$
$= 1 - sin^2x$

Is this what you mean?

----

Edit:
I didn't see there was an additional identity in the equation...

So it would be:
LS:
$= 1 - 2sin^2x + 2sin^2x$

How did I get $1 - 2sin^2x$ from $cos^2x - sin^2x$ ? From the double angle identities...

**Macleef** · November 24th 2007, 06:34 PM

For number 6:

RS:
$= 1 - \frac {cos^2x}{sin^2x} - 1$

$= \frac {sin^2x - cos^2x - sin^2x}{sin^2x}$

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

$= \frac {sin^2x - 1 + sin^2x - sin^2x}{sin^2x}$

$= \frac {2sin^2x - 1 - sin^2x}{sin^2x}$

$= \frac {2sin^2x}{sin^2x} - \frac {1}{sin^2x} - \frac {sin^2x}{sin^2x}$

$= 2 - csc^2x - 1$

Therefore, RS = LS

**Macleef** · November 24th 2007, 06:43 PM

For number 7:

RS:
$= \frac {1}{cos^2x} + 4$

$= \frac {1 + 4 cos^2x}{cos^2x}$

$= \frac {cos^2x + sin^2x + 4cos^2x}{cos^2x}$

$= \frac {5cos^2x + sin^2x}{cos^2x}$

$= \frac {5cos^2x}{cos^2x} + \frac {sin^2x}{cos^2x}$

$= 5 + tan^2x$

$= tan^2x + 5$

Therefore, RS = LS

**Macleef** · November 24th 2007, 06:45 PM

For number 8:

RS:
$= cos^2x - sin^2x$

Therefore, RS = LS <<< again, double angle identities

**aikenfan** · November 25th 2007, 05:57 AM

For number 5 and number 8, I am a little confused about how you did them...I am not familiar with double angle identities...I know quotient id, pythag id, and reciprocal id

**topsquark** · November 25th 2007, 06:12 AM

Originally Posted by aikenfan

I'm sorry, I should have said...I did numbers 5,6, and 7 the best I could, but i needed someone to look over them...I do not think they are right because number 5 is not done...I was stuck on number 5....It is supposed to equal 1 - sin^2x but i only get 1 and I don't know why...

The point of "solving" an identity is that both sides of the equation can be shown to be equal to one another. Now, the typical method is to take one side of the equation and manipulate it such that it becomes identical in form to the other side. But there is nothing wrong with taking standard solution steps to do the same thing. For example, in #5 you have:
$cos^2(x) - sin^2(x) = 1 - 2sin^2(x)$

$cos^2(x) - sin^2(x) + 2sin^2(x) = 1 - 2sin^2(x) + 2sin^2(x)$

$cos^2(x) + sin^2(x) = 1$

The only "missing" step is to use the Pythagorean identity to say
$1 = 1$

Both sides are the same, hence you have shown that this is an identity.

Numbers 6 and 7 work the same way. Using the steps that you have written you are essentially done.

-Dan

**aikenfan** · November 25th 2007, 09:38 AM

This is what I have got...

I'm sorry, but I am still a bit confused about something...number 8, i get 2 - csc^2x - 1
Isn't it supposed to be equal to 2 - csc^2x?

And also, number 3, i get cosx secx
it is supposed to be equal to cscx secx

I appreciate everyone being very patient with me...this is very new to me and I am having a hard time understanding it.

**topsquark** · November 25th 2007, 09:47 AM

Originally Posted by aikenfan

This is what I have got...

I'm sorry, but I am still a bit confused about something...number 8, i get 2 - csc^2x - 1
Isn't it supposed to be equal to 2 - csc^2x?

And also, number 3, i get cosx secx
it is supposed to be equal to cscx secx

I appreciate everyone being very patient with me...this is very new to me and I am having a hard time understanding it.

I am rather confused by what you are doing in the second line of #8.
$2 - csc^2(x) = 1 - cot^2(x)$

Then you have beneath it:
$1 - \frac{cos^2(x)}{sin^2(x)} - 1$

Which side of the equation is this supposed to be? It is equal to neither.

For #3 I have a similar problem making out what the third line is.
$\frac{csc^2(x)}{cot(x)} = csc(x)sec(x)$

$\frac{csc(x)~csc(x)}{\frac{cos(x)}{sin(x)}}$

Then the cosine factor disappears and the "1/s" is supposed to be 1/sin(x)? It's a bit of a mess.

Some suggestions:
1) Immediately convert everything into sines and cosines. It works for me anyways.

2) NEVER make simplifying abbreviations such as "sin" or even "s" for "sin(x)", etc. The clearer you write it the easier it is to find mistakes while you are checking your work.

-Dan

**aikenfan** · November 25th 2007, 01:08 PM

Now I am just really confused...Number 8 was already done in this thread by someone, but I need help understanding these...I am having a hard time understanding....I don't know if what I have got is correct for these problems...I appreciate your patience...On number 3, the RS and LS aren't =

**Macleef** · November 25th 2007, 01:43 PM

For number 8:

I have no idea where you got the extra "-sinx" from...

The equation you want to solve:

$2 - csc^2x = 1 -cost^2x$

I worked with the right side:

$= 1 - \frac {cos^2x}{sin^2x}$

Find common denominator
$= \frac {sin^2x - cos^2x}{sin^2x}$

Put everything in one term, in this case, sinx
$= \frac {sin^2x - (1 - sin^2x)}{sin^2x}$

Use the FOIL or distribution method
$= \frac {sin^2x - 1 + sin^2x}{sin^2x}$

Separate each numerator
$= \frac {sin^2x}{sin^2x} - \frac {1}{sin^2x} + \frac {sin^2x}{sin^2x}$

Simplify
$= 1 - csc^2x + 1$

Simplify
$= 2 - csc^2x$

Therefore, right side equals left side

**Macleef** · November 25th 2007, 02:08 PM

For number 3:

I worked with the left side:

$= \frac {csc^2x}{cotx}$

Put all terms to sinx and/or cosx
$= \frac {\frac{1}{sin^2x}}{\frac{cosx}{sinx}}$

Simplify the denominator by cross multiplying
$= \frac {1}{sin^2x} \times \frac {sinx}{cosx}$

Factor (factor the sinx when cross multiplying)
$= \frac {1}{sinxcosx}$

Simplify
$= \frac {1}{sinx} \times \frac {1}{cosx}$

$= cscxsecx$

Therefore, left side equals the right side

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