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

Question

1 + sin(x) = cos(x)

How can I solve this...none of the squaring stuff...my teacher won't have it!

germano · Accepted 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!

Scythian1950 · Answer

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.

Anonymous · Answer

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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

Anonymous · Answer

1 Sinx

John · Answer

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

Como · Answer

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°

Cam · Answer

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

1+six=7

asds · Answer

multiply both sides by (1- sin x) and take it from t here johnny boy

Anonymous · Answer

sin3x*sinx

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

9 Answers