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

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

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sin^2(φ)+cos^2(φ)=2

Solution

sin^{2} (φ) + cos^{2} (φ) = 2

Solution

No Solution for φ \in R

Solution steps

sin^{2} (φ) + cos^{2} (φ) = 2

Subtract

cos^{2} (φ)

from both sides

sin^{2} (φ) = 2 - cos^{2} (φ)

Square both sides

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

Subtract

(2 - cos^{2} (φ))^{2}

from both sides

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

Apply exponent rule:

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

(1 - cos^{2} (φ)) (1 - cos^{2} (φ))

Apply exponent rule:

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

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

= (1 - cos^{2} (φ))^{1 + 1}

Add the numbers:

1 + 1 = 2

= (1 - cos^{2} (φ))^{2}

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

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

Apply Perfect Square Formula:

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

a = 1, b = cos^{2} (φ)

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

2 \cdot 1 \cdot cos^{2} (φ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos^{2} (φ)

(cos^{2} (φ))^{2} = cos^{4} (φ)

(cos^{2} (φ))^{2}

Apply exponent rule:

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

= cos^{2 \cdot 2} (φ)

Multiply the numbers:

2 \cdot 2 = 4

= cos^{4} (φ)

= 1 - 2 cos^{2} (φ) + cos^{4} (φ)

= 1 - 2 cos^{2} (φ) + cos^{4} (φ)

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add similar elements:

- cos^{4} (φ) + cos^{4} (φ) = 0

= 2 cos^{2} (φ) - 4 + 1

Add/Subtract the numbers:

- 4 + 1 = - 3

= 2 cos^{2} (φ) - 3

= 2 cos^{2} (φ) - 3

= 2 cos^{2} (φ) - 3

- 3 + 2 cos^{2} (φ) = 0

Solve by substitution

- 3 + 2 cos^{2} (φ) = 0

Let:

cos (φ) = 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 (φ)

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

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

cos (φ) = \frac{3}{2} :

No Solution

cos (φ) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (φ) = - \frac{3}{2} :

No Solution

cos (φ) = - \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} (φ) + cos^{2} (φ) = 2

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

No Solution for φ \in R

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Frequently Asked Questions (FAQ)

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