What is the general solution for cos^2(a)+2cos(a)=sin^2(a/2)-2 ?

The general solution for cos^2(a)+2cos(a)=sin^2(a/2)-2 is a=pi+2pin

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Popular Trigonometry >

cos^2(a)+2cos(a)=sin^2(a/2)-2

Solution

cos^{2} (a) + 2 cos (a) = sin^{2} (\frac{a}{2}) - 2

Solution

a = π + 2 πn

Degrees

a = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (a) + 2 cos (a) = sin^{2} (\frac{a}{2}) - 2

Subtract

sin^{2} (\frac{a}{2}) - 2

from both sides

cos^{2} (a) + 2 cos (a) - sin^{2} (\frac{a}{2}) + 2 = 0

Rewrite using trig identities

2 + cos^{2} (a) - sin^{2} (\frac{a}{2}) + 2 cos (a)

Use the following identity:

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

Use the Double Angle identity

cos (2 θ) = 1 - 2 sin^{2} (θ)

Switch sides

2 sin^{2} (θ) - 1 = - cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 - cos (2 θ)

Divide both sides by

2

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

= 2 + cos^{2} (a) - \frac{1 - cos ( 2 \cdot \frac{a}{2} )}{2} + 2 cos (a)

Simplify

2 + cos^{2} (a) - \frac{1 - cos ( 2 \cdot \frac{a}{2} )}{2} + 2 cos (a) : \frac{5 cos ( a ) + 2 cos ^{2} ( a ) - 1}{2} + 2

2 + cos^{2} (a) - \frac{1 - cos ( 2 \cdot \frac{a}{2} )}{2} + 2 cos (a)

Multiply

2 \cdot \frac{a}{2} : a

2 \cdot \frac{a}{2}

Multiply fractions:

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

= \frac{a \cdot 2}{2}

Cancel the common factor:

2

= a

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

Combine the fractions

- \frac{- cos ( a ) + 1}{2} + cos^{2} (a) + 2 cos (a) : \frac{5 cos ( a ) + 2 cos ^{2} ( a ) - 1}{2}

- \frac{- cos ( a ) + 1}{2} + cos^{2} (a) + 2 cos (a)

Convert element to fraction:

cos^{2} (a) = \frac{cos ^{2} ( a ) 2}{2}, 2 cos (a) = \frac{2 cos ( a ) 2}{2}

= - \frac{1 - cos ( a )}{2} + \frac{cos ^{2} ( a ) \cdot 2}{2} + \frac{2 cos ( a ) \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{- ( 1 - cos ( a ) ) + cos ^{2} ( a ) \cdot 2 + 2 cos ( a ) \cdot 2}{2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- ( - cos ( a ) + 1 ) + 2 cos ^{2} ( a ) + 4 cos ( a )}{2}

Expand

- (1 - cos (a)) + cos^{2} (a) \cdot 2 + 4 cos (a) : 5 cos (a) + 2 cos^{2} (a) - 1

- (1 - cos (a)) + cos^{2} (a) \cdot 2 + 4 cos (a)

= - (1 - cos (a)) + 2 cos^{2} (a) + 4 cos (a)

- (1 - cos (a)) : - 1 + cos (a)

- (1 - cos (a))

Distribute parentheses

= - (1) - (- cos (a))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + cos (a)

= - 1 + cos (a) + cos^{2} (a) \cdot 2 + 4 cos (a)

Simplify

- 1 + cos (a) + cos^{2} (a) \cdot 2 + 4 cos (a) : 5 cos (a) + 2 cos^{2} (a) - 1

- 1 + cos (a) + cos^{2} (a) \cdot 2 + 4 cos (a)

Group like terms

= cos (a) + 2 cos^{2} (a) + 4 cos (a) - 1

Add similar elements:

cos (a) + 4 cos (a) = 5 cos (a)

= 5 cos (a) + 2 cos^{2} (a) - 1

= 5 cos (a) + 2 cos^{2} (a) - 1

= \frac{5 cos ( a ) + 2 cos ^{2} ( a ) - 1}{2}

= \frac{2 cos ^{2} ( a ) + 5 cos ( a ) - 1}{2} + 2

= \frac{5 cos ( a ) + 2 cos ^{2} ( a ) - 1}{2} + 2

2 + \frac{- 1 + 2 cos ^{2} ( a ) + 5 cos ( a )}{2} = 0

Solve by substitution

2 + \frac{- 1 + 2 cos ^{2} ( a ) + 5 cos ( a )}{2} = 0

Let:

cos (a) = u

2 + \frac{- 1 + 2 u ^{2} + 5 u}{2} = 0

2 + \frac{- 1 + 2 u ^{2} + 5 u}{2} = 0 : u = - 1, u = - \frac{3}{2}

2 + \frac{- 1 + 2 u ^{2} + 5 u}{2} = 0

Multiply both sides by

2

2 + \frac{- 1 + 2 u ^{2} + 5 u}{2} = 0

Multiply both sides by

2

2 \cdot 2 + \frac{- 1 + 2 u ^{2} + 5 u}{2} \cdot 2 = 0 \cdot 2

Simplify

4 - 1 + 2 u^{2} + 5 u = 0

4 - 1 + 2 u^{2} + 5 u = 0

2 u^{2} + 5 u + 3 = 0

Solve with the quadratic formula

2 u^{2} + 5 u + 3 = 0

Quadratic Equation Formula:

For

a = 2, b = 5, c = 3

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

5^{2} - 4 \cdot 2 \cdot 3 = 1

5^{2} - 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 5 \pm 1}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 5 + 1}{2 \cdot 2}, u_{2} = \frac{- 5 - 1}{2 \cdot 2}

u = \frac{- 5 + 1}{2 \cdot 2} : - 1

\frac{- 5 + 1}{2 \cdot 2}

Add/Subtract the numbers:

- 5 + 1 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- 5 - 1}{2 \cdot 2} : - \frac{3}{2}

\frac{- 5 - 1}{2 \cdot 2}

Subtract the numbers:

- 5 - 1 = - 6

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 6}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{6}{4}

Cancel the common factor:

2

= - \frac{3}{2}

The solutions to the quadratic equation are:

u = - 1, u = - \frac{3}{2}

Substitute back

u = cos (a)

cos (a) = - 1, cos (a) = - \frac{3}{2}

cos (a) = - 1, cos (a) = - \frac{3}{2}

cos (a) = - 1 : a = π + 2 πn

cos (a) = - 1

General solutions for

cos (a) = - 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}

a = π + 2 πn

a = π + 2 πn

cos (a) = - \frac{3}{2} :

No Solution

cos (a) = - \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

a = π + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(a)+2cos(a)=sin^2(a/2)-2 ?
The general solution for cos^2(a)+2cos(a)=sin^2(a/2)-2 is a=pi+2pin