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Question Number 44028 by pieroo last updated on 20/Sep/18

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

Commented by maxmathsup by imad last updated on 20/Sep/18

b y s c a l a r p r o d u c t w e c a n p r o v e t h a t c o s (a + b) = c o s a c o s b - s i n a s i n b

l e t t a k e a = b = x \Rightarrow c o s (2 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 1 + c o s (2 x) = 2 {c o s}^{2} x \Rightarrow {c o s}^{2} x = \frac{1 + c o s (2 x)}{2} .

Answered by MJS last updated on 20/Sep/18

f (x) = \cos^{2} x

f (0) = 1

f (45 °) = \frac{1}{2}

f (90 °) = 0

f (135 °) = \frac{1}{2}

f (180 °) = 1

f (225 °) = \frac{1}{2}

f (270 °) = 0

f (315 °) = \frac{1}{2}

so we have a periodic function

period = 180 ° or π

minima = 0; maxima = 1

looks like a cosinus twice as fast and half as

wide, shifted up so the \min & \max are 0 & 1

instead of \pm \frac{1}{2}

twice as fast = \cos 2 x

half as wide = \frac{1}{2} \cos 2 x

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

Commented by MJS last updated on 20/Sep/18

I thought it might be of interest to show it

without using any formula, because the

given equation is considered to be a formula

itself \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Sep/18

R H S

c o s (α + β) = c o s α c o s β - s i n α s i n β

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

c o z 2 x = {c o s}^{2} x - (1 - {c o s}^{2} x)

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

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

Answered by $@ty@m last updated on 20/Sep/18

W e h a v e

\cos (a + b) = \cos a . \cos b - \sin a . \sin b

P u t a = b = x

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

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

= 2 \cos^{2} x - 1

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

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