What is the general solution for sin^4(θ)-cos^4(θ)=0 ?

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

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Popular >

sin^4(θ)-cos^4(θ)=0

Solution

sin^{4} (θ) - cos^{4} (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^{4} (θ) - cos^{4} (θ) = 0

Factor

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

sin^{4} (θ) - cos^{4} (θ)

Rewrite

sin^{4} (θ) - cos^{4} (θ)

(sin^{2} (θ))^{2} - (cos^{2} (θ))^{2}

sin^{4} (θ) - cos^{4} (θ)

Apply exponent rule:

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

sin^{4} (θ) = (sin^{2} (θ))^{2}

= (sin^{2} (θ))^{2} - cos^{4} (θ)

Apply exponent rule:

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

cos^{4} (θ) = (cos^{2} (θ))^{2}

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

Factor

sin^{2} (θ) - cos^{2} (θ) : (sin (θ) + cos (θ)) (sin (θ) - cos (θ))

sin^{2} (θ) - cos^{2} (θ)

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (sin (θ) + cos (θ)) (sin (θ) - cos (θ))

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

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

Rewrite using trig identities

(sin^{2} (θ) + cos^{2} (θ)) (sin (θ) + cos (θ)) (sin (θ) - cos (θ))

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= (- cos (θ) + sin (θ)) (cos (θ) + sin (θ)) \cdot 1

Simplify

(- cos (θ) + sin (θ)) (cos (θ) + sin (θ)) \cdot 1 : (- cos (θ) + sin (θ)) (cos (θ) + sin (θ))

(- cos (θ) + sin (θ)) (cos (θ) + sin (θ)) \cdot 1

Multiply:

(cos (θ) + sin (θ)) \cdot 1 = (cos (θ) + sin (θ))

= (sin (θ) - cos (θ)) (cos (θ) + sin (θ))

= (- cos (θ) + sin (θ)) (cos (θ) + sin (θ))

(- cos (θ) + sin (θ)) (cos (θ) + sin (θ)) = 0

Solving each part separately

- cos (θ) + sin (θ) = 0 or cos (θ) + sin (θ) = 0

- cos (θ) + sin (θ) = 0 : θ = \frac{π}{4} + πn

- cos (θ) + sin (θ) = 0

Rewrite using trig identities

- cos (θ) + sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{- cos ( θ ) + sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

- 1 + \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

- 1 + tan (θ) = 0

- 1 + tan (θ) = 0

Move

1

to the right side

- 1 + tan (θ) = 0

Add

1

to both sides

- 1 + tan (θ) + 1 = 0 + 1

Simplify

tan (θ) = 1

tan (θ) = 1

General solutions for

tan (θ) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

cos (θ) + sin (θ) = 0 : θ = \frac{3 π}{4} + πn

cos (θ) + sin (θ) = 0

Rewrite using trig identities

cos (θ) + sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{cos ( θ ) + sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

1 + \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

1 + tan (θ) = 0

1 + tan (θ) = 0

Move

1

to the right side

1 + tan (θ) = 0

Subtract

1

from both sides

1 + tan (θ) - 1 = 0 - 1

Simplify

tan (θ) = - 1

tan (θ) = - 1

General solutions for

tan (θ) = - 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Combine all the solutions

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