• Jan 1st 2006, 02:52 PM
lizzy20
I am trying to verify some trig identities but I am running into some problems.
If you could help me at all with any problem, I would really appreciate it, thanks!

http://i7.photobucket.com/albums/y29...y/67d13cf4.jpg
• Jan 1st 2006, 04:35 PM
ticbol
Quote:

Originally Posted by lizzy20
I am trying to verify some trig identities but I am running into some problems.
If you could help me at all with any problem, I would really appreciate it, thanks!

http://i7.photobucket.com/albums/y29...y/67d13cf4.jpg

It so happens that all 4 problems are easy and "short" to solve/verify, otherwise you should have posted them separtately. Asking for many problems in only one posting is too much if the problems are "hard" or "long" to solve.
If one doesn't have time to solve all of the problems, he might ignore all of the problems outright, and you'd get no answer from him. Or if it is me, even if I have enough time, if I cannot solve all of the problems, I will ignore all of them and I will bypass your question.

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[cscX] /[cot +tanX] = cosX

This kind, we convert all functions into their sine or cosine equivalents.
Lefthand Side, (LFS) =
= [cscX] /[cotX +tanX]
= [1 /sinX] /[(cosX/sinX) +(sinX/cosX)]
= [1 /sinX] /[(cosX*cosX +sinX*sinX) /(sinX*cosX)]
= [1 /sinX] /[(cos2(X) +sin^2(x)) /(sinXcosX)]
Since sin^2(X) +cos^2(x) = 1, then,
= [1 /sinX] /[1 /(sinXcosX)]
= [1 /sinX]*[(sinXcosX) /1]
= cosX
= Righthand Side, (RHS)

Verified.

-----------------------------------
[1 /(sinY -1)] -[1 /(sinY +1)] = -2sec^2(Y)

LFS =
= [1 /(sinY -1)] -[1 /(sinY +1)]
Combine them into one fcraction only,
= [(sinY +1)*1 -(sinY -1)*1] /[(sinY -1)(sinY +1)]
= [sinY +1 -sinY +1] /[(sinY)^2 -1^2]
= [2] /[sin^2(Y) -1]
From sin^2(Y) +cos^2(Y) = 1, we can get sin^2(Y) = 1 -cos^2(Y), so,
= 2 /[(1 -cos^2(Y)) -1]
= 2 /[-cos^2(y)]
= -2sec^2(Y)
= RHS

Verified.

-----------------------------------
sin^3(X) -cos^3(X) = (1 +sinXcosX)(sinX -cosX)

The LHS is a difference of cubes, in the form a^3 -b^3. Factoring that,
a^3 -b^3
= (a-b)(a^2 +ab +b^2)
So, adopting that to the LHS,
= (sinX -cosX)[sin^2(X) +sinXcosX +cos^2(X)]
= (sinX -cosX)[sin^2(X) +cos^2(X) +sinXcosX]
= (sinX -cosX)[1 +sinXcosX]
= (1 +sinXcosX)(sinX -cosX)
= RHS

Verified.

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[tan(theta)] +[(cos(theta)) /(1 +sin(theta))] = sec(theta)

Let X = theta, for less confusion, and easy typing.

[tanX] +[cosX /(1 +sinX)] = secX

LHS =
= tanX +[cosX /(1 +sinX)]
= [sinX /cosX] +[cosX /(1 +sinX)]
= [(1 +sinX)*sinX +cosX*cosX] /[cosX*(1 +sinX)]
= [sinX +sin^2(X) +cos^2(X)] /[cosX*(1 +sinX)]
= [sinX +1] /[cosX*(1 +sinX)]
The (sinX +1) or (1 +sinX) is called out,
= [1] /[cosX]
= secX
= RHS

Verified.

--------------------------
I assume you know how to combine two fractions into one fraction only.
a/b +c/d
The common denominator is b*d,
= (d*a +b*c) /(b*d).
• Jan 1st 2006, 05:35 PM
ThePerfectHacker
I am not trying to argue with ticbol he is correct. I just want to redo them in LaTeX form because I think it is easier to follow and I also need practice with my LaTeX code. I might do them out of order or not even all of them.
Problem 3 verify that:
$sin^3x-cos^3x=(1+sinxcosx)(sinx-cosx)$
by the difference of 2 cubes identity:
$a^3-b^3=(a-b)(a^2+ab+b^2)$
we have:
$sin^3x-cos^3x=(sinx-cosx)(sin^2x+sinxcosx+cos^2x)$
but: (pythagorean identities)
$sin^2x+cos^2x=1$
thus:
$sin^3x-cos^3x=(sinx-cosx)(1+sinxcosx)$
Verified
• Jan 1st 2006, 05:46 PM
ThePerfectHacker
Problem 1
Verify that
$\frac{cscx}{cotx+tanx}=cosx$
First work with the denominator of the fraction:
$cotx+tanx=\frac{cosx}{sinx}+\frac{sinx}{cosx}$
Adding fraction the common denominator is:
$sinxcosx$
thus:
$\frac{sin^2x+cos^2x}{sinxcosx}$
but:
$sin^2x+cos^2x=1$
thus:
$\frac{1}{sinxcosx}$
is the simplified denominator.

Second the numerator of the fraction is:
$cscx=\frac{1}{sinx}$

Here is the most important step remember when you divide fractions you flip the one on the right and then multiply. Thus, the left hand side is a mess but it simplifies:
$\frac{\frac{1}{sinx}}{\frac{1}{sinxcosx}}$
Thus, you have:
$\frac{1}{sinx}\frac{sinxcosx}{1}$
Cancel $sinx$ and you are left with
$cosx$
Verified
• Jan 1st 2006, 06:01 PM
ThePerfectHacker
Problem 2
Verified that
$\frac{1}{sinx-1}-\frac{1}{sinx+1}=-2sec^2x$
Remember you are adding fractions, the common denominator here is the product of the two fractions. Thus the common denominator is:
$(sinx-1)(sinx+1)=sin^2x+sinx-sinx-1=sin^2x-1$ by FOIL
But $1-sin^2x=cos^2x$ By Pythagorean Identity thus:
$sin^2x-1=-cos^2x$
Now change to common denominator:
$\frac{sinx+1}{(sinx-1)(sinx+1)}-\frac{sinx-1}{(sinx-1)(sinx+1)}$
Bring them together because they have a common denominator:
$\frac{(sinx+1)-(sinx-1)}{(sinx-1)(sinx+1)}$
Open the parantheses ( :eek: watch the negative sign)
$\frac{sinx+1-sinx+1}{(sinx-1)(sinx+1)}$
But in the very beginning I explained that the common denominator
$(sinx-1)(sinx+1)$ is $-cos^2x$
But:
$\frac{1}{cosx}=secx$
Thus:
$\frac{2}{-cos^2x}=-2sec^2x$
Verified.
• Jan 2nd 2006, 12:05 PM
ThePerfectHacker
Problem 4)
Verify that
$tanx+\frac{cos}{1+sinx}=secx$
Express $tanx$ as
$\frac{sinx}{cosx}+\frac{cosx}{1+sinx}$
You are adding fractions again, thus find the common denominator which in this case this the product of the denominatos thus $cosx(1+sinx)$
Change fractions to common denominator
$\frac{sinx(1+sinx)}{cos(1+sinx)}+\frac{cosxcosx}{c osx(1+sinx)}$
Adding fraction because they have a common denominator and remembering that $cosxcosx=cos^2x$ you have:
$\frac{sinx(1+sinx)+cos^2x}{cosx(1+sinx)}$
Open parantheses:
$\frac{sinx+sin^2x+cos^2x}{cox(1+sinx)}$
But $cos^2x+sin^2x=1$
Thus:
$\frac{sinx+1}{cox(1+sinx)}$
Now cancel $1+sinx$
$\frac{1}{cosx}$
But $secx=\frac{1}{cosx}$
Verified.