# 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!

Update:

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.

Relevance

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!

• Log in to reply to the answers
• 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.

• Log in to reply to the answers
• Anonymous
4 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

• Log in to reply to the answers
• Anonymous
4 years ago

1 Sinx

• Log in to reply to the answers
• 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

• Log in to reply to the answers
• 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°

• Log in to reply to the answers
• 1+sin(x)/n

1+sin(x)/n, cross out both n's

1+six=7

• Log in to reply to the answers
• multiply both sides by (1- sin x) and take it from t here johnny boy

• Log in to reply to the answers
• Anonymous
7 years ago

sin3x*sinx

• Log in to reply to the answers