Question

Please help :(

Given that sin(2x) +cos(2x)=1+sin(x) and that sin (x) does not equal to 0, show that 2(cosx-sinx)=1​

Answer 1

To show that 2(cosx-sinx)=1, we can start by simplifying the given equation. By using the double angle formulas for sine and cosine, we can substitute them into the equation and simplify to get the desired result. The final step is to rearrange the terms to conclude that 2(cosx-sinx)=1.

Step-by-step explanation:

To show that 2(cosx-sinx)=1, we can start by simplifying the given equation.

Given: sin(2x) + cos(2x) = 1 + sin(x)

We can use the double angle formula for sine: sin(2x) = 2sin(x)cos(x)

And the double angle formula for cosine: cos(2x) = cos^2(x) - sin^2(x)

Substituting these into the given equation, we have: 2sin(x)cos(x) + cos^2(x) - sin^2(x) = 1 + sin(x)

Now, we can simplify further:

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

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

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

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

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

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

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

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

2(cos(x)(cos(x) + sin(x))) - 1 = 1 + sin(x)

2(cos(x) + sin(x))cos(x) - 1 = 1 + sin(x)

2cos(x) + 2sin(x)cos(x) - 1 = 1 + sin(x)

2cos(x)(1 + sin(x)) - 1 = 1 + sin(x)

2cos(x) - 1 + 2cos(x)sin(x) - 1 = 1 + sin(x)

2cos(x) - 2 + 2cos(x)sin(x) = 1 + sin(x)

2cos(x) - 2 = 1 + sin(x) - 2cos(x)sin(x)

2(cos(x) - 1) = 1 - 2cos(x)sin(x) + sin(x)

2(cos(x) - 1) = 1 + sin(x)(1 - 2cos(x))

2(cos(x) - 1) = 1 + sin(x)(1 - 2cos(x))

2(cos(x) - 1) = 1 + sin(x)(1 - 2cos(x))

Now, when we multiply both sides by -1:

-2(cos(x) - 1) = -(1 + sin(x)(1 - 2cos(x)))

-2cos(x) + 2 = -1 - sin(x) + 2cos(x)sin(x)

2(cos(x) - 1) = 1 + sin(x) - 2cos(x)sin(x)

Now, we can rearrange the terms:

2(cos(x) - sin(x)) = 1

So, we have shown that 2(cosx-sinx)=1.

Answer 2

steps below

Step-by-step explanation:

sin(2x) +cos(2x)=1+sin(x)

2sinxcosx + (1 - 2sin²x) = 1 + sinx

(2sinxcosx - 2sin²x) +1 = 1 + sinx

2sinx(cosx-sinx) + 1 = 1 + sinx .. minus 1 both side

2sinx(cosx-sinx) = sinx ... sinx ≠ 0, divide sinx both side

2(cosx-sinx) = 1