Is cos(2x+pi/3)+2=2cos^2(x+pi/6) ?

The answer to whether cos(2x+pi/3)+2=2cos^2(x+pi/6) is False

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prove cos(2x+pi/3)+2=2cos^2(x+pi/6)

Solution

prove

cos (2 x + \frac{π}{3}) + 2 = 2 cos^{2} (x + \frac{π}{6})

Solution

False

Solution steps

cos (2 x + \frac{π}{3}) + 2 = 2 cos^{2} (x + \frac{π}{6})

Show that the two sides are not equal

For

cos (2 x + \frac{π}{3}) + 2 = 2 cos^{2} (x + \frac{π}{6})

plug in

x = 0

cos (2 \cdot 0 + \frac{π}{3}) + 2 = \frac{5}{2} (Decimal : 2.5)

cos (2 \cdot 0 + \frac{π}{3}) + 2

Simplify

= cos (\frac{π}{3}) + 2

Use the following trivial identity:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

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}

= \frac{1}{2}

= \frac{1}{2} + 2

Simplify

\frac{1}{2} + 2 : \frac{5}{2}

\frac{1}{2} + 2

Convert element to fraction:

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

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

2 \cdot 2 + 1 = 5

2 \cdot 2 + 1

Multiply the numbers:

2 \cdot 2 = 4

= 4 + 1

Add the numbers:

4 + 1 = 5

= 5

= \frac{5}{2}

= \frac{5}{2}

2 cos^{2} (0 + \frac{π}{6}) = \frac{3}{2} (Decimal : 1.5)

2 cos^{2} (0 + \frac{π}{6})

Simplify

= 2 cos^{2} (\frac{π}{6})

Use the following trivial identity:

cos (\frac{π}{6}) = \frac{3}{2}

cos (\frac{π}{6})

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}

= \frac{3}{2}

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

Simplify

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

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

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

= \frac{3}{2}

The two sides are not equal

\Rightarrow False

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Is cos(2x+pi/3)+2=2cos^2(x+pi/6) ?
The answer to whether cos(2x+pi/3)+2=2cos^2(x+pi/6) is False