Is prove-2sin^2(x/2)=cos(x) ?

The answer to whether prove-2sin^2(x/2)=cos(x) is False

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prove-2sin^2(x/2)=cos(x)

prove

- 2 sin^{2} (\frac{x}{2}) = cos (x)

False

Solution steps

- 2 sin^{2} (\frac{x}{2}) = cos (x)

Show that the two sides are not equal

For

- 2 sin^{2} (\frac{x}{2}) = cos (x)

plug in

x = 0

- 2 sin^{2} (\frac{0}{2}) = 0

- 2 sin^{2} (\frac{0}{2})

Rewrite using trig identities:

sin (\frac{0}{2}) = 0

sin (\frac{0}{2})

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

= - 2 \cdot 0^{2}

Simplify

= 0

cos (0) = 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

The two sides are not equal

\Rightarrow False

sin^{2} (u) (cot^{2} (u) + 1) = 1

\frac{1 + csc ( x )}{cos ( x ) + cot ( x )} = sec (x)

cos (x + \frac{π}{6}) = - sin (x - \frac{π}{3})

4 \cdot cos (2 \cdot x) = 1 - 3 \cdot cos (x)

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

Is prove-2sin^2(x/2)=cos(x) ?
The answer to whether prove-2sin^2(x/2)=cos(x) is False