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[Kwokuen]

We've been tinkering with identities in my summer classes, and they're a bitch to work with. If anyone could lend me some help on verifying them, I'd appreciate it.

The first one is: (tan u + cot u)(cos u + sin u) = csc u + sec u

I believe that starting with the left side is the best option, so converting tan and cot to sines and cosines is the first step.

(sin u / cos u + cos u / sin u) (cos u + sin u)

At this point however, I'm lost. Do I square the left side, then factor it, or do I cross-multiply?

Some other fun-filled problems:

2. cot B - tan B / sin B + cos B = csc B - sec B

3. 1 / csc y - cot y = csc y + cot y

 

[sefskillz]

(sin u / cos u + cos u / sin u) (cos u + sin u)

start with the left part and get a common denominator of ((cos u)(sin u)) this gives you:

sin^2 u + cos^2 u / 2 (cos u sin u) which = 1 / 2 (cos u * sin u) ..

doh, i just realized i made a mistake. ill keep playin with it.. maybe what i did will help you

 

[Dilbert]

We've been tinkering with identities in my summer classes, and they're a bitch to work with. If anyone could lend me some help on verifying them, I'd appreciate it.

The first one is: (tan u + cot u)(cos u + sin u) = csc u + sec u

I believe that starting with the left side is the best option, so converting tan and cot to sines and cosines is the first step.

(sin u / cos u + cos u / sin u) (cos u + sin u)

At this point however, I'm lost. Do I square the left side, then factor it, or do I cross-multiply?

The next step is to multiply the left side using FOIL (First, Outer, Inner, Last) since it's the product of binomials:

tan u cos u + tan u sin u + cot u cos u + cot u sin u = csc u + sec u

Using the fact that tan u = (sin u / cos u) and cot u = (cos u / sin u):

sin u + (sin²u / cos u) + (cos²u / sin u) + cos u = csc u + sec u

Looks ugly, right? We can proceed in a couple of ways, but a neat trick is to rewrite the sin u and cos u terms so that they have a denominator:

(sin²u / sin u) + (sin²u / cos u) + (cos²u / sin u) + (cos²u / cos u) = csc u + sec u

(Notice that all we've done is multiply them by unity to make our lives easier.) Now, group the terms on the left side by common denominator:

(sin²u / sin u) + (cos²u / sin u) + (sin²u / cos u) + (cos²u / cos u) = csc u + sec u

Regrouping the numerators:

(sin²u + cos²u / sin u) + (sin²u + cos²u / cos u) = csc u + sec u

Using the identity sin²u + cos²u = 1 (definition of the unit circle):

(1 / sin u) + (1 / cos u) = csc u + sec u

csc u + sec u = csc u + sec u

...and the identity is proved.

Hope this helped. Have fun with #2 and #3...I need to get to bed.

 

[Cooper]

You can answer all three of those pretty quickly if you remember

cos^2 u + sin^2 u = 1

and

(a + b) (a - b) = a^2 - b^2

 

[fart]

h to the o to the u to the izzy

 

[nitewulf]

jinx is my hero, i'd have nether thought out the multiplying by unity trick.

for #2,

(cotB - tanB)/(sinB + cosB) = cscB - secB

just the left side:

[cosB/sinB - sinB/cosB]/[sinB+cosB]
note:cot and tan has been converted
[(cos^2B - sin^2B)/(sinBcosB)]/[sinB+cosB]
note:numerator has been combined
[cos^2B - sin^2B]/[sinBcosB] * 1/[sinB+cosB]
note:denominator re-written as the reciprocal, such that the whole term can be represented as a product
[cos^2B - sin^2B]/[sinBcosB][sinB+cosB]
note: multiplication performed, keeping the factors
[(cosB + sinB)(cosB - sinB)]/[sinBcosB][sinB+cosB]
note: (a + b) (a - b) = a^2 - b^2, thanx cooper
(cosB - sinB)/sinBcosB
note: term cancelled out
cosB/sinBcosB - sinB/sinBcosB
note: each term of the numerator has been divided by the denominator
1/sinB - 1/cosB
note: term cancelled
cscB - secB
note: voila.

 

[Escape Goat]

Trig help at 1:40 in the morning?

 

[AniHawk]

And the headaches come rushing back to me.

 

[Dilbert]

jinx is my hero, i'd have nether thought out the multiplying by unity trick.

Hah...thanks for the props, but it isn't necessary to solve the problem. It's just a little more elegant. The brute force way of solving ALL of these identities is to change both sides into sines and cosines and play algebra games until they are identical. When you do that, though, it can sometimes be hard to see what's REALLY going on with the identity and get lost somewhere.

By the way, I think that #3 (1 / csc y - cot y = csc y + cot y) has a typo in it. Consider the first-quadrant value of pi/4 for y -- when you look at the "identity," you find that the left side is negative and the right half is positive! (1 / csc y (also known as sin y) = (SQRT(2))/2, csc y = SQRT(2), and cot y = 1...you can do the math.)

So, don't waste any more time on #3 until you find out where the mistake is.

 

[Cooper]

No typo in #3. He just wasn't very helpful with his parenthesis.

1 / (csc y - cot y)

not

(1 / csc y ) - cot y

Clue is the same problem in #2, but a little more obvious there.

 

[Diablos]

I wish the world was kind of like the matrix. Then I could upload jinx's math knowledge to my head in like two minutes.

 

[Kwokuen]

Shite. My bad on 2 & 3. Cooper nailed it. On another note, thanks for all the help guys, I appreciate it.

 

[bachikarn]

#3

1 /(csc y - cot y) = csc y + cot y
1/(1/sin y - cos y/siny) = csc y + cot y
1/( (1 - cosy)/sin y ) = cscy + cot y
sin y / (1 - cosy) = csc y + cot y
sin y / (1 - cosy) * (1 + cosy)/(1+cosy) = csc y + cot y
( siny + sin y * cos y ) / (1 - cos^2 y) = csc y + cot y
( siny + sin y * cos y ) / sin^2 y = csc y + cot y
1/ sin y + cos y / sin y = csc y + cot y
csc y + cot y = csc y + cot y

 

[Dilbert]

No typo in #3. He just wasn't very helpful with his parenthesis.

Ahhhhh...<LIGHT CLICKS ON>

Now that you're responding to these threads, Cooper, I'm going to retire!