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.

We will work on the right side of the equal sign and rewrite the expression until it matches the left side.

[latex]\begin{align}{\left(\sin \theta +\cos \theta \right)}^{2}&={\sin }^{2}\theta +2\sin \theta \cos \theta +{\cos }^{2}\theta \\ &=\left({\sin }^{2}\theta +{\cos }^{2}\theta \right)+2\sin \theta \cos \theta \\ &=1+2\sin \theta \cos \theta \\ &=1+\sin \left(2\theta \right)\end{align}[/latex]

Analysis of the Solution

This process is not complicated, as long as we recall the perfect square formula from algebra:

[latex]{\left(a\pm b\right)}^{2}={a}^{2}\pm 2ab+{b}^{2}[/latex]

where [latex]a=\sin \theta [/latex] and [latex]b=\cos \theta [/latex]. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.

In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.

[latex]\begin{align}\tan \left(2\theta \right)&=\frac{2\tan \theta }{1-{\tan }^{2}\theta }&& \text{Double-angle formula} \\ &=\frac{2\tan \theta \left(\frac{1}{\tan \theta }\right)}{\left(1-{\tan }^{2}\theta \right)\left(\frac{1}{\tan \theta }\right)}&& \text{Multiply by a term that results in desired numerator}. \\ &=\frac{2}{\frac{1}{\tan \theta }-\frac{{\tan }^{2}\theta }{\tan \theta }} \\ &=\frac{2}{\cot \theta -\tan \theta }&& \text{Use reciprocal identity for }\frac{1}{\tan \theta }.\end{align}[/latex]

Analysis of the Solution

Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show

[latex]\frac{2\tan \theta }{1-{\tan }^{2}\theta }=\frac{2}{\cot \theta -\tan \theta }[/latex]

Let’s work on the right side.

[latex]\begin{align}\frac{2}{\cot \theta -\tan \theta }&=\frac{2}{\frac{1}{\tan \theta }-\tan \theta }\left(\frac{\tan \theta }{\tan \theta }\right) \\ &=\frac{2\tan \theta }{\frac{1}{\cancel{\tan \theta }}\left(\cancel{\tan \theta }\right)-\tan \theta \left(\tan \theta \right)} \\ &=\frac{2\tan \theta }{1-{\tan }^{2}\theta } \end{align}[/latex]

When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.