need help for test monday!!!

sam

Apr 2006
4
0
can't do trig! stuff like sec^2(x)-1/sec^2(x).
Help me!
more to come
 

CaptainBlack

MHF Hall of Fame
Nov 2005
14,975
5,273
erehwon
sam said:
can't do trig! stuff like sec^2(x)-1/sec^2(x).
Help me!
more to come
This is not a question, what do you want to do with

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

RonL
 

sam

Apr 2006
4
0
trig

that's the way the prob is in my textbook
 

sam

Apr 2006
4
0
it says to simplify the trigonometric expression
 

CaptainBlack

MHF Hall of Fame
Nov 2005
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5,273
erehwon
sam said:
can't do trig! stuff like sec^2(x)-1/sec^2(x).
Help me!
more to come
Simplify:


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

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

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

If that is simpler.

RonL
 
Last edited:
Dec 2005
117
0
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?
 
Last edited:
Dec 2005
117
0
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.
 
Last edited:
Dec 2005
117
0
by the way, there are no unique answers, there are many different correct answers. CaptainBlack's or my answers are just fine.
 

CaptainBlack

MHF Hall of Fame
Nov 2005
14,975
5,273
erehwon
sam said:
also 1+cos(y)/1+sec(y)
This is ambiguous, it could mean:

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

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

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