What is the general solution for cos^2(β)-cot^2(β)=cos^2(β)cot^2(β) ?

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

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Popular Trigonometry >

cos^2(β)-cot^2(β)=cos^2(β)cot^2(β)

Solution

cos^{2} (β) - cot^{2} (β) = cos^{2} (β) cot^{2} (β)

Solution

β = \frac{π}{2} + πn

Degrees

β = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

cos^{2} (β) - cot^{2} (β) = cos^{2} (β) cot^{2} (β)

Subtract

cos^{2} (β) cot^{2} (β)

from both sides

cos^{2} (β) - cot^{2} (β) - cos^{2} (β) cot^{2} (β) = 0

Express with sin, cos

cos^{2} (β) - cot^{2} (β) - cos^{2} (β) cot^{2} (β)

Use the basic trigonometric identity:

cot (x) = \frac{cos ( x )}{sin ( x )}

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

Simplify

cos^{2} (β) - (\frac{cos ( β )}{sin ( β )})^{2} - cos^{2} (β) (\frac{cos ( β )}{sin ( β )})^{2} : \frac{cos ^{2} ( β ) sin ^{2} ( β ) - cos ^{2} ( β ) - cos ^{4} ( β )}{sin ^{2} ( β )}

cos^{2} (β) - (\frac{cos ( β )}{sin ( β )})^{2} - cos^{2} (β) (\frac{cos ( β )}{sin ( β )})^{2}

(\frac{cos ( β )}{sin ( β )})^{2} = \frac{cos ^{2} ( β )}{sin ^{2} ( β )}

(\frac{cos ( β )}{sin ( β )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{cos ^{2} ( β )}{sin ^{2} ( β )}

cos^{2} (β) (\frac{cos ( β )}{sin ( β )})^{2} = \frac{cos ^{4} ( β )}{sin ^{2} ( β )}

cos^{2} (β) (\frac{cos ( β )}{sin ( β )})^{2}

(\frac{cos ( β )}{sin ( β )})^{2} = \frac{cos ^{2} ( β )}{sin ^{2} ( β )}

(\frac{cos ( β )}{sin ( β )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{cos ^{2} ( β )}{sin ^{2} ( β )}

= \frac{cos ^{2} ( β )}{sin ^{2} ( β )} cos^{2} (β)

Multiply fractions:

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

= \frac{cos ^{2} ( β ) cos ^{2} ( β )}{sin ^{2} ( β )}

cos^{2} (β) cos^{2} (β) = cos^{4} (β)

cos^{2} (β) cos^{2} (β)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos^{2} (β) cos^{2} (β) = cos^{2 + 2} (β)

= cos^{2 + 2} (β)

Add the numbers:

2 + 2 = 4

= cos^{4} (β)

= \frac{cos ^{4} ( β )}{sin ^{2} ( β )}

= cos^{2} (β) - \frac{cos ^{2} ( β )}{sin ^{2} ( β )} - \frac{cos ^{4} ( β )}{sin ^{2} ( β )}

Combine the fractions

- \frac{cos ^{2} ( β )}{sin ^{2} ( β )} - \frac{cos ^{4} ( β )}{sin ^{2} ( β )} : \frac{- cos ^{2} ( β ) - cos ^{4} ( β )}{sin ^{2} ( β )}

Apply rule

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

= \frac{- cos ^{2} ( β ) - cos ^{4} ( β )}{sin ^{2} ( β )}

= cos^{2} (β) + \frac{- cos ^{4} ( β ) - cos ^{2} ( β )}{sin ^{2} ( β )}

Convert element to fraction:

cos^{2} (β) = \frac{cos ^{2} ( β ) sin ^{2} ( β )}{sin ^{2} ( β )}

= \frac{cos ^{2} ( β ) sin ^{2} ( β )}{sin ^{2} ( β )} + \frac{- cos ^{2} ( β ) - cos ^{4} ( β )}{sin ^{2} ( β )}

Since the denominators are equal, combine the fractions:

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

= \frac{cos ^{2} ( β ) sin ^{2} ( β ) - cos ^{2} ( β ) - cos ^{4} ( β )}{sin ^{2} ( β )}

= \frac{cos ^{2} ( β ) sin ^{2} ( β ) - cos ^{2} ( β ) - cos ^{4} ( β )}{sin ^{2} ( β )}

\frac{- cos ^{2} ( β ) - cos ^{4} ( β ) + cos ^{2} ( β ) sin ^{2} ( β )}{sin ^{2} ( β )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- cos^{2} (β) - cos^{4} (β) + cos^{2} (β) sin^{2} (β) = 0

Factor

- cos^{2} (β) - cos^{4} (β) + cos^{2} (β) sin^{2} (β) : - cos^{2} (β) (1 + cos^{2} (β) - sin^{2} (β))

- cos^{2} (β) - cos^{4} (β) + cos^{2} (β) sin^{2} (β)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

cos^{4} (β) = cos^{2} (β) cos^{2} (β)

= - cos^{2} (β) - cos^{2} (β) cos^{2} (β) + sin^{2} (β) cos^{2} (β)

Factor out common term

- cos^{2} (β)

= - cos^{2} (β) (1 + cos^{2} (β) - sin^{2} (β))

- cos^{2} (β) (1 + cos^{2} (β) - sin^{2} (β)) = 0

Solving each part separately

cos^{2} (β) = 0 or 1 + cos^{2} (β) - sin^{2} (β) = 0

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

cos^{2} (β) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (β) = 0

General solutions for

cos (β) = 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}

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

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

1 + cos^{2} (β) - sin^{2} (β) = 0 : β = \frac{π}{2} + πn

1 + cos^{2} (β) - sin^{2} (β) = 0

Rewrite using trig identities

1 + cos^{2} (β) - sin^{2} (β)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

= 1 + cos (2 β)

1 + cos (2 β) = 0

Move

1

to the right side

1 + cos (2 β) = 0

Subtract

1

from both sides

1 + cos (2 β) - 1 = 0 - 1

Simplify

cos (2 β) = - 1

cos (2 β) = - 1

General solutions for

cos (2 β) = - 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 β = π + 2 πn

2 β = π + 2 πn

Solve

2 β = π + 2 πn : β = \frac{π}{2} + πn

2 β = π + 2 πn

Divide both sides by

2

2 β = π + 2 πn

Divide both sides by

2

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

Simplify

β = \frac{π}{2} + πn

β = \frac{π}{2} + πn

β = \frac{π}{2} + πn

Combine all the solutions

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

Merge Overlapping Intervals

β = \frac{π}{2} + πn

Graph

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Frequently Asked Questions (FAQ)

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