Question

If 450° < θ < 540° and sin θ = $\frac{12}{13}$ then calculate sin($\frac{\theta}{2}$) and cos($\frac{\theta}{2}$).

Answer 1

it is given that , sinθ = 12/13 = p/h

so, p = 12 and h = 13

then, b = √(h²- p²) = √(13² - 12²) = ±5

so, cosθ = ±5/13 but θ lies between 450° to 540°
e.g., 360° + 90° < θ < 360° + 180°

or, 90° < θ < 180° [ because 360° is the period of sine and cosine function ]

now, θ lies in 2nd quadrant but we know, cosine will be negative in 2nd quadrant.
so, cosθ = - 5/13

now, sin²(θ/2) = (1 - cosθ)/2 [ from formula]

= (1 + 5/13)/2 = 18/26 = 9/13

sin(θ/2) = ± 3/√13 , but 225° < θ < 270° [ e.g., 3rd quadrant ] and we know, sine function will be negative in 3rd quadrant.

so, sin(θ/2) = -3/√13


again, cos²(θ/2) = (1 + cosθ)/2 [ from formula]

= (1 - 5/13)/2 = 8/26 = 4/13

cos(θ/2) = ±2/√(13), but 225° < θ < 270° [ e.g., 3rd quadrant ] and we know, cosine function will be negative in 3rd quadrant.

so, cos(θ/2) = -2/√13

Answer 2

Step-by-step explanation:

her we go:

sinA=√1+cos2A/2

and

cosA=√1-cos2A/2