Calculus Book: Simplify Trigonometric Expression, Trig Identities Help

[Tickle]

I find it difficult to proceed. I think I have a gap in my knowledge. This question comes from my calculus book.
In which; is why I posted it under the Calculus subforum.

Can someone explain the steps for solving this problem. I have been trying to solve it now for over 48 hours.

Question with answer from back of book:
( 1 + sinu ) / ( cosu ) + ( cosu ) / ( 1 + sinu ) = 2secu

I need help in finding the solution to why it is 2secu ?

Kind Regards,
-Tickle

 

[Otis]

Can someone explain the steps for [simplifying] this problem.

(1 + sinu ) / ( cosu ) + ( cosu ) / ( 1 + sinu )

Hi. For this exercise, you need to remember some basic trigonometric identities, like these two:

sin(u)^2 + cos(u)^2 = 1

1/cos(u) = sec(u)

The first step toward simplifying your expression is to add the two fractions algebraically. (You will need to multiply out the resulting numerator, but leave the resulting denominator in factored form.)

For review, here's a symbolic guide for adding two fractions.

A/B + C/D = (A*D + B*C)/(B*D)

After you multiply out the resulting numerator, look for a way to simplify it, by using a basic identity.

Next, try other simplifications, like factoring and then looking for a cancellation with the denominator.

If you get that far, using another basic identity yields 2*sec(u).

But, the first step is to add the fractions. If you get stuck, please post what you tried, and we can go from there.

Cheers :cool:

 

[Tickle]

I've tried a lot of things with it, I just want to be walked through solution please thanks. I have a chalk board here at home a very big one.
But I've spent countless hours rearranging the equation and I can't get it to equal 2secu for the life of me.

So if someone can write out the steps in sequence that would nice.

Kind Regards,
-Tickle

 

[Deleted member 4993]

I've tried a lot of things with it, I just want to be walked through solution please thanks. I have a chalk board here at home a very big one.
But I've spent countless hours rearranging the equation and I can't get it to equal 2secu for the life of me.

So if someone can write out the steps in sequence that would nice.

Kind Regards,
-Tickle

It requires a bit of algebra - so - can you simplify the following (do the addition):

\(\displaystyle \displaystyle{\frac{a+b}{c} \ + \ \frac{c}{a+b} \ = \ ??}\)

 

[Otis]

I think I have a gap in my knowledge.

How long has it been, since you passed your algebra and trigonometry courses?

If it's been a long time, then it's natural to forget some basics.

This question comes from my calculus book.
In which; is why I posted it under the Calculus subforum.

The question comes from the first chapter, in your calculus book. The entire chapter is review, but you don't need to understand it all -- just the parts covering basic lessons from previous courses.

I've tried a lot of things with it

If you're willing to show some of that here, I'm willing to help you. I am not willing to post a step-by-step answer to your homework.

The first step is to add the fractions. If you know how to add two algebraic fractions, then you ought to understand why the numerator in their sum is

(1 + sin)*(1 + sin) + cos*cos

Multiply out this expression using the FOIL algorithm, from algebra.

Please show me what you get. (I'm thinking that i'ts already on your blackboard somewhere.)

But, if you don't remember FOIL, then let me know; I'm willing to post a step-by-step solution to a similar exercise, if you think it might help.

Thank you

 

[Steven G]

I find it difficult to proceed. I think I have a gap in my knowledge. This question comes from my calculus book.
In which; is why I posted it under the Calculus subforum.

Can someone explain the steps for solving this problem. I have been trying to solve it now for over 48 hours.

Question with answer from back of book:
( 1 + sinu ) / ( cosu ) + ( cosu ) / ( 1 + sinu ) = 2secu

I need help in finding the solution to why it is 2secu ?

Kind Regards,
-Tickle

[( 1 + sinu ) / ( cosu )]*[(1+sinu)/(1+sinu] + [( cosu ) / ( 1 + sinu)][cosu/cosu]
See what you can do with this.
EDIT: The equal sign should be +

 

[Tickle]

I'm actually 31 years old. This is by no way my homework. I'm not even in school.
I'm just learning calculus on my own. Mostly for curiosity than anything.

This doesn't make any sense:
[( 1 + sinu ) / ( cosu )]*[(1+sinu)/(1+sinu] = [( cosu ) / ( 1 + sinu)][cosu/cosu]

Shouldn't it be:
( 1 + sinu / cosu ) * ( 1 + sinu / 1 + sinu ) = 1 + sin^2u / cosu( 1 + sinu )

not:
[( cosu ) / ( 1 + sinu)][cosu/cosu]

 

[soroban]

Hello, Tickle!

Be careful . . .
. . \(\displaystyle (1+ \sin x)^2\) is NOT equal to \(\displaystyle 1+ \sin^2\!x.\)

Prove: \(\displaystyle \dfrac{1 + \sin x}{\cos x} + \dfrac{\cos x}{1 + \sin x}\;=\; 2\sec x\)

Multiply the first fraction by \(\displaystyle \dfrac{1+\sin x}{1+\sin x}.\)
. . . .. Multiply the second fraction by \(\displaystyle \dfrac{\cos x}{\cos x}.\)

\(\displaystyle \displaystyle \frac{1+\sin x}{\cos x}\cdot \color{red}{\frac{1+\sin x}{1+\sin x}} + \frac{\cos x}{1+\sin x}\cdot \color{red}{\frac{\cos x}{\cos x}} \)

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

. . \(\displaystyle \displaystyle =\;\frac{(1 + \sin x)^2 + \cos^2\!x}{\cos x(1+\sin x)} \;=\; \frac{1+ 2\sin x + \overbrace{\sin^2\!x + \cos^2\!x}^{\text{This is 1}}}{\cos x(1 + \sin x)} \)

. . \(\displaystyle \displaystyle =\;\frac{2 + 2\sin x}{\cos x(1+\sin x)} \;=\;\frac{2(1+\sin x)}{\cos(1+\sin x)} \;=\;\frac{2}{\cos x} \;=\;2\sec x\)

 

[Steven G]

I'm actually 31 years old. This is by no way my homework. I'm not even in school.
I'm just learning calculus on my own. Mostly for curiosity than anything.

This doesn't make any sense:
[( 1 + sinu ) / ( cosu )]*[(1+sinu)/(1+sinu] = [( cosu ) / ( 1 + sinu)][cosu/cosu]

Shouldn't it be:
( 1 + sinu / cosu ) * ( 1 + sinu / 1 + sinu ) = 1 + sin^2u / cosu( 1 + sinu )

not:
[( cosu ) / ( 1 + sinu)][cosu/cosu]

Before adding two fractions the denominators 1st need to be made the same. The denominators are now both cosu(1+sinu). Just simplify the numerators and add them.

 

[mmm4444bot]

I'm actually 31 years old. This is by no way my homework. I'm not even in school.

There are many 31-year-old students enrolled in calculus courses.

There are also many students who come here to cheat. Your first two posts look similar to theirs' (i.e., no effort shown, just want answers).

Now that you have provided an explanation, we understand your situation. There is a link to the complete forum guidelines at the following page; please also read the forum rules.

http://www.freemathhelp.com/forum/threads/86071-Read-Before-Posting

I'm just learning calculus on my own. Mostly for curiosity than anything.

Good for you!

Do you also have access to algebra and trigonometry texts? If you try to learn algebra, trignonometry, and calculus simultaneously, a review chapter in a calculus text may not be sufficient for such a monumental undertaking.

There are many free algebra and trigonometry courses (self-paced with video lectures) on the Internet. Some of these provide printable texts. If you're interested, check out khanacademy.com and coursera.org.

Cheers :cool: