How do you solve for these identities:

Question

How do you solve for these identities:

1-sin^2è/cosè= cosè

sin^4è-cos^4è=2sin^2è-1

Answer 1

that e symbol is suppose to be theta.

Answer 2

LS

= 1-sin^2Ø/cosØ
= cos^2 Ø/cos
= cos Ø = RS

Ls
= sin^4Ø-cos^4Ø
= (sin^2Ø + cos^2Ø)(sin^2Ø - cos^2Ø)
= 1( sin^2Ø - (1-sin^2Ø))
= 2sin^2Ø - 1
= RS

Answer 3

To solve the given identities, we'll break down the steps for each one:

1) 1 - sin^2θ / cosθ = cosθ

To solve this identity, we'll start by simplifying the left-hand side (LHS) of the equation:

1 - sin^2θ / cosθ = cosθ

Multiplying both the numerator and the denominator of the fraction by cosθ, we have:

(cosθ - sin^2θ) / cosθ = cosθ

Expanding the numerator, we get:

cosθ - sin^2θ = cosθ * cosθ

Using the Pythagorean identity sin^2θ + cos^2θ = 1, we substitute sin^2θ with 1 - cos^2θ:

cosθ - (1 - cos^2θ) = cosθ * cosθ

Rearranging the terms, we have:

cosθ - 1 + cos^2θ = cos^2θ

Moving all terms to one side of the equation, we get:

2cos^2θ - cosθ - 1 = 0

Now, we have a quadratic equation in terms of cosθ. We can factorize it or use the quadratic formula to solve for cosθ. Once we find the values of cosθ, we can substitute them back into the original equation to check if they satisfy the identity.

2) sin^4θ - cos^4θ = 2sin^2θ - 1

To solve this identity, we'll apply the identity x^4 - y^4 = (x^2 + y^2)(x^2 - y^2):

(sin^2θ + cos^2θ)(sin^2θ - cos^2θ) = 2sin^2θ - 1

Using the Pythagorean identity sin^2θ + cos^2θ = 1, the equation simplifies to:

1(sin^2θ - cos^2θ) = 2sin^2θ - 1

Expanding the left-hand side (LHS) and rearranging the terms, we get:

sin^2θ - cos^2θ = 2sin^2θ - 1

Moving all terms to one side of the equation, we have:

3sin^2θ - cos^2θ - 1 = 0

Again, we have a quadratic equation, this time in terms of sin^2θ and cos^2θ. We can solve it by factoring or using the quadratic formula. Once we find the solutions, we can substitute them back into the original equation to check if they satisfy the identity.

Remember to always check the solutions to ensure they satisfy the original equation, as sometimes there could be extraneous solutions or restrictions on the variables.