Precal help

[danny_l_2005]

I did bad in my test and our professor was kind enough to let us correct the answers. I am NOT asking you to give me the answer. I am only asking to teach me how it works.

Ok.

I have this question : Verify the following: cscx - sinx= cosxcotx

Can someone make an example and show me how to do it. Please don't do the one I put there just show me the steps what I need to do to get the answer. Thanks

 

[snowborder714]

Does this help out at all?
Verifying Trig Identities? - Yahoo! Answers

 

[danny_l_2005]

Helps a bit. Thanks. I don't understand this part. Maybe I do. I'm a bit confused.

"Make a common denominator (to make one fraction):
= 1/sin(x) - sin²(x)/sin(x)"

How did they get that from 1/sin(x) - sin(x) but most importantly this part "sin²(x)/sin(x)"

 

[danny_l_2005]

Nevermind I understood it. sinx is the same thing as sin²(x)/sin(x) . Thanks

 

[snowborder714]

Glad you figured it out because I was of no more help. Google helped me with that link

 

[danny_l_2005]

I'm lost again. What is another way to say 2sec²x?

I have this question 2sec²x-2sec²xsin²x-sin²x-cos²x=1 (i don't want my thread deleted so please use another example to help me to understand)

I know that sec²x is the same as secx * secx or the same as 1/cosx * 1/cosx but where does the 2 before the sec²x fit in?

I also understand that sin²x+cos²x=1 .

 

[95redturboawd]

Umm... just a guess but maybe you could break up 2sec^2x to 2 * sec^2x... so then you could do 2 * ((1/cosx) * (1/cosx)). Not sure though, I hate calculus.

 

[danny_l_2005]

I figured it out earlier. sec²x = tan²x+1 and secxsinx=tanx

so its 2(tan²x+1) -2tan²x-1=1
2tan²x+2-2tan²x-1=1
2-1=1
1=1


Does anyone know Trig Substitution? or how to use Sum and Difference Formula?

 

[VETTE_50_TH]

cos(A+B) = cos A cos B − sin A sin B

sin(A+B) = sin A cos B + cos A sin B

James

 

[danny_l_2005]

Thanks James but I got the formulas.So how are they used?

 

[VETTE_50_TH]

Say you want to find the angle of 60*. Which is a non standard angle that you dont know what the sin (60). But you know what sin of 30 is. You use the second formula to find the sin of 60*.

What is this class for? College, high school? Just wait until you take Calc 1 or 2 and do integration, derivatives and partial fraction decomposition. FUN FUN FUN.

James

 

[danny_l_2005]

This is for College, all other chapters are pretty easy for now, It is just this Chapter. I need to do a problem of which is Cos(165*). Can you show me an example such as like Cos(105*)

do I start like this : =cos(60) cos(45)-sin(60)sin(45) ?

What do I need to do after that? Do I get those values from the right triangle with angles 45* and the right triangle with angles 60* and 30*?

 

[95redturboawd]

Haha just a guess again but couldn't you do cos(120) + cos(45)?

 

[VETTE_50_TH]

Haha just a guess again but couldn't you do cos(120) + cos(45)?

No.

Cos (A+B) /= cos(A) + cos(B)

James

 

[danny_l_2005]

I turn my work already but isn't cos(165) =cos(120) cos(45)-sin(120)sin(45)
cos(165) = (-1/2)(sqrt(2)/2)-(sqrt(3)/2)(sqrt(2)/2)
cos(165) = (sqrt(2)/4) - ((sqrt(2))(sqrt(3)))/4
cos(165) = ((sqrt(2)(1+sqrt(3)))/4

 

[95redturboawd]

Beats me. Check it on your calculator. Just make sure you're in the right mode.

 

[VETTE_50_TH]

I turn my work already but isn't cos(165) =cos(120) cos(45)-sin(120)sin(45)
cos(165) = (-1/2)(sqrt(2)/2)-(sqrt(3)/2)(sqrt(2)/2)
cos(165) = (sqrt(2)/4) - ((sqrt(2))(sqrt(3)))/4
cos(165) = ((sqrt(2)(1+sqrt(3)))/4

Yes. That looks good.

James