Precalc help

[Oberon.Smite]

Hi guys, I have a Precalc quiz tomorrow and I kind of understand the concept behind cofunction trig identities, but these kind of problems are the type I don't get.

 

Anyways here's the problem

 

Edit:

 

if 

 

 

find 

 

[AnonymousGuy]

what's the question?  you've put a statement of fact...sin is cos shifted right by pi/2 radians.

[Oberon.Smite]

what's the question?  you've put a statement of fact...sin is cos shifted right by pi/2 radians.

Oh crap, edited. Sorry.

[TopDollar]

It's been a very long time since I have done any sort of proof but I would think (substituted theta for "x" for clarity)

 

cos(x - pi/2) = -3.56 since

 

cos(x - pi/2) = cos(-1*(pi/2 - x)) 

 

cos(-1*(pi/2 - x)) = cos(pi/2 - x) due to even/odd identity

 

sin(x) = cos(pi/2 - x) due to co-function identity therefore

 

cos(pi/2 - x) = -3.56

[Oberon.Smite]

It's been a very long time since I have done any sort of proof but I would think (substituted theta for "x" for clarity)

 

cos(x - pi/2) = -3.56 since

 

cos(x - pi/2) = cos(-1*(pi/2 - x)) 

 

cos(-1*(pi/2 - x)) = cos(pi/2 - x) due to even/odd identity

 

sin(x) = cos(pi/2 - x) due to co-function identity therefore

 

cos(pi/2 - x) = -3.56

Thank you, and it's only even/odd when it's cos right?

[TopDollar]

sin(-x) = -sin(x)

cos(-x) = cos(x)

tan(-x) = -tan(x)

csc(-x) = csc(x)

sec(-x) = sec(x)

cot(-x) = cot(x)

Solving these problems is just sort of manipulating the functions and applying identities.

[Oberon.Smite]

sin(-x) = -sin(x)

cos(-x) = cos(x)

tan(-x) = -tan(x)

csc(-x) = csc(x)

sec(-x) = sec(x)

cot(-x) = cot(x)

Solving these problems is just sort of manipulating the functions and applying identities.

Got you, thanks for the help.