A)
The Pythagorean theorem states that the sum of the squared legs is the squared hypothenuse:
[tex] a^2+b^2=c^2 [/tex]
If we divide the whole expression by [tex] c^2 [/tex], we have
[tex] \dfrac{a^2}{c^2} + \dfrac{b^2}{c^2} = \dfrac{c^2}{c^2} [/tex]
B)
The sine of x is the ratio between the opposite leg and the hypothenuse, while the cosine of x is the ratio between the adjacent leg and the hypothenues:
[tex] \sin(x) = \dfrac{a}{c},\quad \cos(x) = \dfrac{b}{c} [/tex]
This means that
[tex] \sin^2(x) = \left(\dfrac{a}{c}\right)^2 = \dfrac{a^2}{c^2},\quad \cos^2(x) = \left(\dfrac{b}{c}\right)^2 = \dfrac{b^2}{c^2}[/tex]
C)
From part B, we know that
[tex] \sin^2(x)+\cos^2(x) = \dfrac{a^2}{c^2} + \dfrac{b^2}{c^2} [/tex]
From part A, we know that this sum equals
[tex] \sin^2(x)+\cos^2(x) = \dfrac{a^2}{c^2} + \dfrac{b^2}{c^2}=\dfrac{c^2}{c^2}=1[/tex]
Which terminates the proof.
