Trig Identities - Angle Addition - Maple Help

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Main Concept

There are several trig identities used in mathematics, among which are the angle addition and subtraction formulas. These formulas are summarized as follows:

 Angle Addition or Angle Subtraction $\mathrm{sin}\left(A+B\right)$ $\mathrm{sin}\left(A\right)\mathrm{cos}\left(B\right)+\mathrm{cos}\left(A\right)\mathrm{sin}\left(B\right)$ $\mathrm{sin}\left(A-B\right)$ $\mathrm{sin}\left(A\right)\mathrm{cos}\left(B\right)-\mathrm{cos}\left(A\right)\mathrm{sin}\left(B\right)$ $\mathrm{cos}\left(A+B\right)$ $\mathrm{cos}\left(A\right)\mathrm{cos}\left(B\right)-\mathrm{sin}\left(A\right)\mathrm{sin}\left(B\right)$ $\mathrm{cos}\left(A-B\right)$ $\mathrm{cos}\left(A\right)\mathrm{cos}\left(B\right)+\mathrm{sin}\left(A\right)\mathrm{sin}\left(B\right)$ $\mathrm{tan}\left(A+B\right)$ $\frac{\mathrm{tan}\left(A\right)+\mathrm{tan}\left(B\right)}{1-\mathrm{tan}\left(A\right)\mathrm{tan}\left(B\right)}$ $\mathrm{tan}\left(A-B\right)$

The following questions focus on angle addition. Both the identities for $\mathrm{sin}\left(A+B\right)$ and $\mathrm{cos}\left(A+B\right)$ can be derived using geometry as shown below.

Select the appropriate radio button and click "Next" to see, step by step, how the identity is derived.





Questions

1. Using the identities in the above table, determine the value of sin(75°).

2. Prove that $\mathrm{tan}\left(A+B\right)=\frac{\mathrm{tan}\left(A\right)+\mathrm{tan}\left(B\right)}{1-\mathrm{tan}\left(A\right)\mathrm{tan}\left(B\right)}.$

Solutions

 Solution 1 You can use the identity $\mathrm{sin}\left(A+B\right)=\mathrm{sin}\left(A\right)\mathrm{cos}\left(B\right)+\mathrm{cos}\left(A\right)\mathrm{sin}\left(B\right)$ to determine the value of sin(75°):   First, note that 45° + 30° = 75°. This is important, as these angles form one of the special triangles:     From this, you can substitute the two angles into the identity for $\mathrm{sin}\left(A+B\right)$:   $\mathrm{sin}\left(A+B\right)=\mathrm{sin}\left(A\right)\mathrm{cos}\left(B\right)+\mathrm{cos}\left(A\right)\mathrm{sin}\left(B\right)$ $\mathrm{sin}\left(30°+45°\right)=\mathrm{sin}\left(30°\right)\mathrm{cos}\left(45°\right)+\mathrm{cos}\left(30°\right)\mathrm{sin}\left(45°\right)$ $\mathrm{sin}\left(75°\right)=\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{2}}\right)$ $\mathrm{sin}\left(75°\right)=\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}$ $\mathrm{sin}\left(75°\right)=\frac{\sqrt{2}+\sqrt{6}}{4}$
 Solution 2 First, note that:   $\mathrm{tan}\left(A+B\right)=\frac{\mathrm{sin}\left(A+B\right)}{\mathrm{cos}\left(A+B\right)}$   Keeping this in mind, you can substitute the trig identities for $\mathrm{sin}\left(A+B\right)$ and $\mathrm{cos}\left(A+B\right)$ and get:   $\frac{\mathrm{sin}\left(A\right)\mathrm{cos}\left(B\right)+\mathrm{cos}\left(A\right)\mathrm{sin}\left(B\right)}{\mathrm{cos}\left(A\right)\mathrm{cos}\left(B\right)-\mathrm{sin}\left(A\right)\mathrm{sin}\left(B\right)}$   Now, divide every term by $\mathrm{cos}\left(A\right)\mathrm{cos}\left(B\right)$:   $\frac{\frac{\mathrm{sin}\left(A\right)\mathrm{cos}\left(B\right)}{\mathrm{cos}\left(A\right)\mathrm{cos}\left(B\right)}+\frac{\mathrm{cos}\left(A\right)\mathrm{sin}\left(B\right)}{\mathrm{cos}\left(A\right)\mathrm{cos}\left(B\right)}}{\frac{\mathrm{cos}\left(A\right)\mathrm{cos}\left(B\right)}{\mathrm{cos}\left(A\right)\mathrm{cos}\left(B\right)}-\frac{\mathrm{sin}\left(A\right)\mathrm{sin}\left(B\right)}{\mathrm{cos}\left(A\right)\mathrm{cos}\left(B\right)}}$   Simplify this expression as follows:   $\frac{\frac{\mathrm{sin}\left(A\right)}{\mathrm{cos}\left(A\right)}+\frac{\mathrm{sin}\left(B\right)}{\mathrm{cos}\left(B\right)}}{1-\frac{\mathrm{sin}\left(A\right)\mathrm{sin}\left(B\right)}{\mathrm{cos}\left(A\right)\mathrm{cos}\left(B\right)}}$   Finally, you can simplify this expression further by recalling that $\mathrm{tan}\left(Q\right)=\frac{\mathrm{sin}\left(Q\right)}{\mathrm{cos}\left(Q\right)}$:   $\frac{\mathrm{tan}\left(A\right)+\mathrm{tan}\left(B\right)}{1-\mathrm{tan}\left(A\right)\mathrm{tan}\left(B\right)}$





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