Help me with precalc homework

[duragezic]

I'm generally pretty good with it but this analytical trigonometry or whatever it is called, I'm stoopid in it! I just can't seem to grasp it, although it is fairly simple. And if the teacher helps me through it it seems so easy but when I sit down and try to do it I just don't know how to start or what to do next.

So if you guys could do these problems and show the steps I'll understand it better and finish my hw.

Oh yeah you're suppose to "verify the identities" which means to kinda like substitute and sh*t on the left side so it equals the right side.

29: (sec^2)(csc^2) = sec^2 + csc^2

32: tan^2 - sin^2 = (tan^2)(sin^2)

34: sec + csc sin + cos
----------- = ------------
sec - csc sin - cos

36: 1 1
---------- + ----------- = 2csc^2
1-cos 1+cos

Yah... um if anyone knows what's up but doesn't know the identities (like 1=sin^2+cos^2) then I'll post those.

Thanks!

edit: Man these things turned out like complete crap... (in formatting)

 

[TheVrolok]

gimme a few but I'll get back to you.. gotta get the brain working again

 

[Martin]

I'm too lazy to do them ( I have my own calculus homework to do ), but a good rule of thumb is to simplify everything to sin and cos only and then look for ways to combine/simplify them using trig identities.....

 

[Soccer55]

I don't think there are many people here that will do these problems for you because it's your homework. But I'm sure there will be people that will be more than happy to help you get in the right direction. Here are some of my suggestions:

29) use the definitions of csc and sec and get a common denominator on the right side, then simplify

32) use the definition of tan^2 to get an expression on the left that you can simplify and use the distributive law and trig identities on

34) use the definitions of csc and sec to simplify the left side

36) get a common denominator on the left side and simplify, then apply one of the trig identities


-Tom

 

[duragezic]

Um I tried all those and I can't get very far...

29)
sin^2 cos^2
---------- + -----------
(sincos)^2 (sincos)^2

Don't know what to do from there.

32)
sin^2
------- - sin^2
cos^2

Distributive? How?

34)
sin + cos
-------- ------------
cossin cossin
---------------------------------------
sin - cos
--------- ------------
cossin cossin

??

36) Don't know how to get common denominator there.

 

[Soccer55]

<< 29)
sin^2 cos^2
---------- + -----------
(sincos)^2 (sincos)^2 >>



Try writing the denominator as sin^2 * cos^2 instead of (sincos)^2. Then, what is sin^2 + cos^2? You should be able to do it from there



<< 32)
sin^2
------- - sin^2
cos^2

Distributive? How? >>



a(b+c) = ab + ac, so you can factor out a sin^2. Then it should be a simple trig identity to finish it off.



<< 34)
sin + cos
-------- ------------
cossin cossin
---------------------------------------
sin - cos
--------- ------------
cossin cossin >>



The denominator of the fraction on the top cancels with the denominator of the fraction on the bottom, then you're done.



<< 36: 1 1
---------- + ----------- = 2csc^2
1-cos 1+cos
>>



Try (1-cos)*(1+cos) as your common denominator for the fractions on the left side

-Tom

 

[duragezic]

Still don't get it. God this sucks got a test tomorrow and I can't do a single one.

29: so i'm left with
sin^2/1 & cos^2/1. It'd work if those were flipped but they're not....

32: Still don't get how it's distributive... this one is like (a/b) - c...

34: Ok.

36: left with
1
--------------
(1-cos)(1+cos)
+
1
-------------
(1+cos)(1-cos)

 

[Soccer55]

For 29) sin^2 + cos^2 should be your numerator and (sin^2)*(cos^2) ishould be your denominator. So when you said it would work if it was flipped, you were right and it should be flipped

For 32) let a = sin^2, b = 1/(cos^2), and c = 1. Then by substituting, sin^2/cos^2 - sin^2 ----> ab-ac = a*(b-c).

For 36) Do the multiplication in the denominator and it kind of falls into place after that

-Tom

 

[duragezic]

Well still have 29 (ok so they need to be flipped but I don't know how) and 32 I still have no idea. But whatever I don't care. Time for bed. Thanks a lot for the help though. You get a 10.

 

[notfred]

TRIGONOMETRY IS EVIL!

Calculus is much better. I HATE trig. To hell with trig. substitution.....

 

[20_MuleTeam_Borax]

Calculus is much better until you have to do calculus w/ trig functions. You can't escape them :disgust:

 

[goodoptics]

<<Oh yeah you're suppose to "verify the identities" which means to kinda like substitute and sh*t on the left side so it equals the right side. >>

or substitute and sh*t on the right side so it equals the right side.

 

[goodoptics]

29. (sec^2)(csc^2) = 1/((cos^2)(sin^2)) = (sin^2+cos^2)/((cos^2)(sin^2)) = 1/cos^2 + 1/sin^2 = sec^2 + csc^2

 

[goodoptics]

32. tan^2-sin^2 = sin^2/cos^2 - sin^2 = (sin^2-(cos^2)(sin^2))/cos^2= sin^2 * (1-cos^2)/cos^2 = sin^2* sin^2/cos^2 = (sin^2)(tan^2) = (tan^2)(sin^2).

 

[Unsickle]

dude, i'm about to graduate with a double engineering major and I don't even know that stuff. haha.

 

[goodoptics]

Hi, Unsickle,

What ME and MSE courses are you currently taking at Berkeley?

 

[goodoptics]

34. (sec+csc)/(sec-csc) = (1/cos+1/sin)/(1/cos-1/sin) = ((sin+cos)/(cos*sin))/((sin-cos)/(cos*sin)) = ((sin+cos)/(cos*sin)) * (cos*sin)/(sin-cos) = (sin+cos)/(sin-cos).

 

[goodoptics]

36. 1/(1+cos) + 1/(1-cos) = (1+1)/((1+cos)(1-cos)) = 2/(1-cos^2)= 2/sin^2 = 2 csc^2.

 

[cdan]

Ahh Trigonometry. Or as I call it trippin' on a tree. I'm in that now with a teacher that doesn't like talking. Next semester I'll have Anal Geometry. We just finished that chapter and BTW I'm not doing them for you.

 

[goodoptics]

<<Anal Geometry>>

LOL