#1
cos(x+20) = sin(90 - (x+20) ) , one of the basic rules
but sinx = cos(x+20)
sinx = sin(70-x)
arcsin both sides
x = 70-x
2x=70
x = 35
check:
LS = sin 35
RS = cos(35+20) = cos 55
but cos 55 = sin 35
so all is good
tanx = sinx/cosx
= cos(x+20)/cosx
= (cosxcos20 - sinxsin20)/cosx
= cos20 - tanxsin20
tanx = cos20 - tanxsin20
tanx + tanxsin20 = cos20
tanx(1 + sin20) = cos20
tanx = cos20/(1 + sin20)
Given that sinx = cos(x+20o) write tanx in terms of sin20o and cos 20o. a) Hence find the values of tan x.
b) Find the general solution of 3sec2y=5(1+tany)
2 answers
You did not specify whether you want degrees or radians. I will do it for degrees.
Repeat for radians, make sure to set your calculator to radians
3sec^2 y = 5(1+tany)
We know sec^2 A = 1 + tan^2 A
3(1 + tan^2 y) = 5(1 + tany)
3 + 3tan^2 y = 5 + 5tany
3tan^2 y - 5tany - 2 = 0
(tany - 2)(3tany + 1) = 0
tany = 2 or tany = -1/3
I will do the 2nd one, you do the first.
if tany = -1/3 , y is in quads II or IV
we know tan 18.435°
y = 180-18.435 = 161.565°
y = 360-18.435 =341.565
general solution:
y = 161.565+180k, where k is an integer
(Notice we hit 341.565 when k=1)
repeat for the tany = 2 solution , repeat in radians if needed
Repeat for radians, make sure to set your calculator to radians
3sec^2 y = 5(1+tany)
We know sec^2 A = 1 + tan^2 A
3(1 + tan^2 y) = 5(1 + tany)
3 + 3tan^2 y = 5 + 5tany
3tan^2 y - 5tany - 2 = 0
(tany - 2)(3tany + 1) = 0
tany = 2 or tany = -1/3
I will do the 2nd one, you do the first.
if tany = -1/3 , y is in quads II or IV
we know tan 18.435°
y = 180-18.435 = 161.565°
y = 360-18.435 =341.565
general solution:
y = 161.565+180k, where k is an integer
(Notice we hit 341.565 when k=1)
repeat for the tany = 2 solution , repeat in radians if needed
0 answers