Why-does-cos-2-x-1-2-cos-2x-1-

Question Number 5495 by FilupSmith last updated on 16/May/16

Why does :

\cos^{2} x = \frac{1}{2} (\cos (2 x) + 1)

Commented by Yozzii last updated on 16/May/16

L e t a = (\begin{matrix} c o s A \\ s i n A \end{matrix}), b = (\begin{matrix} c o s B \\ - s i n B \end{matrix}) .

F o r a l l 0 < A, B < 2 π, t h e a n g l e b e t w e e n a a n d b i s A + B

∴ f r o m d o t p r o d u c t a ∙ b =∣ a ∣∣ b ∣ c o s (A + B)

\Rightarrow c o s A c o s B - s i n A s i n B = \sqrt{{c o s}^{2} A + {s i n}^{2} A} \times \sqrt{{c o s}^{2} B + {s i n}^{2} B} c o s (A + B)

S i n c e {c o s}^{2} A + {s i n}^{2} A = {c o s}^{2} B + {s i n}^{2} B = 1

\Rightarrow c o s (A + B) = c o s A c o s B - s i n A s i n B .

∴ c o s (x + x) = c o s x c o s x - s i n x s i n x

= {c o s}^{2} x - {s i n}^{2} x

= {c o s}^{2} x - 1 + {c o s}^{2} x

= 2 {c o s}^{2} x - 1

\Rightarrow c o s 2 x = 2 {c o s}^{2} x - 1

\Rightarrow {c o s}^{2} x = \frac{c o s 2 x + 1}{2}

Answered by Rasheed Soomro last updated on 16/May/16

We know that

\cos (α + β) = \cos α \cos β - \sin α \sin β

Let α = β = x

\cos (x + x) = \cos x \cos x - \sin x \sin x

\cos 2 x = \cos^{2} x - \sin^{2} x

But \sin^{2} x = 1 - \cos^{2} x

So \cos 2 x = \cos^{2} x - (1 - \cos^{2} x) = \cos^{2} x - 1 + \cos^{2} x

\cos 2 x = {2 \cos}^{2} x - 1

{2 \cos}^{2} x = \cos 2 x + 1

or \cos^{2} x = \frac{1}{2} (\cos (2 x) + 1)