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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

cos (6 0^{\circ})

Simplify

= \frac{1}{2}

Manipulating right side

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

Simplify

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

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

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

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

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

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

Simplify

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

cos (3 0^{\circ})

Use the following trivial identity:

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

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}

= \frac{3}{2}

= (\frac{3}{2})^{2}

Apply exponent rule:

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

= \frac{( 3 ) ^{2}}{2 ^{2}}

(3)^{2} : 3

Apply radical rule:

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

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 3^{\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

= 3

= \frac{3}{2 ^{2}}

= 2 \cdot \frac{3}{2 ^{2}}

Multiply fractions:

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

= \frac{3 \cdot 2}{2 ^{2}}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{6}{2 ^{2}}

Factor

6 : 2 \cdot 3

Factor

6 = 2 \cdot 3

= \frac{2 \cdot 3}{2 ^{2}}

Cancel the common factor:

2

= \frac{3}{2}

= - 1 + \frac{3}{2}

Simplify

- 1 + \frac{3}{2}

Convert element to fraction:

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

= - \frac{1 \cdot 2}{2} + \frac{3}{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 + 3}{2}

- 1 \cdot 2 + 3 = 1

- 1 \cdot 2 + 3

Multiply the numbers:

1 \cdot 2 = 2

= - 2 + 3

Add/Subtract the numbers:

- 2 + 3 = 1

= 1

= \frac{1}{2}

= \frac{1}{2}

= \frac{1}{2}

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove (sin(t)+cos(t))^2=1+sin(2t)prove

(sin (t) + cos (t))^{2} = 1 + sin (2 t)

prove sec^2(θ)= 1/(cos^2(θ))prove

sec^{2} (θ) = \frac{1}{cos ^{2} ( θ )}

prove csc^2(θ)= 1/(sin^2(θ))prove

csc^{2} (θ) = \frac{1}{sin ^{2} ( θ )}

prove sin(x)= 1/2 prove

sin (x) = \frac{1}{2}

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

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

Frequently Asked Questions (FAQ)

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