Question

Sin(x-п/4)+cos(x-п/4)=sin2x

Answer 1

Sin(x-п/4)+cos(x-п/4)=sin2x
sinxcosπ/4-cosxsinπ/4+cosxcosπ/4+sinxsinπ/4=sin2x
√2/2*sinx-√2/2*cosx+√2/2*cosx+√2/2*sinx=sin2x
√2*sinx-2sinxcosx=0
√2sinx(1-√2cosx)=0
sinx=0⇒x=πk,k∈x
cosx=1/√2⇒x=+-π/4+2πk,k∈z

Answer 2

To solve the trigonometric equation sin(x - π/4) + cos(x - π/4) = sin(2x), we'll start by using trigonometric identities to simplify the left-hand side of the equation.

1. Trigonometric identities: sin(a - b) = sin(a) * cos(b) - cos(a) * sin(b) cos(a - b) = cos(a) * cos(b) + sin(a) * sin(b)

2. Applying the identities: sin(x - π/4) + cos(x - π/4) = sin(x) * cos(π/4) - cos(x) * sin(π/4) + cos(x) * cos(π/4) + sin(x) * sin(π/4)

3. Simplifying further: sin(x) * (1/√2) - cos(x) * (1/√2) + cos(x) * (1/√2) + sin(x) * (1/√2) = sin(x) * (1/√2) + sin(x) * (1/√2) = √2 * sin(x)

So, the equation becomes: √2 * sin(x) = sin(2x)

Now, we can try to solve for x. Let's do that:

Divide both sides by √2: sin(x) = (1/√2) * sin(2x)

Now, we can use a double-angle trigonometric identity: sin(2x) = 2 * sin(x) * cos(x)

Replace sin(2x) in the equation: sin(x) = (1/√2) * (2 * sin(x) * cos(x))

Next, multiply both sides by √2 to get rid of the denominator: √2 * sin(x) = 2 * sin(x) * cos(x)

Now, there are two possibilities for the equation to be satisfied:

1. sin(x) = 0 (one solution): If sin(x) = 0, then x can be any multiple of π. So, one set of solutions is x = n * π, where n is an integer.

2. cos(x) = 1/√2 (another set of solutions): If cos(x) = 1/√2, then x must be either π/4 or 7π/4 because cosine is positive in the first and fourth quadrants.

So, the final set of solutions for the given equation is: x = n * π, where n is an integer, or x = π/4 + 2 * π * k or x = 7π/4 + 2 * π * k, where k is an integer.