Find sin x/2, cos x/2 and tan x/2, if tan x = -4/3 , x in quadrant II

Solution:

tan x = -4/3 = Base/Perpendicular (-ve sign is because it is in quadrant II)

Base = 3

Perpendicular = 4

Therefore hypotenuse = √3² + 4² = √9 + 16 = √25 = 5

sinx = 4/5 (+ve because Sin is positive in the second quadrant)

cos x = -3/5 (-ve, because Cos is negative in the second quadrant)

cos(x) = cos2(x/2) = cos²(x/2) - sin²(x/2) = -3/5 --- (1)

We also know that

cos²(x/2) + sin²(x/2) = 1 --- (2)

cos²(x/2) = 1 - sin²(x/2) --- (3)

Substituting (3) in 1 we get,

1 - sin²(x/2) - sin²(x/2) = -3/5

2sin²(x/2) = 1 + 3/5

 2sin²(x/2) = 8/5

sin²(x/2) = 4/5

sin(x/2) = 2/√5

sinx = 2sin(x/2)cos(x/2) = 4/5

2(2/√5)cos(x/2) = 4/5

cos(x/2) = 1/√5

tan(x/2) = sin(x/2)/cos(x/2)

= (2/√5)/(1/√5)

= 2

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Find sin x/2, cos x/2 and tan x/2, if tan x = -4/3 , x in quadrant II

Summary:

The values of sin x/2, cos x/2 and tan x/2 , if tan x = -4/3, (x in quadrant II) are 2/√5, 1/√5 and 2 respectively.