Is cos(x+pi)cos(x-pi)=cos^2(x) ?

The answer to whether cos(x+pi)cos(x-pi)=cos^2(x) is True

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Popular Trigonometry >

prove cos(x+pi)cos(x-pi)=cos^2(x)

Solution

prove

cos (x + π) cos (x - π) = cos^{2} (x)

Solution

True

Solution steps

cos (x + π) cos (x - π) = cos^{2} (x)

Manipulating left side

cos (x + π) cos (x - π)

Rewrite using trig identities

cos (x - π)

Use the Angle Difference identity:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

= cos (x) cos (π) + sin (x) sin (π)

Simplify

cos (x) cos (π) + sin (x) sin (π) : - cos (x)

cos (x) cos (π) + sin (x) sin (π)

cos (x) cos (π) = - cos (x)

cos (x) cos (π)

Simplify

cos (π) : - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= (- 1) cos (x)

Refine

= - cos (x)

= - cos (x) + sin (π) sin (x)

sin (x) sin (π) = 0

sin (x) sin (π)

Simplify

sin (π) : 0

sin (π)

Use the following trivial identity:

sin (π) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

= - cos (x) + 0

- cos (x) + 0 = - cos (x)

= - cos (x)

= - cos (x)

= cos (x + π) (- cos (x))

Rewrite using trig identities

cos (x + π)

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (x) cos (π) - sin (x) sin (π)

Simplify

cos (x) cos (π) - sin (x) sin (π) : - cos (x)

cos (x) cos (π) - sin (x) sin (π)

cos (x) cos (π) = - cos (x)

cos (x) cos (π)

Simplify

cos (π) : - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= (- 1) cos (x)

Refine

= - cos (x)

= - cos (x) - sin (π) sin (x)

sin (x) sin (π) = 0

sin (x) sin (π)

Simplify

sin (π) : 0

sin (π)

Use the following trivial identity:

sin (π) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

= - cos (x) - 0

- cos (x) - 0 = - cos (x)

= - cos (x)

= - cos (x)

= (- cos (x)) (- cos (x))

Simplify

(- cos (x)) (- cos (x)) : cos^{2} (x)

(- cos (x)) (- cos (x))

Remove parentheses:

(- a) = - a, - (- a) = a

= cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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Is cos(x+pi)cos(x-pi)=cos^2(x) ?
The answer to whether cos(x+pi)cos(x-pi)=cos^2(x) is True