Is 3cos(2x+pi/2)-2=6cos^2(x+pi/4) ?

The answer to whether 3cos(2x+pi/2)-2=6cos^2(x+pi/4) is False

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Popular >

prove 3cos(2x+pi/2)-2=6cos^2(x+pi/4)

Solution

prove

3 cos (2 x + \frac{π}{2}) - 2 = 6 cos^{2} (x + \frac{π}{4})

Solution

False

Solution steps

3 cos (2 x + \frac{π}{2}) - 2 = 6 cos^{2} (x + \frac{π}{4})

Show that the two sides are not equal

For

3 cos (2 x + \frac{π}{2}) - 2 = 6 cos^{2} (x + \frac{π}{4})

plug in

x = 0

3 cos (2 \cdot 0 + \frac{π}{2}) - 2 = - 2

3 cos (2 \cdot 0 + \frac{π}{2}) - 2

Simplify

= 3 cos (\frac{π}{2}) - 2

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

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}

= 0

= 3 \cdot 0 - 2

Simplify

= - 2

6 cos^{2} (0 + \frac{π}{4}) = 3

6 cos^{2} (0 + \frac{π}{4})

Rewrite using trig identities:

cos (0 + \frac{π}{4}) = \frac{2}{2}

cos (0 + \frac{π}{4})

Simplify

= cos (\frac{π}{4})

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

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{2}{2}

= \frac{2}{2}

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

Simplify

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

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

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

= 6 \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{1 \cdot 6}{2}

Multiply the numbers:

1 \cdot 6 = 6

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

= 3

The two sides are not equal

\Rightarrow False

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Frequently Asked Questions (FAQ)

Is 3cos(2x+pi/2)-2=6cos^2(x+pi/4) ?
The answer to whether 3cos(2x+pi/2)-2=6cos^2(x+pi/4) is False