Is cos(pi-θ)=cos(θ) ?

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

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prove cos(pi-θ)=cos(θ)

prove

cos (π - θ) = cos (θ)

False

Solution steps

cos (π - θ) = cos (θ)

Show that the two sides are not equal

For

cos (π - θ) = cos (θ)

plug in

θ = 0

cos (π - 0) = - 1

cos (π - 0)

Simplify

= cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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)

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

The two sides are not equal

\Rightarrow False

tan (x) = \frac{4}{5}

tan (2 x) = \frac{2 tan ( x )}{( 1 - tan ^{2} ( x ) )}

\frac{cot ^{2} ( t )}{cos ( t )} = cot (t) csc (t)

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

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