prove cos(1*pi)=-1

Solution

prove

cos (1 \cdot π) = - 1

True

Solution steps

cos (1 π) = - 1

Manipulating left side

cos (1 π)

Simplify

cos (1 π) : - 1

cos (1 π)

Multiply:

1 π = π

= 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

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ^{2} ( x )}{sin ( x ) cos ( x )} = tan (x)

cos (6 0^{\circ} + θ) - cos (6 0^{\circ} - θ) = - 3 sin (θ)

cos (θ) cot (θ) = \frac{1}{sin ( θ )} - sin (θ)

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

sin^{4} (a) - cos^{4} (a) = 2 sin^{2} (a) - 1