Trigonometric Proofs Exercise 4

When the numerator contains terms that are also factors of the denominator, splitting a fraction can be an effective first step in proving a trig identity. In this identity, I can see a cotangent and a cosine in the numerator, both of which are factors of the denominator. So I’m going to split this fraction with a difference in the numerator, into a difference of two fractions with a common denominator. (Remember, you can only split a fraction with a sum/difference in the numerator. You can’t split a denominator.) \[\begin{aligned} \frac{\cot{x}-\cos{x}\sin{x}}{\cot{x}\cos{x}} &= \cos{x} \\ \frac{\cot{x}}{\cot{x}\cos{x}} - \frac{\cos{x}\sin{x}}{\cot{x}\cos{x}} &= \cos{x} \\ \frac{1}{\cos{x}} - \frac{\sin{x}}{\cot{x}} &= \cos{x} \end{aligned}\]

Then we see that everything but that cotangent is in terms of sine and cosine, so I’ll use the ratio identity \(\cot{x}=\frac{\cos{x}}{\sin{x}}\). \[\begin{aligned} \frac{1}{\cos{x}} - \frac{\sin{x}}{\frac{\cos{x}}{\sin{x}}} &= \cos{x} \end{aligned}\] So we copy-dot-flip, \[\begin{aligned} \frac{1}{\cos{x}} - \sin{x} \left( \frac{\sin{x}}{\cos{x}} \right) &= \cos{x} \\ \frac{1}{\cos{x}} - \frac{\sin^2{x}}{\cos{x}} &= \cos{x} \end{aligned}\] Now we have a common denominator again so I’ll combine these fractions, \[\begin{aligned} \frac{1-\sin^2{x}}{\cos{x}} &= \cos{x} \end{aligned}\] And now the numerator is a rearrangement of the Pythagorean identity \[\begin{aligned} \sin^2{x}+\cos^2{x}=1 \\ \cos^2{x} = 1 - \sin^2{x} \end{aligned}\] So we plug that in and simplify. \[\begin{aligned} \frac{\cos^2{x}}{\cos{x}} &= \cos{x} \\ \cos{x} &= \cos{x} \end{aligned}\]