Is sin^2(0)=2-2cos(0) ?

The answer to whether sin^2(0)=2-2cos(0) is True

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prove sin^2(0)=2-2cos(0)

prove

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

True

Solution steps

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

Manipulating left side

sin^{2} (0)

Simplify

sin^{2} (0) : 0

sin^{2} (0)

Simplify

sin (0) : 0

sin (0)

Use the following trivial identity:

sin (0) = 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^{2}

Apply rule

0^{a} = 0

= 0

= 0

Manipulating right side

2 - 2 cos (0)

Simplify

2 - 2 cos (0) : 0

2 - 2 cos (0)

2 cos (0) = 2

2 cos (0)

Simplify

cos (0) : 1

cos (0)

Use the following trivial identity:

cos (0) = 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

= 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2

= 2 - 2

Simplify

= 0

= 0

We showed that the two sides could take the same form

\Rightarrow True

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tan (x) + \frac{cos ( x )}{1 - sin ( x )} = sec (x)

sin (x) cos (x) = tan (x)

cot (\frac{15 π}{8}) = cot (\frac{7 π}{8})