# Trigonometric Equation: 1 + sin(x) = cos(x)?

1 + sin(x) = cos(x)

How can I solve this...none of the squaring stuff...my teacher won't have it!

Unfortunately, I cannot give a solution by inspection. Also, how does multiplying by (1-sinx) help at all? You end up with an even more complex equation: cosx(cosx+sinx-1)=0.

### 9 Answers

- germanoLv 71 decade agoBest answer
1 + sin x = cos x

I suggest to express sin x and cos x as functions of tan(x/2):

sin x = [2 tan(x/2)] / [1 + tan²(x/2)]

cos x = [1 - tan²(x/2)] / [1 + tan²(x/2)]

thus

1 + sin x = cos x →

1 + [2 tan(x/2)] / [1 + tan²(x/2)] = [1 - tan²(x/2)] / [1+ tan²(x/2)] →

let us multiply all terms by [1 + tan²(x/2)]:

[1 + tan²(x/2)] + [2 tan(x/2)] = [1 - tan²(x/2)] →

1 + tan²(x/2) + 2 tan(x/2) = 1 - tan²(x/2) →

1 + tan²(x/2) + 2 tan(x/2) - 1 + tan²(x/2) = 0 →

2 tan²(x/2) + 2 tan(x/2) = 0 →

2 tan(x/2) [ tan(x/2) + 1] = 0 →

tan(x/2)₁= 0 → (x₁/2) = 0 + n180°→ x₁= n360°

tan(x/2)₂= - 1 → (x₂/2) = 135°+ n 180°→ x₂= 270°+ n360°

being n a relative integer

Bye!

- Scythian1950Lv 71 decade ago
Let Sin(x) = s. Then, squaring both sides, we have:

1 + 2s + s² = 1 - s²

2s(1+s) = 0

which has roots s = 0, -1. In other words, 1 + Sin(x) = Cos(x) holds true when Sin(x) = 0, and Sin(x) = -1, or when x = 0 + n π, and x = (3/2) π + n π, where n is any positive or negative integer.

The other way to do it is to just plot the functions out, and see how that the above is true. Maybe that's how your teacher wants you to do it, because doing this by "squaring stuff" is just too much excellence in math.

- SanraLv 44 years ago
For the best answers, search on this site https://shorturl.im/GxbP3

Combine trig functions, their periods and horizontal shifts are the same: sin(x) + cos(x) = 1 √(2) ⋅ sin(x + ¼π) = 1 sin(x + ¼π) = ½√(2) Or you could square both sides: sin(x) + cos(x) = 1 [ sin(x) + cos(x) ]² = 1² sin²(x) + 2sin(x)cos(x) + cos²(x) = 1 1 + 2sin(x)cos(x) = 1 2sin(x)cos(x) = 0 sin(2x) = 0 ...... or ....... sin(x) = 0; cos(x) = 0

- What do you think of the answers? You can sign in to give your opinion on the answer.
- JohnLv 71 decade ago
This is one of those that you just have to think about and look at for a while.

Obviously, x = 0 works since

1 + sin(0) = cos(0) which is the same as (1 + 0 = 1.)

As it turns out, x = 0 is the only solution.

-John

- ComoLv 71 decade ago
cos x - sin x = 1

Let

cos x - sin x = k cos (x - α )

cos x - sin x = (k cos α) cos x + (k sin α) sin x

1 = k cos α

- 1 = k sin α----------- (where k > 0)

tan α = - 1 (α in 4th quadrant)

α = 315°

1² + (-1)² = k²

k² = 2

k = √ 2

cos x - sin x = √ 2 cos (x - 315°)

√ 2 cos (x - 315°) = 1

cos (x - 315°) = 1 /√ 2

x - 315° = 45° or 315°

x = 360° or 630°

x = 360° or 270°

- Anonymous6 years ago
sin3x*sinx