Simplify sin(x/2)-cos(x/2)=1?

Multiply it with 1/sqrt 2. You get
sin x/2 *1/sqrt 2 - cos x/2 1/sqrt 2 = 1/sqrt 2
Now put the values for 1/sqrt 2
sin x/2 cos pi/4- cos x/2 son pi/4= sin (x/2-pi/4) = sin pi/4
This gives
x/2 -pi/4= pi/4
x/2 = pi/2, x=pi

sin(x/2)-cos(x/2)=1
=>
sin(x/2)cos(7pi/4)-cos(x/2)sin(7pi/4)=1
=>
sin(x/2-7pi/4)=1

or

sin(x/2)-cos(x/2)=1
=>
sin(x/2)=1+cos(x/2)
=>
2sin(x/4)cos(x/4)=2cos^2(x/4)
=>
sin(x/4)=cos(x/4)
=>
tan(x/4)=1

Do you know this identity?

sin(a - b) = sin(a).cos(b) - cos(a).sin(b) → suppose that: a = (x/2) and tha: b = φ

sin[(x/2) - φ] = sin(x/2).cos(φ) - cos(x/2).sin(φ) → you multiply by C

C.sin[(x/2) - φ] = C.sin(x/2).cos(φ) - C.cos(x/2).sin(φ) → let: C.cos(φ) = 1 ← equation (1)

C.sin[(x/2) - φ] = sin(x/2) - C.cos(x/2).sin(φ) → let: C.sin(φ) = 1 ← equation (2)

C.sin[(x/2) - φ] = sin(x/2) - cos(x/2) ← memorize this expression


You can get a system of 2 equations:

(1) : C.cos(φ) = 1

(2) : C.sin(φ) = 1


You calculate (2)/(1)

[C.sin(φ)]/[C.cos(φ)] = 1/1

sin(φ)/cos(φ) = 1

tan(φ) = 1

φ = π/4


You calculate (1)² + (2)²

[C.cos(φ)]² + [C.sin(φ)]² = 1² + 1²

C².cos²(φ) + C².sin²(φ) = 2

C².[cos²(φ) + sin²(φ)] = 2 → recall the famous formula: cos²(φ) + sin²(φ) = 1

C² = 2

C = √2


Restart from your equation:

sin(x/2) - cos(x/2) = 1 → recall the memorized expression

C.sin[(x/2) - φ] = 1

sin[(x/2) - φ] = 1/C → we’ve seen that: φ = π/4

sin[(x/2) - (π/4)] = 1/C → we’ve seen that: C = √2

sin[(x/2) - (π/4)] = 1/√2

sin[(x/2) - (π/4)] = (√2)/2 ← the corresponding angle is: (π/4) or [π - (π/4)] i.e. (3π/4)



First case: the corresponding angle is (π/4)

(x/2) - (π/4) = (π/4) + 2kπ → where k is an integer

x/2 = (π/2) + 2kπ

x = π + 4kπ

When: k = 0 → x = π

When: k = 1 → x = 5π

When: k = 2 → x = 9π

…



Second case: the corresponding angle is (3π/4)

(x/2) - (π/4) = (3π/4) + 2kπ → where k is an integer

x/2 = π + 2kπ

x = 2π + 4kπ


When: k = 0 → x = 2π

When: k = 1 → x = 6π

When: k = 2 → x = 10π

…

sin(x/2) - cos(x/2) = 1
cos(x/2) + 1 = sin(x/2)
x ≈ 12.566 n + 3.1416, n element Z

sin(x/2)-cos(x/2) = 1
√2 sin(x/2-π/4) = 1
sin(x/2-π/4) = 1/√2
x/2-π/4 = π/4, 3π/4
x/2 = π/2, π
x = π, 2π

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

sqrt(2)*sin((x/2) - (pi/4)) = 1;

sin((x/2) - (pi/4)) = 1/sqrt(2);

(x/2) - (pi/4) = pi/4;

x/2 = pi/2; x = pi;

I have x = 4n(pi) + pi or 4n(pi) + 2pi; factored out they are x = (4n + 1)pi or 2pi(2n + 1). I'll have to show you the math later because I have to work, but I will show you unless someone else beats me to it. For both solutions, assume that n is an integer.

*EDIT*

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

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

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

-2 sin(x/2)cos(x/2) = 0

-sin(2(x/2)) = 0

-sin(x) = 0 ---> x = n * pi, assuming n is an integer. Right? Well, not exactly.

If n = 0, sin(0/2) - cos(0/2) = 0 - 1 = -1. Fail.
If n = 1, sin(pi/2) - cos(pi/2) = 1 - 0 = 1. Good.
If n = 2, sin(pi) - cos(pi) = 0 - (-1) = 1. Good.
If n = 3, sin(3pi/2) - cos(3pi/2) = -1 - 0 = -1. Fail.

That means you have a pattern of two valid solutions, followed by two extraneous solutions, then two valid, then two extraneous, and so on. That means n = 4 doesn't work, as 4pi/2 = 2pi, making x coterminal with 0. n = 7 also doesn't work, as 7pi/2 is coterminal with 3pi/2. But n = 5 and n = 6 work, as they would end up being coterminal angles with the middle two cases (n = 1 and n = 2). Since you have half-angles in play, 2pi can (and/or must) be expressed as 4pi/2. Therefore, you get pi and every 2pi after it (x = 2pi(2n + 1) = 4n(pi) + 2pi), or every multiple of pi that is one more than a multiple of 4 (x = (4n + 1)pi = 4n(pi) + pi).

I hope this helped somewhat; if you are still confused, by all means, seek additional opinions because there are others who can explain or work this better than me. Good luck to you!!!! :)

It's an equation not an identity.
By inspection, at sin 0 = 0 and cos 0 = 1., so x = 0 (and 360)

Let u = x/2 =>
sin(u) = 1 + cos(u), or
sin^2(u) = 1 + 2*cos(u) + cos^2(u), or
1 - cos^2(u) = 1 + 2*cos(u) + cos^2(u), or
0 = 2*cos(u) + 3*cos^2(u).
Hence, cos(u) = 0 or else -2/3 = cos(u).
If cos(u) = 0, then u = +/-pi/2, and x = +/- pi/4, +/- 3pi/4, etc.
If cos(u) = -2/3, then x = (1/2)*arccos(-2/3), which has many approximate decimal solutions, such as x = 1.1503 radians.