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Answer 1 · 2017-04-10T21:15:02

$sin(theta)=(-2sqrt(2))/3$
$sin(2theta) = (-4sqrt(2))/9$
$cos(2theta) = 7/9$
$sin(pi/6+theta) = (1-2sqrt(6))/6$
$tan(x/2) = -sqrt(2)/2$

Given: $sec(theta)=3$

$cos(theta) = 1/sec(theta) = 1/3$

$sin(theta) = -sqrt(1-cos^2(theta)) larr$ we use the negative because we are told that it is the 4th quadrant.

$sin(theta)=-sqrt(1-(1/3)^2)$

$sin(theta)=-sqrt(9/9-1/9)$

$sin(theta)=-sqrt(8/9)$

$sin(theta)=(-2sqrt(2))/3$

$sin(2theta) = 2sin(theta)cos(theta)$

$sin(2theta) = 2(-2sqrt(2))/3(1/3)$

$sin(2theta) = (-4sqrt(2))/9$

$cos(2theta) = 1-2cos^2(theta)$

$cos(2theta) = 1-2(1/3)^2$

$cos(2theta) = 7/9$

$sin(pi/6+theta) = sin(pi/6)cos(theta)+cos(pi/6)sin(theta)$

$sin(pi/6+theta) = (1/2)(1/3)+(sqrt3/2)((-2sqrt(2))/3)$

$sin(pi/6+theta) = (1-2sqrt(6))/6$

$tan(x/2) = sin(x)/(1+cos(x)$

$tan(x/2) = ((-2sqrt(2))/3)/(1+ 1/3)$

$tan(x/2) = -sqrt(2)/2$