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

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

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

prove

cos (x) + 1 = sin (2 x)

False

Solution steps

cos (x) + 1 = sin (2 x)

Show that the two sides are not equal

For

cos (x) + 1 = sin (2 x)

plug in

x = 0

cos (0) + 1 = 2

cos (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

= 1 + 1

Simplify

= 2

sin (2 \cdot 0) = 0

sin (2 \cdot 0)

Simplify

= sin (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0

The two sides are not equal

\Rightarrow False

p ro v e \frac{( sec ^{2} ( x ) )}{sec ^{2} ( x ) - 1} = csc^{2} (x)

p ro v e (sin (x)) (tan (x) cos (x) - cot (x) cos (x)) = 1 - 2 cos (2 x)

p ro v e sin (2 0^{\circ}) = 2 cos (1 0^{\circ}) \cdot sin (1 0^{\circ})

p ro v e sin^{2} (x) = sec (x) cos (x) - cos^{2} (x)

p ro v e 1 - 2 sin^{2} (2 θ) = 8 cos^{4} (θ) - 8 cos^{2} (θ) + 1