Is sin^2(0)=4(cos(-0)-1) ?

The answer to whether sin^2(0)=4(cos(-0)-1) is True

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prove sin^2(0)=4(cos(-0)-1)

prove

sin^{2} (0) = 4 (cos (- 0) - 1)

True

Solution steps

sin^{2} (0) = 4 (cos (- 0) - 1)

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

4 (cos (- 0) - 1)

Use the negative angle identity:

cos (- x) = cos (x)

= 4 (- 1 + cos (0))

Simplify

(- 1 + cos (0)) 4 : 0

(- 1 + cos (0)) \cdot 4

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}

= 4 (1 - 1)

Simplify

(- 1 + 1) \cdot 4

Add/Subtract the numbers:

- 1 + 1 = 0

= 0 \cdot 4

Apply rule

0 \cdot a = 0

= 0

= 0

= 0

We showed that the two sides could take the same form

\Rightarrow True

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

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

sec (2 a) + tan (2 a) = \frac{cos ( a ) + sin ( a )}{cos ( a ) - sin ( a )}

sec^{2} (x) = \frac{1}{1 - sin ^{2} ( x )}

cos (\frac{π}{2} - x) sec (x) = tan (x)