Is cos(20)=cos(θ)+cos(θ) ?

The answer to whether cos(20)=cos(θ)+cos(θ) is False

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prove cos(20)=cos(θ)+cos(θ)

prove

cos (2 0^{\circ}) = cos (θ) + cos (θ)

False

Solution steps

cos (2 0^{\circ}) = cos (θ) + cos (θ)

Show that the two sides are not equal

For

cos (2 0^{\circ}) = cos (θ) + cos (θ)

plug in

θ = 0

cos (2 0^{\circ}) = 0.93969 \dots

cos (2 0^{\circ})

Refine to a decimal form

= 0.93969 \dots

cos (0) + cos (0) = 2

cos (0) + cos (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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

cos^{2} (x) - sin^{2} (x) = tan^{2} (x)

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

(1 - cos (x)) (1 + sec (x)) = sin (x) tan (x)

cos (2 β) = \frac{1 - tan ^{2} ( β )}{1 + tan ^{2} ( β )}

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