If sin a = 0.25 \sin a = 0.25 sin a = 0.25 and cos b = 0.25 \cos b = 0.25 cos b = 0.25 , which of the following is sin b / 2 + cos a / 2 \sin^b /_2 + \cos^a /_2 sin b / 2 + cos a / 2 ?
A. 1.60
B. 1.32
C. 0.26
D. 1.06
**Solution.**
Since sin a = 0.25 > 0 \sin a = 0.25 > 0 sin a = 0.25 > 0 then a ∈ ( 0 , π ) a \in (0, \pi) a ∈ ( 0 , π ) where sine is a positive value.
Since cos b = 0.25 > 0 \cos b = 0.25 > 0 cos b = 0.25 > 0 then b ∈ ( 0 , π ) b \in (0, \pi) b ∈ ( 0 , π ) where cosine is a positive value.
So sin b 2 > 0 \sin \frac{b}{2} > 0 sin 2 b > 0 and cos a 2 > 0 \cos \frac{a}{2} > 0 cos 2 a > 0 .
Express sin b / 2 = f ( cos b ) \sin^b /_2 = f(\cos b) sin b / 2 = f ( cos b ) :
We have the formula:
sin 2 α = 1 − cos 2 α 2 \sin^2 \alpha = \frac{1 - \cos 2\alpha}{2} sin 2 α = 2 1 − cos 2 α
Use it:
sin 2 b 2 = 1 − cos b 2 \sin^2 \frac{b}{2} = \frac{1 - \cos b}{2} sin 2 2 b = 2 1 − cos b
Then
sin b 2 = 1 − cos b 2 = 1 − 0.25 2 = 0.375 ≈ 0.61 \sin \frac{b}{2} = \sqrt{\frac{1 - \cos b}{2}} = \sqrt{\frac{1 - 0.25}{2}} = \sqrt{0.375} \approx 0.61 sin 2 b = 2 1 − cos b = 2 1 − 0.25 = 0.375 ≈ 0.61
Similarly, express cos a / 2 = f ( sin a ) \cos^a /_2 = f(\sin a) cos a / 2 = f ( sin a ) :
Use this formula cos 2 α = 1 + cos 2 α 2 \cos^2 \alpha = \frac{1 + \cos 2\alpha}{2} cos 2 α = 2 1 + c o s 2 α :
cos 2 a 2 = 1 + cos a 2 = 1 + 1 − sin 2 a 2 = 1 + 1 − 0.0625 2 ≈ 0.99 \cos^2 \frac{a}{2} = \frac{1 + \cos a}{2} = \frac{1 + \sqrt{1 - \sin^2 a}}{2} = \frac{1 + \sqrt{1 - 0.0625}}{2} \approx 0.99 cos 2 2 a = 2 1 + cos a = 2 1 + 1 − sin 2 a = 2 1 + 1 − 0.0625 ≈ 0.99
So calculate sin b / 2 + cos a / 2 \sin^b /_2 + \cos^a /_2 sin b / 2 + cos a / 2 :
sin b / 2 + cos a / 2 = 0.61 + 0.99 = 1.6 \sin^b /_2 + \cos^a /_2 = 0.61 + 0.99 = 1.6 sin b / 2 + cos a / 2 = 0.61 + 0.99 = 1.6
**Answer:** A. 1.60