Is cos(θ)=-cos(-θ) ?

The answer to whether cos(θ)=-cos(-θ) is False

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove cos(θ)=-cos(-θ)

prove

cos (θ) = - cos (- θ)

False

Solution steps

cos (θ) = - cos (- θ)

Show that the two sides are not equal

For

cos (θ) = - cos (- θ)

plug in

θ = 0

cos (0) = 1

cos (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

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 (- 0) = - 1

- cos (- 0)

Rewrite using trig identities:

cos (- 0) = 1

cos (- 0)

Simplify

= cos (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

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

= - 1

The two sides are not equal

\Rightarrow False

sin (θ) + cos (θ) = sin (θ + 4 5^{\circ})

cot (t) - sin (2 t) = cot (t) cos (2 t)

\frac{sin ( 2 θ )}{sin ( θ )} = 2 cos (θ)

81 - 9 (3 sin (x))^{2} = 1

\frac{1 + sec ( x )}{csc ( x )} = sin (x) + tan (x)