development of trig identities problems?? 10 points best answer?

Question

im so confused with my homework !
How is it that
sec( - theta) = - csc theta ? ? ?
and how does
cot ( pi + pi/4) = 1??
i am really confused!! can anyone help??

Answer 1

sec( - theta) = - csc theta is not an identity.

cot ( pi + pi/4) = 1 is true because the period of cot is pi.
cot ( pi + pi/4) = cot(pi/4) = 1

Answer 2

sec(-theta) does not = - csc(theta)

cot pi/4 = 1
So cot(pi/4 +pi) also = 1 because they both have same reference angle of pi/4

Answer 3

you mean csc(-θ) = -csc(θ)

since csc(-θ) = 1/sin(-θ) = 1/-sin(θ) = -csc(θ) [ since sin is a odd

function ]

cot(pi + pi/4) = 1/tan(pi + pi/4)

=>1/(tan(pi) + tan(pi/4)) / (1 - tan(pi)*tan(pi/4))

=>1/(0 + 1)/(1 -0) = 1

Answer 4

Ninja girl
just remember
sin(x) -- up down down up
-- 0 -to - 90 - 180 - 270 - 360
cos(x) -- down down up up
-- 0 -to - 90 - 180 - 270 - 360

and tan =sin(x)/cos(x)
and cotan =cos(x)/sin(x) = 1/tan(x)
http://i55.photobucket.com/albums/g149/delta_bravo_1kilo/yahoo/cossin.jpg

u probably already know this but just to give u an idea

Answer 5

cot(Pi/4 + Pi)= 1/tan(Pi + Pi/4)=(1-tan(Pi)tan(Pi/4))/(tan(Pi)+tan(Pi/4))

tan(pi) =0 and tan(pi/4)=1
so (1-(0)(1))/(0+1)=1

Answer 6

a million) 2cotxsin^2x=sin(2x) RHS = sin(2x) = *2(sin x) * (cos x) = 2 * (cos x)* (sin x) {sin x/sinx) = 2*{(cos x )/(sinx)} * (sin^2 x) = 2*(cot x) * (sin^2 x) = LHS 2)RHS = (sinxcosx/a million+cosx) - (sinx/a million-cosx) = [{(sin x cos x)*( a million-cos x)} - {(sin x)*(a million+cos x)}]/(a million -cos^2 x) = [{(sin x cos x)*( a million-cos x)} - {(sin x)*(a million+cos x)}]/(sin^2 x) = [{sin x cos x - sin x cos^2 x)} - sin x - sin x cos x)}]/(sin^2 x) [{-sin x cos^2 x- sin x}]/(sin x) = -(cotx cosx+cscx) = LHS 3) RHS = (cosx/secx-a million) - (cosx/tan^2x) = (cosx/secx-a million) - (cosx{sec^2 x -a million} = [{cos x *(sec x +a million) - cos x}/{sec^2 x -a million}] = a million /(tan^2 x) = cot^2 x = LHS