What is the general solution for cos^2(θ)+sin^2(θ)=4 ?

The general solution for cos^2(θ)+sin^2(θ)=4 is No Solution for θ\in\mathbb{R}

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cos^2(θ)+sin^2(θ)=4

Solution

cos^{2} (θ) + sin^{2} (θ) = 4

Solution

No Solution for θ \in R

Solution steps

cos^{2} (θ) + sin^{2} (θ) = 4

Subtract

sin^{2} (θ)

from both sides

cos^{2} (θ) = 4 - sin^{2} (θ)

Square both sides

(cos^{2} (θ))^{2} = (4 - sin^{2} (θ))^{2}

Subtract

(4 - sin^{2} (θ))^{2}

from both sides

cos^{4} (θ) - 16 + 8 sin^{2} (θ) - sin^{4} (θ) = 0

Apply exponent rule:

a^{b} = a^{2} a^{b - 2}

- 16 - sin^{4} (θ) + 8 sin^{2} (θ) + cos^{2} (θ) cos^{2} (θ) = 0

Rewrite using trig identities

- 16 - sin^{4} (θ) + 8 sin^{2} (θ) + cos^{2} (θ) cos^{2} (θ)

Use the Pythagorean identity:

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

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

= - 16 - sin^{4} (θ) + 8 sin^{2} (θ) + (1 - sin^{2} (θ)) (1 - sin^{2} (θ))

Simplify

- 16 - sin^{4} (θ) + 8 sin^{2} (θ) + (1 - sin^{2} (θ)) (1 - sin^{2} (θ)) : 6 sin^{2} (θ) - 15

- 16 - sin^{4} (θ) + 8 sin^{2} (θ) + (1 - sin^{2} (θ)) (1 - sin^{2} (θ))

(1 - sin^{2} (θ)) (1 - sin^{2} (θ)) = (1 - sin^{2} (θ))^{2}

(1 - sin^{2} (θ)) (1 - sin^{2} (θ))

Apply exponent rule:

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

(1 - sin^{2} (θ)) (1 - sin^{2} (θ)) = (1 - sin^{2} (θ))^{1 + 1}

= (1 - sin^{2} (θ))^{1 + 1}

Add the numbers:

1 + 1 = 2

= (1 - sin^{2} (θ))^{2}

= - 16 - sin^{4} (θ) + 8 sin^{2} (θ) + (- sin^{2} (θ) + 1)^{2}

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

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = sin^{2} (θ)

= 1^{2} - 2 \cdot 1 \cdot sin^{2} (θ) + (sin^{2} (θ))^{2}

Simplify

1^{2} - 2 \cdot 1 \cdot sin^{2} (θ) + (sin^{2} (θ))^{2} : 1 - 2 sin^{2} (θ) + sin^{4} (θ)

1^{2} - 2 \cdot 1 \cdot sin^{2} (θ) + (sin^{2} (θ))^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 2 \cdot 1 \cdot sin^{2} (θ) + (sin^{2} (θ))^{2}

2 \cdot 1 \cdot sin^{2} (θ) = 2 sin^{2} (θ)

2 \cdot 1 \cdot sin^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 sin^{2} (θ)

(sin^{2} (θ))^{2} = sin^{4} (θ)

(sin^{2} (θ))^{2}

Apply exponent rule:

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

= sin^{2 \cdot 2} (θ)

Multiply the numbers:

2 \cdot 2 = 4

= sin^{4} (θ)

= 1 - 2 sin^{2} (θ) + sin^{4} (θ)

= 1 - 2 sin^{2} (θ) + sin^{4} (θ)

= - 16 - sin^{4} (θ) + 8 sin^{2} (θ) + 1 - 2 sin^{2} (θ) + sin^{4} (θ)

Simplify

- 16 - sin^{4} (θ) + 8 sin^{2} (θ) + 1 - 2 sin^{2} (θ) + sin^{4} (θ) : 6 sin^{2} (θ) - 15

- 16 - sin^{4} (θ) + 8 sin^{2} (θ) + 1 - 2 sin^{2} (θ) + sin^{4} (θ)

Group like terms

= - sin^{4} (θ) + 8 sin^{2} (θ) - 2 sin^{2} (θ) + sin^{4} (θ) - 16 + 1

Add similar elements:

8 sin^{2} (θ) - 2 sin^{2} (θ) = 6 sin^{2} (θ)

= - sin^{4} (θ) + 6 sin^{2} (θ) + sin^{4} (θ) - 16 + 1

Add similar elements:

- sin^{4} (θ) + sin^{4} (θ) = 0

= 6 sin^{2} (θ) - 16 + 1

Add/Subtract the numbers:

- 16 + 1 = - 15

= 6 sin^{2} (θ) - 15

= 6 sin^{2} (θ) - 15

= 6 sin^{2} (θ) - 15

- 15 + 6 sin^{2} (θ) = 0

Solve by substitution

- 15 + 6 sin^{2} (θ) = 0

Let:

sin (θ) = u

- 15 + 6 u^{2} = 0

- 15 + 6 u^{2} = 0 : u = \frac{5}{2}, u = - \frac{5}{2}

- 15 + 6 u^{2} = 0

Move

15

to the right side

- 15 + 6 u^{2} = 0

Add

15

to both sides

- 15 + 6 u^{2} + 15 = 0 + 15

Simplify

6 u^{2} = 15

6 u^{2} = 15

Divide both sides by

6

6 u^{2} = 15

Divide both sides by

6

\frac{6 u ^{2}}{6} = \frac{15}{6}

Simplify

u^{2} = \frac{5}{2}

u^{2} = \frac{5}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{5}{2}, u = - \frac{5}{2}

Substitute back

u = sin (θ)

sin (θ) = \frac{5}{2}, sin (θ) = - \frac{5}{2}

sin (θ) = \frac{5}{2}, sin (θ) = - \frac{5}{2}

sin (θ) = \frac{5}{2} :

No Solution

sin (θ) = \frac{5}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = - \frac{5}{2} :

No Solution

sin (θ) = - \frac{5}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

cos^{2} (θ) + sin^{2} (θ) = 4

Remove the ones that don't agree with the equation.

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(θ)+sin^2(θ)=4 ?
The general solution for cos^2(θ)+sin^2(θ)=4 is No Solution for θ\in\mathbb{R}