Example 3 from Algebra and Trigonometry, 9.2 Sum and Difference Identities

Using Sum and Difference Identities to Evaluate the Difference of Angles

Use the sum and difference identities to evaluate the difference of the angles and show that part a equals part b.

ⓐ $\sin (45° - 30°)$
ⓑ $\sin (135° - 120°)$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
1. ⓐ Let’s begin by writing the formula and substitute the given angles.
  $\begin{matrix} \sin (α - β) & = & \sin α \cos β - \cos α \sin β \\ \sin (45° - 30°) & = & \sin (45°) \cos (30°) - \cos (45°) \sin (30°) \end{matrix}$
  Next, we need to find the values of the trigonometric expressions.
  
  $\sin (45°) = \frac{\sqrt{2}}{2}, \cos (30°) = \frac{\sqrt{3}}{2}, \cos (45°) = \frac{\sqrt{2}}{2}, \sin (30°) = \frac{1}{2}$
  Now we can substitute these values into the equation and simplify.
  
  $\begin{matrix} \sin (45° - 30°) & = & \frac{\sqrt{2}}{2} (\frac{\sqrt{3}}{2}) - \frac{\sqrt{2}}{2} (\frac{1}{2}) \\ = & \frac{\sqrt{6} - \sqrt{2}}{4} \end{matrix}$
2. ⓑ Again, we write the formula and substitute the given angles.
  $\begin{matrix} \sin (α - β) & = & \sin α \cos β - \cos α \sin β \\ \sin (135° - 120°) & = & \sin (135°) \cos (120°) - \cos (135°) \sin (120°) \end{matrix}$
  Next, we find the values of the trigonometric expressions.
  
  $\sin (135°) = \frac{\sqrt{2}}{2}, \cos (120°) = - \frac{1}{2}, \cos (135°) = \frac{\sqrt{2}}{2}, \sin (120°) = \frac{\sqrt{3}}{2}$
  
  Now we can substitute these values into the equation and simplify.
  
  $\begin{matrix} \sin (135° - 120°) & = & \frac{\sqrt{2}}{2} (- \frac{1}{2}) - (- \frac{\sqrt{2}}{2}) (\frac{\sqrt{3}}{2}) \\ = & \frac{- \sqrt{2} + \sqrt{6}}{4} \\ = & \frac{\sqrt{6} - \sqrt{2}}{4} \\ \sin (135° - 120°) & = & \frac{\sqrt{2}}{2} (- \frac{1}{2}) - (- \frac{\sqrt{2}}{2}) (\frac{\sqrt{3}}{2}) \\ = & \frac{- \sqrt{2} + \sqrt{6}}{4} \\ = & \frac{\sqrt{6} - \sqrt{2}}{4} \end{matrix}$

How did it go?

Solution