What is the general solution for cos(4y)=2cos(2y)-1 ?

The general solution for cos(4y)=2cos(2y)-1 is y=pin,y= pi/4+pin,y=(3pi)/4+pin

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cos(4y)=2cos(2y)-1

Solution

cos (4 y) = 2 cos (2 y) - 1

Solution

y = πn, y = \frac{π}{4} + πn, y = \frac{3 π}{4} + πn

Degrees

y = 0^{\circ} + 18 0^{\circ} n, y = 4 5^{\circ} + 18 0^{\circ} n, y = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

cos (4 y) = 2 cos (2 y) - 1

Subtract

2 cos (2 y) - 1

from both sides

cos (4 y) - 2 cos (2 y) + 1 = 0

Rewrite using trig identities

1 + cos (4 y) - 2 cos (2 y)

cos (4 y) = 2 cos^{2} (2 y) - 1

cos (4 y)

Rewrite as

= cos (2 \cdot 2 y)

Use the Double Angle identity:

cos (2 y) = 2 cos^{2} (y) - 1

cos (2 \cdot 2 y) = 2 cos^{2} (2 y) - 1

= 2 cos^{2} (2 y) - 1

= 1 + 2 cos^{2} (2 y) - 1 - 2 cos (2 y)

Simplify

1 + 2 cos^{2} (2 y) - 1 - 2 cos (2 y) : 2 cos^{2} (2 y) - 2 cos (2 y)

1 + 2 cos^{2} (2 y) - 1 - 2 cos (2 y)

Group like terms

= 2 cos^{2} (2 y) - 2 cos (2 y) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (2 y) - 2 cos (2 y)

= 2 cos^{2} (2 y) - 2 cos (2 y)

- 2 cos (2 y) + 2 cos^{2} (2 y) = 0

Solve by substitution

- 2 cos (2 y) + 2 cos^{2} (2 y) = 0

Let:

cos (2 y) = u

- 2 u + 2 u^{2} = 0

- 2 u + 2 u^{2} = 0 : u = 1, u = 0

- 2 u + 2 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

2 u^{2} - 2 u = 0

Solve with the quadratic formula

2 u^{2} - 2 u = 0

Quadratic Equation Formula:

For

a = 2, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 2)^{2} - 4 \cdot 2 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 2 \cdot 0

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 2^{2} - 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

u_{1, 2} = \frac{- ( - 2 ) \pm 2}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2}{2 \cdot 2}, u_{2} = \frac{- ( - 2 ) - 2}{2 \cdot 2}

u = \frac{- ( - 2 ) + 2}{2 \cdot 2} : 1

\frac{- ( - 2 ) + 2}{2 \cdot 2}

Apply rule

- (- a) = a

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

Add the numbers:

2 + 2 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 2}{2 \cdot 2} : 0

\frac{- ( - 2 ) - 2}{2 \cdot 2}

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 2 = 0

= \frac{0}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 1, u = 0

Substitute back

u = cos (2 y)

cos (2 y) = 1, cos (2 y) = 0

cos (2 y) = 1, cos (2 y) = 0

cos (2 y) = 1 : y = πn

cos (2 y) = 1

General solutions for

cos (2 y) = 1

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 y = 0 + 2 πn

2 y = 0 + 2 πn

Solve

2 y = 0 + 2 πn : y = πn

2 y = 0 + 2 πn

0 + 2 πn = 2 πn

2 y = 2 πn

Divide both sides by

2

2 y = 2 πn

Divide both sides by

2

\frac{2 y}{2} = \frac{2 πn}{2}

Simplify

y = πn

y = πn

y = πn

cos (2 y) = 0 : y = \frac{π}{4} + πn, y = \frac{3 π}{4} + πn

cos (2 y) = 0

General solutions for

cos (2 y) = 0

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 y = \frac{π}{2} + 2 πn, 2 y = \frac{3 π}{2} + 2 πn

2 y = \frac{π}{2} + 2 πn, 2 y = \frac{3 π}{2} + 2 πn

Solve

2 y = \frac{π}{2} + 2 πn : y = \frac{π}{4} + πn

2 y = \frac{π}{2} + 2 πn

Divide both sides by

2

2 y = \frac{π}{2} + 2 πn

Divide both sides by

2

\frac{2 y}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 y}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 y}{2} : y

\frac{2 y}{2}

Divide the numbers:

\frac{2}{2} = 1

= y

Simplify

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

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

y = \frac{π}{4} + πn

y = \frac{π}{4} + πn

y = \frac{π}{4} + πn

Solve

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

2 y = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 y = \frac{3 π}{2} + 2 πn

Divide both sides by

2

\frac{2 y}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 y}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 y}{2} : y

\frac{2 y}{2}

Divide the numbers:

\frac{2}{2} = 1

= y

Simplify

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

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

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

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

y = \frac{3 π}{4} + πn

y = \frac{3 π}{4} + πn

y = \frac{3 π}{4} + πn

y = \frac{π}{4} + πn, y = \frac{3 π}{4} + πn

Combine all the solutions

y = πn, y = \frac{π}{4} + πn, y = \frac{3 π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(4y)=2cos(2y)-1 ?
The general solution for cos(4y)=2cos(2y)-1 is y=pin,y= pi/4+pin,y=(3pi)/4+pin