Is cos(θ+pi)=-cos(θ) ?

The answer to whether cos(θ+pi)=-cos(θ) is True

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prove cos(θ+pi)=-cos(θ)

prove

cos (θ + π) = - cos (θ)

True

Solution steps

cos (θ + π) = - cos (θ)

Manipulating left side

cos (θ + π)

Rewrite using trig identities

cos (θ + π)

Use the Angle Sum identity:

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

= cos (θ) cos (π) - sin (θ) sin (π)

Simplify

cos (θ) cos (π) - sin (θ) sin (π) : - cos (θ)

cos (θ) cos (π) - sin (θ) sin (π)

cos (θ) cos (π) = - cos (θ)

cos (θ) 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 (θ)

Refine

= - cos (θ)

= - cos (θ) - sin (π) sin (θ)

sin (θ) sin (π) = 0

sin (θ) 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 (θ)

Apply rule

0 \cdot a = 0

= 0

= - cos (θ) - 0

- cos (θ) - 0 = - cos (θ)

= - cos (θ)

= - cos (θ)

= - cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ^{2} ( θ ) - 1}{sin ( θ )} = tan (θ) sec (θ)

(1 + csc (x)) (1 + csc (- x)) = - cot^{2} (x)

cos (0) - sec (0) = - sin (0) tan (0)

cot^{2} (x) sin^{2} (x) csc^{2} (x) tan^{2} (x) = 1

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