find sin x/2, cos x/2, tan x/2, if cos x= -1/4, x in quadrant iii
Answers
Given info : cosx = -1/4 in iii quadrant.
To find : we know, In 3rd quadrant,
Only ‘tan’ and ‘cot’ will be positive and rest will be negative.
Let's solve it.
cosx = -1/4
we know, cos2Φ = 2cos²Φ - 1
so, cosx = 2cos²x/2 - 1 = -1/4
⇒2cos²x/2 = 1 - 1/4 = 3/4
⇒cos²x/2 = 3/8
⇒cosx/2 = ±√(3/8)
as 180° < x < 270° [ x is in 3rd quadrant]
⇒90° < x/2 < 135° [ x/2 is in 2nd quadrant ]
In 2nd quadrant, sin and cosec are positive and all remaining identities are negative.
so, cosx/2 = -√(3/8)
now sinx/2 = √(1 - cos²x/2)
= √(1 - 3/8) = √(5/8)
and tanx/2 = (sinx/2)/(cosx/2)
= (√5/8)/-√(3/8)
= -√5
Therefore sinx/2 = √(5/8) , cosx/2 = -√(3/8) and tanx/2 = -√5
Step-by-step explanation:
ANSWER
Given that x is in second quadrant
90<x<180
45<
2
x
<90
2
x
lies in first quadrant
So,
sinx,cosx,tanx are positive in first quadrant
tanx=
1−tan
2
2
x
2tan(
2
x
)
3
−4
=
1−tan
2
2
x
2tan(
2
x
)
Solving the above quadratic equation, we get,
tan(
2
x
)=
2
−1
or tan(
2
x
)=2
∴tan(
2
x
)=2(∵
2
x
) lies in first quadrant
1+tan
2
(
2
x
)=sec
2
(
2
x
)
⇒sec
2
(
2
x
)=5
⇒sec(
2
x
)=
5
cos(
2
x
)=
sec(
2
x
)
1
⇒cos(
2
x
)=
5
1
or cos(
2
x
)=
5
5
sin
2
x+cos
2
x=1
⇒sin
2
x=1−cos
2
x
⇒sin
2
x=
5
4
⇒sinx=±
5
4