Given sin(b)= 0.2183 and b is in quadrant 2, find approx. values of sin(2b) and cos(b/2)

Answer 1

Using the double angle identity:

sin(2x) = 2sin(x)cos(x)

So, we need to know cos(b) to use this.

Sketch the unit circle and since sin(b) would be the y-value and the hypotenuse of the unit circle is 1, you can use the Pythagorean Theorem to solve for "x":

x = √(1² - 0.2183²) = 0.97588 however, remember that we are in Q2 so this would be a negative.

Therefore cos(b) = -0.97588

So sin(2b) = 2sin(b)cos(b) = 2(0.2183)(-0.97588) = -0.4261

Using the half angle identity:

cos(b/2) = ±√[(1+cos(b))/2]

using cos(b) = -0.97588:

cos(b/2) = ±√[(1-0.97588)/2]

cos(b/2) = ±√[0.024118/2]

cos(b/2) = ±√0.012059

cos(b/2) = ± 0.1098

Since b is in Q2 then half of b would be in Q1 therefore the value of cos(b/2) would be positive therefore:

cos(b/2) = 0.1098