Chapter 7: Q. 75 (page 504)

Solve the equation on the interval $0 \leq θ \leq 2 π$ .
$\sin θ - \cos θ = 1$

Short Answer

Expert verified

On solving the equation, we get,

$θ \in \{\frac{π}{2}, π\}$

Step by step solution

Step 1. Given information.

Consider the given question,

$\sin θ - \cos θ = 1 0 \leq θ \leq 2 π$

Squaring both sides of the equation,

${(\sin θ - \cos θ)}^{2} = 1^{2} \sin^{2} θ + \cos^{2} θ - 2 \sin θ \cdot \cos θ = 1 1 - 2 \sin θ \cdot \cos θ = 1 2 \sin θ \cdot \cos θ = 0$

Then,

$2 \sin θ = 0 \sin θ = 0$

Also,

$2 \cos θ = 0 \cos θ = 0$

Step 2. Solve the equation for the given interval.

Consider the given question,

$\sin θ = 0 \cos θ = 0 0 \leq θ \leq 2 π$

Solving $\sin θ = 0$ for the given interval,

$\sin θ = 0 θ = \sin^{- 1} (0) θ = 0, π$

Solving $\cos θ = 0$ for the given interval,

$\cos θ = 0 θ = \cos^{- 1} = 0 θ = \frac{π}{2}, \frac{3 π}{2}$

Step 3. Check the solutions to see if any are extraneous.

As we squared both sides, we must check the solutions to see if any are extraneous,

If $θ = 0$ , then,

$= \sin 0 - \cos 0 = 0 - 1 = - 1$

If $θ = π$ , then,

$= \sin π - \cos π = 0 - (- 1) = 1$

If $θ = \frac{π}{2}$ , then,

$= \sin \frac{π}{2} - \cos \frac{π}{2} = 1 - 0 = 0$

If $θ = \frac{3 π}{2}$ , then,

$= \sin \frac{3 π}{2} - \cos \frac{3 π}{2} = - 1 - 0 = - 1$

Hence, we see $θ = 0, \frac{3 π}{2}$ are extraneous.

Therefore, the solution set is $θ \in \{\frac{π}{2}, π\}$

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Solve the equation on the interval $0 \leq θ \leq 2 π$ .
$\sin θ - \cos θ = 1$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Solve the equation for the given interval.

Step 3. Check the solutions to see if any are extraneous.

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Solve the equation on the interval 0≤θ≤2π.sinθ-cosθ=1

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Solve the equation for the given interval.

Step 3. Check the solutions to see if any are extraneous.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Calculus

Decision Maths

Discrete Mathematics

Geometry

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Solve the equation on the interval $0 \leq θ \leq 2 π$ .
$\sin θ - \cos θ = 1$