can't do trig! stuff like sec^2(x)-1/sec^2(x).

Help me!

more to come

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- Apr 29th 2006, 09:32 AMsamneed help for test monday!!!
can't do trig! stuff like sec^2(x)-1/sec^2(x).

Help me!

more to come - Apr 29th 2006, 09:52 AMCaptainBlackQuote:

Originally Posted by**sam**

$\displaystyle

\sec^2(x)-1/\sec^2(x)?

$

RonL - Apr 29th 2006, 10:45 AMsamtrig
that's the way the prob is in my textbook

- Apr 29th 2006, 10:54 AMsam
it says to simplify the trigonometric expression

- Apr 29th 2006, 10:54 AMsam
also 1+cos(y)/1+sec(y)

- Apr 29th 2006, 11:22 AMCaptainBlackQuote:

Originally Posted by**sam**

$\displaystyle

\sec^2(x)-1/\sec^2(x)=1/\cos^2(x)-\cos^2(x)=\frac{1-\cos^4(x)}{\cos^2(x)}

$

$\displaystyle

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

$

$\displaystyle

=\tan^2(x)(1+\cos^2(x))

$

If that is simpler.

RonL - Apr 29th 2006, 02:41 PMc_323_h
is it $\displaystyle \frac{\sec^2x-1}{\sec^2x}$ or

$\displaystyle \sec^2x-\frac{1}{sec^2x}$

this is what i got for the first one

$\displaystyle \frac{\sec^2x-1}{\sec^2x}$

$\displaystyle \frac{\sec^2x-1}{1} \bullet \frac{1}{sec^2x}$

since

$\displaystyle \frac{1}{secx}=cosx$ you can square both sides to get

$\displaystyle \frac{1}{sec^2x}=cos^2x$

substitute this in to get

$\displaystyle (sec^2x-1)(cos^2x)$=

$\displaystyle (tan^2x)(cos^2x)$

can someone tell me if that dot is the correct multiplication operator on the second line? - Apr 29th 2006, 03:08 PMc_323_h
and for the second:

$\displaystyle

\sec^2x-\frac{1}{sec^2x}

$

since

$\displaystyle \frac{1}{\sec}=\cos$ we have,

$\displaystyle \sec^2x-\cos^2x$

there are trig identities you must understand and know how to manipulate them. - Apr 29th 2006, 05:57 PMc_323_h
by the way, there are no unique answers, there are many different correct answers. CaptainBlack's or my answers are just fine.

- Apr 29th 2006, 10:41 PMCaptainBlackQuote:

Originally Posted by**sam**

$\displaystyle

(1)\ \ \frac{1+\cos(y)}{1+\sec(y)}

$

$\displaystyle

(2)\ \ 1+\frac{\cos(y)}{1+\sec(y)}

$

$\displaystyle

(3)\ \ 1+\frac{\cos(y)}{1}+\sec(y)

$ (not too likely)

Now the first of these seems the most likely, but you are leaving us guessing.

If you are writing this using plain ASCII then write it as:

(1+cos(y))/(1+sec(y))

The rule is: when something is ambiguous when written in plain ASCII then

add brackets until only the intended meaning is possible.

RonL