prove 2cos^2(x)-1=0

Solution

prove

2 cos^{2} (x) - 1 = 0

False

Solution steps

2 cos^{2} (x) - 1 = 0

Show that the two sides are not equal

For

2 cos^{2} (x) - 1 = 0

plug in

x = 0

2 cos^{2} (0) - 1 = 1

2 cos^{2} (0) - 1

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

= 2 \cdot 1^{2} - 1

Simplify

= 1

The two sides are not equal

\Rightarrow False

(1 + cos (2 θ)) (1 - cos (2 θ)) = sin^{2} (2 θ)

sin (x) - cos (x) = 0

sin (2 θ) + cos (2 θ) = 2 sin (θ) cos (θ) + 2 cos^{2} (θ) - 1

\frac{tan ^{4} ( x ) - 1}{tan ^{2} ( x ) - 1} = sec^{2} (x)

csc (β) - csc (β) cos (2 β) = 2 \cdot sin (β)