What is the general solution for arccos(x)=(7pi)/8 ?

The general solution for arccos(x)=(7pi)/8 is x=-(sqrt(2+\sqrt{2)})/2

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Popular Trigonometry >

arccos(x)=(7pi)/8

Solution

arccos (x) = \frac{7 π}{8}

Solution

x = - \frac{2 + 2}{2}

Solution steps

arccos (x) = \frac{7 π}{8}

Apply trig inverse properties

arccos (x) = \frac{7 π}{8}

arccos (x) = a \Rightarrow x = cos (a)

x = cos (\frac{7 π}{8})

cos (\frac{7 π}{8}) = - \frac{2 + 2}{2}

cos (\frac{7 π}{8})

Rewrite using trig identities:

- cos (\frac{π}{8})

cos (\frac{7 π}{8})

Use the basic trigonometric identity:

cos (x) = - cos (π - x)

= - cos (π - \frac{7 π}{8})

Simplify:

π - \frac{7 π}{8} = \frac{π}{8}

π - \frac{7 π}{8}

Convert element to fraction:

π = \frac{π 8}{8}

= \frac{π 8}{8} - \frac{7 π}{8}

Since the denominators are equal, combine the fractions:

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

= \frac{π 8 - 7 π}{8}

Add similar elements:

8 π - 7 π = π

= \frac{π}{8}

= - cos (\frac{π}{8})

= - cos (\frac{π}{8})

Rewrite using trig identities:

cos (\frac{π}{8}) = \frac{2 + 2}{2}

cos (\frac{π}{8})

Rewrite using trig identities:

\frac{1 + cos ( \frac{π}{4} )}{2}

cos (\frac{π}{8})

Write

cos (\frac{π}{8})

cos (\frac{\frac{π}{4}}{2})

= cos (\frac{\frac{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{π}{4} )}{2}

= \frac{1 + cos ( \frac{π}{4} )}{2}

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{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:

n \frac{a}{b} = \frac{n a}{n b},

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:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2 + 2}{2}

= \frac{2 + 2}{2}

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

x = - \frac{2 + 2}{2}

x = - \frac{2 + 2}{2}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for arccos(x)=(7pi)/8 ?
The general solution for arccos(x)=(7pi)/8 is x=-(sqrt(2+\sqrt{2)})/2