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tags:
Fundamental Trigonometric Identity Trigonometric Ratios Trigonometry Trigonometry Series

Trigonometric Proofs Exercise 4

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Prove the following identity. 
\[\cos{x} = \frac{\cot{x}-\cos{x}\sin{x}}{\cot{x}\cos{x}}\]

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. \[\begin{aligned} \frac{\cos^2{x}}{\cos{x}} &= \cos{x} \\ \cos{x} &= \cos{x} \end{aligned}\]

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