Is cos((3pi)/8)=(sqrt(2-\sqrt{2)})/2 ?

The answer to whether cos((3pi)/8)=(sqrt(2-\sqrt{2)})/2 is True

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prove cos((3pi)/8)=(sqrt(2-\sqrt{2)})/2

Solution

prove

cos (\frac{3 π}{8}) = \frac{2 - 2}{2}

Solution

True

Solution steps

cos (\frac{3 π}{8}) = \frac{2 - 2}{2}

Manipulating left side

cos (\frac{3 π}{8})

Simplify

cos (\frac{3 π}{8}) : \frac{2 - 2}{2}

cos (\frac{3 π}{8})

Rewrite using trig identities:

\frac{1 + cos ( \frac{3 π}{4} )}{2}

cos (\frac{3 π}{8})

Write

cos (\frac{3 π}{8})

cos (\frac{\frac{3 π}{4}}{2})

= cos (\frac{\frac{3 π}{4}}{2})

Use the Half Angle identity:

cos (\frac{θ}{2}) = \frac{1 + cos ( θ )}{2}

Use the Double Angle identity

cos (2 θ) = 2 cos^{2} (θ) - 1

Substitute

θ

with

\frac{θ}{2}

cos (θ) = 2 cos^{2} (\frac{θ}{2}) - 1

Switch sides

2 cos^{2} (\frac{θ}{2}) = 1 + cos (θ)

Divide both sides by

2

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

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

range [0, \frac{π}{2}] [\frac{π}{2}, π] [π, \frac{3 π}{2}] [\frac{3 π}{2}, 2 π] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

cos (\frac{θ}{2}) = \frac{( 1 + cos ( θ ) )}{2}

= \frac{1 + cos ( \frac{3 π}{4} )}{2}

= \frac{1 + cos ( \frac{3 π}{4} )}{2}

Use the following trivial identity:

cos (\frac{3 π}{4}) = - \frac{2}{2}

cos (\frac{3 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{1 - \frac{2}{2}}{2}

Simplify

\frac{1 - \frac{2}{2}}{2} : \frac{2 - 2}{2}

\frac{1 - \frac{2}{2}}{2}

\frac{1 - \frac{2}{2}}{2} = \frac{2 - 2}{4}

\frac{1 - \frac{2}{2}}{2}

Join

1 - \frac{2}{2} : \frac{2 - 2}{2}

1 - \frac{2}{2}

Convert element to fraction:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 - 2}{2}

= \frac{\frac{2 - 2}{2}}{2}

Apply the fraction rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 - 2}{4}

= \frac{2 - 2}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{2 - 2}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2 - 2}{2}

= \frac{2 - 2}{2}

= \frac{2 - 2}{2}

= \frac{2 - 2}{2}

We showed that the two sides could take the same form

\Rightarrow True

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Is cos((3pi)/8)=(sqrt(2-\sqrt{2)})/2 ?
The answer to whether cos((3pi)/8)=(sqrt(2-\sqrt{2)})/2 is True