Prove-that-cos-4-sin-4-cos-2-sin-2-

Question Number 83036 by otchereabdullai@gmail.com last updated on 27/Feb/20

Prove that \cos^{4} θ - \sin^{4} θ = \cos^{2} θ - \sin^{2} θ

Commented by Tony Lin last updated on 27/Feb/20

{c o s}^{4} θ - {s i n}^{4} θ

= {({c o s}^{2} θ - {s i n}^{2} θ)}^{2} + 2 {s i n}^{2} θ {c o s}^{2} θ

= \frac{c o s 4 θ + s i n 4 θ + 1}{2} \neq c o s 2 θ

Commented by MJS last updated on 27/Feb/20

\cos θ = c \land \sin θ = s

c^{4} - s^{4} = c^{2} - s^{2}

(c^{2} - s^{2}) (c^{2} + s^{2}) = c^{2} - s^{2}

but c^{2} + s^{2} = 1

\Rightarrow true

Commented by otchereabdullai@gmail.com last updated on 27/Feb/20

thanks prof mjs