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

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

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Popular >

sin^2(θ)+cos^2(θ)=2

Solution

sin^{2} (θ) + cos^{2} (θ) = 2

Solution

No Solution for θ \in R

Solution steps

sin^{2} (θ) + cos^{2} (θ) = 2

Subtract

cos^{2} (θ)

from both sides

sin^{2} (θ) = 2 - cos^{2} (θ)

Square both sides

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

Subtract

(2 - cos^{2} (θ))^{2}

from both sides

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

Apply exponent rule:

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- 4 - cos^{4} (θ) + 4 cos^{2} (θ) + (1 - cos^{2} (θ)) (1 - cos^{2} (θ)) : 2 cos^{2} (θ) - 3

- 4 - cos^{4} (θ) + 4 cos^{2} (θ) + (1 - cos^{2} (θ)) (1 - cos^{2} (θ))

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

(1 - cos^{2} (θ)) (1 - cos^{2} (θ))

Apply exponent rule:

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

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

= (1 - cos^{2} (θ))^{1 + 1}

Add the numbers:

1 + 1 = 2

= (1 - cos^{2} (θ))^{2}

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

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

Apply Perfect Square Formula:

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

a = 1, b = cos^{2} (θ)

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

2 \cdot 1 \cdot cos^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos^{2} (θ)

(cos^{2} (θ))^{2} = cos^{4} (θ)

(cos^{2} (θ))^{2}

Apply exponent rule:

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

= cos^{2 \cdot 2} (θ)

Multiply the numbers:

2 \cdot 2 = 4

= cos^{4} (θ)

= 1 - 2 cos^{2} (θ) + cos^{4} (θ)

= 1 - 2 cos^{2} (θ) + cos^{4} (θ)

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

Simplify

- 4 - cos^{4} (θ) + 4 cos^{2} (θ) + 1 - 2 cos^{2} (θ) + cos^{4} (θ) : 2 cos^{2} (θ) - 3

- 4 - cos^{4} (θ) + 4 cos^{2} (θ) + 1 - 2 cos^{2} (θ) + cos^{4} (θ)

Group like terms

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

Add similar elements:

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

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

Add similar elements:

- cos^{4} (θ) + cos^{4} (θ) = 0

= 2 cos^{2} (θ) - 4 + 1

Add/Subtract the numbers:

- 4 + 1 = - 3

= 2 cos^{2} (θ) - 3

= 2 cos^{2} (θ) - 3

= 2 cos^{2} (θ) - 3

- 3 + 2 cos^{2} (θ) = 0

Solve by substitution

- 3 + 2 cos^{2} (θ) = 0

Let:

cos (θ) = u

- 3 + 2 u^{2} = 0

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

- 3 + 2 u^{2} = 0

Move

3

to the right side

- 3 + 2 u^{2} = 0

Add

3

to both sides

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

Simplify

2 u^{2} = 3

2 u^{2} = 3

Divide both sides by

2

2 u^{2} = 3

Divide both sides by

2

\frac{2 u ^{2}}{2} = \frac{3}{2}

Simplify

u^{2} = \frac{3}{2}

u^{2} = \frac{3}{2}

For

x^{2} = f (a)

the solutions are

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

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

Substitute back

u = cos (θ)

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

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

cos (θ) = \frac{3}{2} :

No Solution

cos (θ) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = - \frac{3}{2} :

No Solution

cos (θ) = - \frac{3}{2}

- 1 \leq cos (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

sin^{2} (θ) + cos^{2} (θ) = 2

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