(3 points) Suppose 𝜃 and 𝜙 are in the first quadrant and cos(𝜃)=5/11 and tan(𝜙)=13/1. Find cos(𝜃+2𝜋)= , cos(𝜃−2𝜋)=, cos(𝜋−𝜃)=, tan(𝜃+2𝜋)=, tan(𝜃−𝜋)= , tan(𝜋−𝜃)

Question

I understand how to do the less complicated versions of these but not sure how to do this version.

Answer 1

cosx = 5/11

cos(x+2pi) = cosx = 5/11 the period for cosines is 2pi

cos(x-2pi) = cosx = 5/11

cosx = 5/11 = adjacent side over hypotenuse. opposite side = sqr(11^2 -5^2) =sqr(121-25) =sqr(96)= 4sqr6

tanx=opposite side over adjacent side = 4sqr6/5

tan(x+2pi) = tanx = 4sqr6/5 period for tangents is pi

cos(pi-x) = cospicosx + sinpisinx = (-1)cosx + 0(sinx) = -cosx= -5/11

sin(pi-x) = sinpicosx - sinxcospi = 0(cosx) - sinx(-1) = sinx = 4sqr6/11

tan(pi-x) = sin(pi-x)/cos(pi-x) = 4sqr6/11 over 5/11 = 4sqr6/5

cos(x-pi) = cosxcospi+ sinxsinpi = -cosx = -5/11

sin(x-pi) = sinxcospi - sinpicosx = -sinx = -4sqr6/11

tain(x-pi) = sin(x-pi)/cos(x-pi)= -4sqr6/11 over -5/11 = 4sqr6/5

for tanx = 13/1

x= 85.6

13/1 = opposite side over adjacent side. hypotenuse = square root of 13^2 +1^2 = sqr170

sinx = 13/sqr170

cosx = 1/sqr170