Question

http://onlineexamhelp.com/wp-content/uploads/2014/08/9709_s14_qp_12.pdf
can someone explain question 3

Answer 1

what do you knolw about reflex angle ?

Answer 2

reflex angle is more than 180 and less than 360 so the range of value of Ѳ is pi<Ѳ<2pi

Answer 3

that means the angle is in III or IVth quadrant, yes ?

Answer 4

yes

Answer 5

@ganeshie8 what does k represent.

Answer 6

k is the value of cosine,
notice that k is positive

Answer 7

in which of the two quadrants is cosine positive ?

Answer 8

4th quadrant

Answer 9

so do we agree that the reflex angle in question is somewhere in 4th quadrant ?

Answer 10

yes i do agree

Answer 11

use below to find sine :
\[\large \sin^2\theta + \cos^2\theta = 1\]

Answer 12

you knw that cos = k,
plug in and solve sine

Answer 13

\[\large \sin^2\theta + k^2 = 1\]

Answer 14

sinѲ=+-(1-k)^1/2

Answer 15

sinѲ=+-(1-k^2)^1/2

Answer 16

yes but whats the sign of sine in 4th quadrant ?

Answer 17

negative

Answer 18

\[\large \sin \theta = -\sqrt{1-k^2}\]

Answer 19

tanѲ=sinѲ/cosѲ
tanѲ=-(1-k^2)^1/2/k=-(1-k^2)^1/2/k

Answer 20

part ii)

Answer 21

sin2Ѳ=2sinѲcosѲ=-k(1-k^2)^1/2

Answer 22

-k(1-k^2)^1/2<0

Answer 23

thanks ganeshie i got the answer if we took in account k>0 and sin2Ѳ<0 values of Ѳ lie between 270 and 360 are negative.

Answer 24

yes

Answer 25

\[\large 270 \lt \theta\lt 360\]

Answer 26

multiply through out :

\[\large 2\times270 \lt 2\theta\lt 2\times 360\]

Answer 27

that gives you \(\large 2\theta \) in 3rd quadrant,
so sine is clearly negative