Is cos(60)=1-2sin^2(30) ?

The answer to whether cos(60)=1-2sin^2(30) is True

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prove cos(60)=1-2sin^2(30)

prove

cos (6 0^{\circ}) = 1 - 2 sin^{2} (3 0^{\circ})

True

Solution steps

cos (6 0^{\circ}) = 1 - 2 sin^{2} (3 0^{\circ})

Manipulating left side

cos (6 0^{\circ})

Simplify

= \frac{1}{2}

Manipulating right side

1 - 2 sin^{2} (3 0^{\circ})

Simplify

1 - 2 sin^{2} (3 0^{\circ}) : \frac{1}{2}

1 - 2 sin^{2} (3 0^{\circ})

2 sin^{2} (3 0^{\circ}) = \frac{1}{2}

2 sin^{2} (3 0^{\circ})

sin^{2} (3 0^{\circ}) = \frac{1}{2 ^{2}}

sin^{2} (3 0^{\circ})

Simplify

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

Use the following trivial identity:

sin (3 0^{\circ}) = \frac{1}{2}

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

= \frac{1}{2}

= (\frac{1}{2})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{2 ^{2}}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{2 ^{2}}

= 2 \cdot \frac{1}{2 ^{2}}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2}{2 ^{2}}

Cancel the common factor:

2

= \frac{1}{2}

= 1 - \frac{1}{2}

Simplify

1 - \frac{1}{2}

Convert element to fraction:

1 = \frac{1 \cdot 2}{2}

= \frac{1 \cdot 2}{2} - \frac{1}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot 2 - 1}{2}

1 \cdot 2 - 1 = 1

1 \cdot 2 - 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= \frac{1}{2}

= \frac{1}{2}

= \frac{1}{2}

We showed that the two sides could take the same form

\Rightarrow True