Let y = sin t and x = cos t. Then we are given that y - x = 1 or equivalently y = x + 1. From the Pythagorean identity, we must also have y^{2} + x^{2} = 1. Substituting y = x + 1 gives (x+1)^{2} + x^{2} = 1, which simplifies to 2x^{2} + 2x = 0. The last equation holds only when x is either 0 or -1. That is, when cos t = 0 or cos t = -1. These further occur when t = (2k+1)π/2 and t = (2k+1)π, respectively, taking k to be any integer. However, we must reject the values t = (4k+3)π/2 since they do not satisfy the given equation. This leaves t = (4k+1)π/2 or t = (2k+1)π.

Nataly M.

asked • 02/02/20# Find all solutions to equation sin t - cos t = 1

express results in radians

## 2 Answers By Expert Tutors

The range of both sin(t) and cos(t) is [-1, 1], meaning the largest either get is 1 and the smallest either get is -1.

That means that sin(t) - cos(t) can only be 1 if sin(t) = 1 at the same time that cos(t) = 0 ( 1 - 0 = 1) or when sin(t) = 0 at the same time that cos(t) = -1 (0 - -1 = 1).

** Case 1**. sin(t) = 1 AND cos(t) = 0

This only occurs at π/2

** Case 2** sin(t) = 0 AND cos(t) = -1

This only occurs at π

However, the problem gives no restriction to the domain. That means that we need to include both those values and also every 2π increment. So the answer is:

t = π/2 + 2πk and t = π + 2πk where k is any integer

William W.

03/03/20

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Marc N.

03/03/20