Is 2sin^2(45)=1-cos(90) ?

The answer to whether 2sin^2(45)=1-cos(90) is True

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prove 2sin^2(45)=1-cos(90)

Solution

prove

2 sin^{2} (4 5^{\circ}) = 1 - cos (9 0^{\circ})

Solution

True

Solution steps

2 sin^{2} (4 5^{\circ}) = 1 - cos (9 0^{\circ})

Manipulating left side

2 sin^{2} (4 5^{\circ})

Simplify

2 sin^{2} (4 5^{\circ}) : 1

2 sin^{2} (4 5^{\circ})

sin^{2} (4 5^{\circ}) = \frac{1}{2}

sin^{2} (4 5^{\circ})

Simplify

sin (4 5^{\circ}) : \frac{2}{2}

sin (4 5^{\circ})

Use the following trivial identity:

sin (4 5^{\circ}) = \frac{2}{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{2}{2}

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

Apply exponent rule:

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

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

(2)^{2} : 2

Apply radical rule:

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

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

Apply exponent rule:

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

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 2

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

Cancel the common factor:

2

= \frac{1}{2}

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

Simplify

2 \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{2}{2}

Cancel the common factor:

2

= 1

= 1

= 1

Manipulating right side

1 - cos (9 0^{\circ})

Simplify

1 - cos (9 0^{\circ}) : 1

1 - cos (9 0^{\circ})

Use the following trivial identity:

cos (9 0^{\circ}) = 0

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

= 1 - 0

Simplify

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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Is 2sin^2(45)=1-cos(90) ?
The answer to whether 2sin^2(45)=1-cos(90) is True