Calculus

[Lalalori1]

I need help to solve this problem; don't even know where to start
what are the steps to solve sec y sin (pi/2-y)=1

 

[chrisr]

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

\(\displaystyle [sec(y)]sin(\frac{\pi}{2}-y)=1\)

\(\displaystyle \frac{sin(\frac{\pi}{2}-y)}{cos(y)}=1\)

Since

\(\displaystyle sin[\frac{\pi}{2}-x]=cosx\)

the problem is just an identity restated.

 

[Lalalori1]

thanks but could you also help with this one
1+tan^2x/csc^2x

 

[chrisr]

\(\displaystyle cosec(x)=\frac{1}{sinx}\)

\(\displaystyle cosec^{2}(x)=\frac{1}{sin^{2}x}\).

Is it

\(\displaystyle \frac{1+tan^{2}2x}{cosec^{2}2x}\)

or
\(\displaystyle 1+\frac{tan^{2}2x}{cosec^{2}2x}\ ?\)

 

[Lalalori1]

the answer is supposed to be tan^2x but im not sure how to get there

 

[Lalalori1]

i didnt get what you were asking but the problem is the first one

 

[chrisr]

Do you mean

\(\displaystyle \frac{1+tan^{2}2x}{cosec^{2}2x}\)

should equal

\(\displaystyle tan^{2}2x\ ?\)

 

[Lalalori1]

yes

 

[chrisr]

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

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

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

yes, it's true.

 

[Lalalori1]

im not quite following your steps on how you solved it

 

[chrisr]

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

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

 

[chrisr]

Are the trigonometric identities I'm using familiar?

Sorry, the first time i did it,
i was just checking quickly that the answer was \(\displaystyle tan^{2}2x.\)

Second time around was the derivation, using

\(\displaystyle cosec^{2}2x=\frac{1}{sin^{2}2x}\)

\(\displaystyle \frac{1}{cosec^{2}2x}=sin^{2}2x\)

Also

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

and

\(\displaystyle \frac{sin^{2}2x}{cos^{2}2x}=tan^{2}2x\)

 

[Lalalori1]

yea they are familiar but im just not sure when to use each identity

 

[Lalalori1]

Is there something I should look for in a problem to know which identity?

 

[chrisr]

You start by writing what cosec is.
Then you write what tan is.
This simplifies the expression.

Then you add the parts in brackets using

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

Now you can make a single fraction in brackets.

The terms in brackets sum to one, leaving you with the result.

You need to try writing it out, get familiar with toying with the identities,
they get much easier with practice.

 

[chrisr]

Is there something I should look for in a problem to know which identity?

\(\displaystyle Sin^{2}x+cos^{2}x=1\)

is the most fundamental identity after

\(\displaystyle \frac{sinx}{cosx}=tanx\)

The sec and cosec are just cosine and sine inverted.

The rest is adding fractions, getting used to working with other identities and so on.

This was a good example to start with.
You should practice with it until it's really clear.
Then you will be more fluent at more complex ones.

 

[Lalalori1]

Thanks so much. I appreciate your help.

 

[chelsearooo]

I REALLY Need help with my functions I have two questions i need help with.

1. Find (f/g)(x) f(x)=7x^2+3x g(x)=-8-x



2. f of g (x) f(x)=3x^2+2x g(x)=2x+1



PLEASE I REALLY NEED HELP!!!!!!!

Thank You

 

[galactus]

\(\displaystyle f(g(x)), \;\ f(x)=3x^{2}+2x, \;\ g(x)=2x+1\)

It would be a good idea to start your own thread. Folks are more apt to see it that way.

These are actually easy once you catch on to what is going on. This means 'f compose g'.

Just take g(x) and plug it in for x in f(x). Like so,

\(\displaystyle 3(2x+1)^{2}+2(2x+1)\). See there?. Now simplify.