Is cos^2(x)=1+cos(2x) ?

The answer to whether cos^2(x)=1+cos(2x) is False

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prove cos^2(x)=1+cos(2x)

prove

cos^{2} (x) = 1 + cos (2 x)

False

Solution steps

cos^{2} (x) = 1 + cos (2 x)

Show that the two sides are not equal

For

cos^{2} (x) = 1 + cos (2 x)

plug in

x = 0

cos^{2} (0) = 1

cos^{2} (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^{2}

Simplify

= 1

1 + cos (2 \cdot 0) = 2

1 + cos (2 \cdot 0)

Simplify

= 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 + 1

Simplify

= 2

The two sides are not equal

\Rightarrow False

sin (θ - \frac{π}{2}) + sin (θ + \frac{π}{2}) = 0

1 = cos^{2} (x) + sin^{2} (x)

(sin^{2} (x) + cos^{2} (x))^{3} = 1

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

sec (\frac{π}{4} + x) sec (\frac{π}{4} - x) = 2 sec (2 x)