What is the general solution for cos(2θ-pi/2)=(sqrt(2))/2 ?

The general solution for cos(2θ-pi/2)=(sqrt(2))/2 is θ=pin+(3pi)/8 ,θ=pin+(9pi)/8

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Popular Trigonometry >

cos(2θ-pi/2)=(sqrt(2))/2

Solution

cos (2 θ - \frac{π}{2}) = \frac{2}{2}

Solution

θ = πn + \frac{3 π}{8}, θ = πn + \frac{9 π}{8}

Degrees

θ = 67. 5^{\circ} + 18 0^{\circ} n, θ = 202. 5^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 θ - \frac{π}{2}) = \frac{2}{2}

General solutions for

cos (2 θ - \frac{π}{2}) = \frac{2}{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}

2 θ - \frac{π}{2} = \frac{π}{4} + 2 πn, 2 θ - \frac{π}{2} = \frac{7 π}{4} + 2 πn

2 θ - \frac{π}{2} = \frac{π}{4} + 2 πn, 2 θ - \frac{π}{2} = \frac{7 π}{4} + 2 πn

Solve

2 θ - \frac{π}{2} = \frac{π}{4} + 2 πn : θ = πn + \frac{3 π}{8}

2 θ - \frac{π}{2} = \frac{π}{4} + 2 πn

Move

\frac{π}{2}

to the right side

2 θ - \frac{π}{2} = \frac{π}{4} + 2 πn

Add

\frac{π}{2}

to both sides

2 θ - \frac{π}{2} + \frac{π}{2} = \frac{π}{4} + 2 πn + \frac{π}{2}

Simplify

2 θ - \frac{π}{2} + \frac{π}{2} = \frac{π}{4} + 2 πn + \frac{π}{2}

Simplify

2 θ - \frac{π}{2} + \frac{π}{2} : 2 θ

2 θ - \frac{π}{2} + \frac{π}{2}

Add similar elements:

- \frac{π}{2} + \frac{π}{2} = 0

= 2 θ

Simplify

\frac{π}{4} + 2 πn + \frac{π}{2} : 2 πn + \frac{3 π}{4}

\frac{π}{4} + 2 πn + \frac{π}{2}

Group like terms

= 2 πn + \frac{π}{4} + \frac{π}{2}

Least Common Multiplier of

4, 2 : 4

4, 2

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

4

2

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{π}{2} :

multiply the denominator and numerator by

2

\frac{π}{2} = \frac{π 2}{2 \cdot 2} = \frac{π 2}{4}

= \frac{π}{4} + \frac{π 2}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{π + π 2}{4}

Add similar elements:

π + 2 π = 3 π

= 2 πn + \frac{3 π}{4}

2 θ = 2 πn + \frac{3 π}{4}

2 θ = 2 πn + \frac{3 π}{4}

2 θ = 2 πn + \frac{3 π}{4}

Divide both sides by

2

2 θ = 2 πn + \frac{3 π}{4}

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 πn}{2} + \frac{\frac{3 π}{4}}{2}

Simplify

\frac{2 θ}{2} = \frac{2 πn}{2} + \frac{\frac{3 π}{4}}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{2 πn}{2} + \frac{\frac{3 π}{4}}{2} : πn + \frac{3 π}{8}

\frac{2 πn}{2} + \frac{\frac{3 π}{4}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{3 π}{8}

= πn + \frac{3 π}{8}

θ = πn + \frac{3 π}{8}

θ = πn + \frac{3 π}{8}

θ = πn + \frac{3 π}{8}

Solve

2 θ - \frac{π}{2} = \frac{7 π}{4} + 2 πn : θ = πn + \frac{9 π}{8}

2 θ - \frac{π}{2} = \frac{7 π}{4} + 2 πn

Move

\frac{π}{2}

to the right side

2 θ - \frac{π}{2} = \frac{7 π}{4} + 2 πn

Add

\frac{π}{2}

to both sides

2 θ - \frac{π}{2} + \frac{π}{2} = \frac{7 π}{4} + 2 πn + \frac{π}{2}

Simplify

2 θ - \frac{π}{2} + \frac{π}{2} = \frac{7 π}{4} + 2 πn + \frac{π}{2}

Simplify

2 θ - \frac{π}{2} + \frac{π}{2} : 2 θ

2 θ - \frac{π}{2} + \frac{π}{2}

Add similar elements:

- \frac{π}{2} + \frac{π}{2} = 0

= 2 θ

Simplify

\frac{7 π}{4} + 2 πn + \frac{π}{2} : 2 πn + \frac{9 π}{4}

\frac{7 π}{4} + 2 πn + \frac{π}{2}

Group like terms

= 2 πn + \frac{π}{2} + \frac{7 π}{4}

Least Common Multiplier of

2, 4 : 4

2, 4

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

2

4

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{π}{2} :

multiply the denominator and numerator by

2

\frac{π}{2} = \frac{π 2}{2 \cdot 2} = \frac{π 2}{4}

= \frac{π 2}{4} + \frac{7 π}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 + 7 π}{4}

Add similar elements:

2 π + 7 π = 9 π

= 2 πn + \frac{9 π}{4}

2 θ = 2 πn + \frac{9 π}{4}

2 θ = 2 πn + \frac{9 π}{4}

2 θ = 2 πn + \frac{9 π}{4}

Divide both sides by

2

2 θ = 2 πn + \frac{9 π}{4}

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 πn}{2} + \frac{\frac{9 π}{4}}{2}

Simplify

\frac{2 θ}{2} = \frac{2 πn}{2} + \frac{\frac{9 π}{4}}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{2 πn}{2} + \frac{\frac{9 π}{4}}{2} : πn + \frac{9 π}{8}

\frac{2 πn}{2} + \frac{\frac{9 π}{4}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{9 π}{4}}{2} = \frac{9 π}{8}

\frac{\frac{9 π}{4}}{2}

Apply the fraction rule:

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

= \frac{9 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{9 π}{8}

= πn + \frac{9 π}{8}

θ = πn + \frac{9 π}{8}

θ = πn + \frac{9 π}{8}

θ = πn + \frac{9 π}{8}

θ = πn + \frac{3 π}{8}, θ = πn + \frac{9 π}{8}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2θ-pi/2)=(sqrt(2))/2 ?
The general solution for cos(2θ-pi/2)=(sqrt(2))/2 is θ=pin+(3pi)/8 ,θ=pin+(9pi)/8