We begin by writing the formula for the product of cosines:

[latex]\cos \alpha \cos \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)+\cos \left(\alpha +\beta \right)\right][/latex]

We can then substitute the given angles into the formula and simplify.

[latex]\begin{align}2\cos \left(\frac{7x}{2}\right)\cos \left(\frac{3x}{2}\right)&=\left(2\right)\left(\frac{1}{2}\right)\left[\cos \left(\frac{7x}{2}-\frac{3x}{2}\right)+\cos \left(\frac{7x}{2}+\frac{3x}{2}\right)\right] \\ &=\left[\cos \left(\frac{4x}{2}\right)+\cos \left(\frac{10x}{2}\right)\right] \\ &=\cos 2x+\cos 5x \end{align}[/latex]

Write the formula for the product of sine and cosine. Then substitute the given values into the formula and simplify.

[latex]\begin{align}\sin \alpha \cos \beta &=\frac{1}{2}\left[\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)\right] \\ \sin \left(4\theta \right)\cos \left(2\theta \right)&=\frac{1}{2}\left[\sin \left(4\theta +2\theta \right)+\sin \left(4\theta -2\theta \right)\right] \\ &=\frac{1}{2}\left[\sin \left(6\theta \right)+\sin \left(2\theta \right)\right] \end{align}[/latex]

We have the product of cosines, so we begin by writing the related formula. Then we substitute the given angles and simplify.

[latex]\begin{align}\cos \alpha \cos \beta &=\frac{1}{2}\left[\cos \left(\alpha -\beta \right)+\cos \left(\alpha +\beta \right)\right] \\ \cos \left(3\theta \right)\cos \left(5\theta \right)&=\frac{1}{2}\left[\cos \left(3\theta -5\theta \right)+\cos \left(3\theta +5\theta \right)\right] \\ &=\frac{1}{2}\left[\cos \left(2\theta \right)+\cos \left(8\theta \right)\right] && \text{Use even-odd identity}. \end{align}[/latex]

We begin by writing the formula for the difference of sines.

[latex]\sin \alpha -\sin \beta =2\sin \left(\frac{\alpha -\beta }{2}\right)\cos \left(\frac{\alpha +\beta }{2}\right)[/latex]

Substitute the values into the formula, and simplify.

[latex]\begin{align}\sin \left(4\theta \right)-\sin \left(2\theta \right)&=2\sin \left(\frac{4\theta -2\theta }{2}\right)\cos \left(\frac{4\theta +2\theta }{2}\right) \\ &=2\sin \left(\frac{2\theta }{2}\right)\cos \left(\frac{6\theta }{2}\right) \\ &=2\sin \theta \cos \left(3\theta \right) \end{align}[/latex]

We begin by writing the formula for the difference of cosines.

[latex]\cos \alpha -\cos \beta =-2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)[/latex]

Then we substitute the given angles and simplify.

[latex]\begin{align}\cos \left({15}^{\circ }\right)-\cos \left({75}^{\circ }\right)&=-2\sin \left(\frac{{15}^{\circ }+{75}^{\circ }}{2}\right)\sin \left(\frac{{15}^{\circ }-{75}^{\circ }}{2}\right) \\ &=-2\sin \left({45}^{\circ }\right)\sin \left(-{30}^{\circ }\right) \\ &=-2\left(\frac{\sqrt{2}}{2}\right)\left(-\frac{1}{2}\right) \\ &=\frac{\sqrt{2}}{2}\end{align}[/latex]

We will start with the left side, the more complicated side of the equation, and rewrite the expression until it matches the right side.

[latex]\begin{align}\frac{\cos \left(4t\right)-\cos \left(2t\right)}{\sin \left(4t\right)+\sin \left(2t\right)}&=\frac{-2\sin \left(\frac{4t+2t}{2}\right)\sin \left(\frac{4t - 2t}{2}\right)}{2\sin \left(\frac{4t+2t}{2}\right)\cos \left(\frac{4t - 2t}{2}\right)} \\ &=\frac{-2\sin \left(3t\right)\sin t}{2\sin \left(3t\right)\cos t} \\ &=\frac{-2\sin \left(3t\right)\sin t}{2\sin \left(3t\right)\cos t} \\ &=-\frac{\sin t}{\cos t} \\ &=-\tan t \end{align}[/latex]

Analysis of the Solution

Recall that verifying trigonometric identities has its own set of rules. The procedures for solving an equation are not the same as the procedures for verifying an identity. When we prove an identity, we pick one side to work on and make substitutions until that side is transformed into the other side.

For verifying this equation, we are bringing together several of the identities. We will use the double-angle formula and the reciprocal identities. We will work with the right side of the equation and rewrite it until it matches the left side.

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