Trigonometry Proofs (No. 1)

Proof, using Pythagoras Theorem: $\sin\theta = \frac{b}{c} , \cos\theta = \frac{a}{c}$ From that, it follows: $\sin^{2}\theta + \cos^{2}\theta = \frac{b^{2}+a^{2}}{c^{2}} = 1$ where the last step applies Pythagoras' Theorem.

Let's draw a $\triangle ACB$ , where:

• $AC$ is the hypotenuse.

• $BC$ is the adjacent side.

• $AB$ is the opposite side.

We know that:

• $\sin \theta = \dfrac{ \text{Opposite side}}{\text{Hypotenuse}} = \dfrac{AB}{AC}$

• $\cos \theta = \dfrac{\text{Adjacent side}}{\text{Hypotenuse}} = \dfrac{BC}{AC}$

• The pythagorean theorem , $AC^2 = AB^2 + BC^2$ .

Now, we need to prove $\sin^2 \theta + \cos^2 \theta = 1$

$\begin{aligned} L.H.S & = \sin^2 \theta + \cos^2 \theta \\ & = \left(\dfrac{AB}{AC}\right)^2 + \left(\dfrac{BC}{AC} \right)^2 \\ & = \dfrac{AB^2}{AC^2} + \dfrac{BC^2}{AC^2} \\ & = \dfrac{AB^2 + BC^2}{AC^2} \\ & = \dfrac{AC^2}{AC^2} \\ & = 1 = R.H.S\\ \end{aligned}$

Proved.

Answer is 1.