Please trigonometry help!!!!!!!!!!!!!!!

Question

sin x = 3/5 , π/2 < x < π

A.cos(x/2)

B. sin(x/2)

C. Tan(x/2)

Answer 1

Given that  sinx = 3/5 , π/2< x<π

Apply pythagorean formula c^2 = a^2 + b^2

b^2 = c^2 - a^2

Sin x = a/c = 3/5 where a = 3 and c = 5                               

b^2 = 5^2 - 3^2

= 25 - 9 = 16

Therefore b = 4

cosx = b/c = 4/5

A) cos(x/2) =±√(1+cosx)/2

= √(1+(4/5))/2

= √9/10 = 3/√10

cos(x/2) = ±3/√10

B)sin(x/2) =±√(1-cosx)/2

=√(1-(4/5))/2

=√1/10

sin(x/2) =±√1/10

C) Tan(x/2) = sinx/(1+cosx)

= (3/5)/(1+(4/5))

= 3/5 (5/9) = 1/3

Tan(x/2) = 1/3.

 

Comments

sin(x/2) = √(9/10).

cos(x/2) = √(1/10).

tan(x/2) = 3.

Answer 2

The value of sin(x) = 3/5 and π/2 < x < π (second quadrant).

By using Pythagorean identity sin²(θ)+cos²(θ)=1, to obtain

(3/5)² + cos²(x) = 1

cos²(x) = 1 - 9/25 = (25-9)/25 = 16/25

cos(x) = ± √(16/25) = ± 4/5

The cosine function is negative in second quadrant.

cos(x) = - 4/5.

 

π/2 < x < π (second quadrant) ⇒ π/4 < x/2 < π/2 (first quadrant).

Half-Angle Formulas :

sin(θ/2) = ± √[ (1 - cos θ)/2], cos(θ/2) = ± √[ (1 + cos θ)/2] and tan(θ/2) = (1 - cos θ)/(sin θ) = (sin θ)/(1 + cos θ)

The all trigonometric functions are positive in first quadrant.

sin(x/2) = √[ (1 - cos x)/2 ] = √[ {1 - (-4/5) }/2 ] = √[ {(5+4)/5 }/2 ] = √(9/10).

cos(x/2) = √[ (1 + cos x)/2 ] = √[ {1 + (-4/5) }/2 ] = √[ {(5-4)/5 }/2 ] = √(1/10).

tan(x/2) = (1 - cos x)/(sin x) = [1 - (-4/5)] / (3/5) = [(5+4)/5] / (3/5) = [9/5] / (3/5) = 3.