Pre-Calc Homework Help

[math_fortress]

Simplify the given expression:
2) (sec^2 x)(csc x)/(csc^2 x)(sec x)

[arildno]

What have you tried?

[math_fortress]

well...i've done this:

(tan^2 x + 1)(csc x)/(cot^2 x + 1)(sec x)

but...i don't know if i'm going in the right direction, for my teacher is horrible, and i don't know where to go from here if i am going in the right direction

any help would be greatly appreciated

[arildno]

OK: Let's make the EASIEST cancellations first:
If you look at the expression like this:
\frac{sec^{2}(x)csc(x)}{sec(x)csc^{2}(x)}

isn't there a couple of cancellations which immediately spring to your mind?

[math_fortress]

so is it (sec x)/(csc x) ???

[arildno]

so is it (sec x)/(csc x) ???
Precisely!
Now, knowing the relation between sec and cos and csc and sin, can you simplify even further?

[math_fortress]

well, since sin/cos = tan....then would sec/csc = 1/tan ?

That's all i can think of

[arildno]

No, you have:
\frac{\frac{1}{\cos(x)}}{\frac{1}{\sin(x)}}=\frac{ 1}{\cos(x)}\frac{1}{\frac{1}{\sin(x)}}=\frac{\sin( x)}{\cos(x)}=tan(x)

[math_fortress]

alright...thanks, i'm starting to get it a little better....

[math_fortress]

Wait...this baffles me...

26.) Find the exact value of sin 5pi/12

Is there any way to do this logically w/ a calculator or anything?

[cepheid]

Of course there is! (hopefully you meant without a calculator) That's why the angle is given to you in radians, as a rational multiple of \pi .

Draw the unit circle: what coordinate points do certain angles represent? \pi, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2} etc.

[math_fortress]

but 5pi/12 isn't on my unit circle...the one's you listed are though...i just don't get how exactly you can find 5pi/12 with information of pi/2, etc...

[math_fortress]

would 1/4 make sense for the answer since i got the sin of 5pi/6 to equal 1/2?

[Sirus]

Try to do this using the trig identity
\cos{2a}=1-2\sin^{2}{a}

[math_fortress]

What? So you're saying 1/4 isn't right then?
Do I even need to use trig identities for this type of question?

(I'm not arguing with that identity..i'm just very confused :bugeye:)

[Sirus]

1/4 is incorrect. Sometimes identities are necessary.

[cepheid]

Yeah, sorry if I gave you the wrong idea. 1/4 is incorrect. You are making the assumption that if I halve the angle, I halve the sine. You can see why that would only be true for a linear relationship right? (which sine is not). I think Sirus has the right technique, since the trig identity involves a term with twice the angle and another with just the angle itself. We know how to work with \frac{\pi}{6} , and multiples of it, so find the cosine of the angle \frac{\5pi}{6} and work from there.

[kreil]

Another way is to convert the radians to degrees and work from there:

sin5pi/12=sin5(180)/12=sin75=sin(30+45)

now you can just use the formula for the sum of angles:
sin (a+b) = cos(b) sin(a) + sin(b) cos(a)

[Leaping antalope]

use the sum formulas of trig:
sin (a+b)=sin(a) cos(b)+cos(a) sin (b)

now, sin 5pi/12=sin (2pi/12 + 3pi/12), so the exact value of sin 5pi/12 is?

[math_fortress]

alright....root 2/4 + root 6/4


Thanks for all help