We see that the left side of the equation includes the sines of the sum and the difference of angles.

[latex]\begin{gathered}\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta\\ \sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta \end{gathered}[/latex]

We can rewrite each using the sum and difference formulas.

[latex]\begin{align}\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)&=\sin \alpha \cos \beta +\cos \alpha \sin \beta +\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ &=2\sin \alpha \cos \beta \end{align}[/latex]

We see that the identity is verified.

We can begin by rewriting the numerator on the left side of the equation.

[latex]\begin{align}\frac{\sin \left(\alpha -\beta \right)}{\cos \alpha \cos \beta }&=\frac{\sin \alpha \cos \beta -\cos \alpha \sin \beta }{\cos \alpha \cos \beta } \\ &=\frac{\sin \alpha \cos \beta}{\cos \alpha \cos \beta}-\frac{\cos \alpha \sin \beta }{\cos \alpha \cos \beta } && \text{Rewrite using a common denominator}. \\ &=\frac{\sin \alpha }{\cos \alpha }-\frac{\sin \beta }{\cos \beta }&& \text{Cancel}. \\ &=\tan \alpha -\tan \beta && \text{Rewrite in terms of tangent}.\end{align}[/latex]

We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine.

Using the difference formula for tangent, this problem does not seem as daunting as it might.

[latex]\begin{align}\tan \theta &=\tan \left({\theta }_{2}-{\theta }_{1}\right) \\ &=\frac{\tan {\theta }_{2}-\tan {\theta }_{1}}{1+\tan {\theta }_{1}\tan {\theta }_{2}} \\ &=\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}\end{align}[/latex]

Let’s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that [latex]\tan \beta =\frac{47}{50}[/latex], and [latex]\tan \left(\beta -\alpha \right)=\frac{40}{50}=\frac{4}{5}[/latex]. We can then use difference formula for tangent.

[latex]\tan \left(\beta -\alpha \right)=\frac{\tan \beta -\tan \alpha }{1+\tan \beta \tan \alpha }[/latex]

Now, substituting the values we know into the formula, we have

[latex]\begin{gathered}\frac{4}{5}=\frac{\frac{47}{50}-\tan \alpha }{1+\frac{47}{50}\tan \alpha } \\ 4\left(1+\frac{47}{50}\tan \alpha \right)=5\left(\frac{47}{50}-\tan \alpha \right) \end{gathered}[/latex]

Use the distributive property, and then simplify the functions.

[latex]\begin{gathered}4\left(1\right)+4\left(\frac{47}{50}\right)\tan \alpha =5\left(\frac{47}{50}\right)-5\tan \alpha\\ 4+3.76\tan \alpha =4.7 - 5\tan \alpha \\ 5\tan \alpha +3.76\tan \alpha =0.7\\ 8.76\tan \alpha =0.7\\ \tan \alpha \approx 0.07991 \\ {\tan }^{-1}\left(0.07991\right)\approx .079741 \end{gathered}[/latex]

Now we can calculate the angle in degrees.

[latex]\alpha \approx 0.079741\left(\frac{180}{\pi }\right)\approx {4.57}^{\circ }[/latex]

Analysis of the Solution

Occasionally, when an application appears that includes a right triangle, we may think that solving is a matter of applying the Pythagorean Theorem. That may be partially true, but it depends on what the problem is asking and what information is given.