What is the general solution for sin^2(θ)-cos^2(θ)=0 ?

The general solution for sin^2(θ)-cos^2(θ)=0 is θ= pi/4+pin,θ=(3pi)/4+pin

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sin^2(θ)-cos^2(θ)=0

Solution

sin^{2} (θ) - cos^{2} (θ) = 0

Solution

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

Degrees

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

Solution steps

sin^{2} (θ) - cos^{2} (θ) = 0

Rewrite using trig identities

sin^{2} (θ) - cos^{2} (θ)

Use the Double Angle identity:

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

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

= - cos (2 θ)

- cos (2 θ) = 0

Divide both sides by

- 1

- cos (2 θ) = 0

Divide both sides by

- 1

\frac{- cos ( 2 θ )}{- 1} = \frac{0}{- 1}

Simplify

cos (2 θ) = 0

cos (2 θ) = 0

General solutions for

cos (2 θ) = 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 θ = \frac{π}{2} + 2 πn, 2 θ = \frac{3 π}{2} + 2 πn

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

Solve

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

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

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

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Solve

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

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

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

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

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

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(θ)-cos^2(θ)=0 ?
The general solution for sin^2(θ)-cos^2(θ)=0 is θ= pi/4+pin,θ=(3pi)/4+pin