Trigonometric Proofs Exercise 1

To prove a new trig identity, we need to show that one side is equivalent to the other using established identities, in this case, the Big 8 (the Ratio, Reciprocal and Pythagorean Identities). You only want to work on one side, so pick the more complicated one and try to simplify it. Looking at this one, the more complicated side looks like the LHS, so I’m going to try manipulating the LHS to make it look like the RHS.

First, I’m going to FOIL out the \((1+\cot{x})^2\), \[\begin{aligned} (1+\cot{x})^2\sin^2{x}&=1+2\sin{x}\cos{x} \\ (1+2\cot{x}+\cot^2{x})\sin^2{x} &= 1+2\sin{x}\cos{x} \end{aligned}\] Then I notice that the RHS is in terms of cosine and sine so I’ll convert the LHS to terms of sine and cosine. So I need to use the ratio identity \(\cot{x}=\frac{\cos{x}}{\sin{x}}\), \[\begin{aligned} (1+2\cot{x}+\cot^2{x})\sin^2{x} &= 1+2\sin{x}\cos{x} \\ (1+2\frac{\cos{x}}{\sin{x}}+\frac{\cos^2{x}}{\sin^2{x}})\sin^2{x} &= 1+2\sin{x}\cos{x} \end{aligned}\] Then I can distribute the \(\sin^2{x}\), \[\begin{aligned} (1+2\frac{\cos{x}}{\sin{x}}+\frac{\cos^2{x}}{\sin^2{x}})\sin^2{x} &= 1+2\sin{x}\cos{x} \\ \sin^2{x}+2\frac{\cos{x} \cdot \sin^2{x}}{\sin{x}}+\frac{\cos^2{x} \cdot \sin^2{x}}{\sin^2{x}} &= 1+2\sin{x}\cos{x} \\ \sin^2{x} +2\sin{x}\cos{x}+\cos^2{x} &= 1+2\sin{x}\cos{x} \end{aligned}\] Now I have a \(\sin^2{x} \text{ and } \cos^2{x}\), so I can use the Pythagorean Identity \(\sin^2{x}+\cos^2{x}=1\) \[\begin{aligned} \sin^2{x} +2\sin{x}\cos{x}+\cos^2{x} &= 1+2\sin{x}\cos{x} \\ 1+2\sin{x}\cos{x} &= 1+2\sin{x}\cos{x} \end{aligned}\]