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Math  /  Trigonometry

Question4. sin⁡(2θ)=2cos⁡θ\sin (2 \theta)=\sqrt{2} \cos \theta

Studdy Solution

STEP 1

What is this asking? We need to find the values of θ\theta that make the equation sin⁡(2θ)=2cos⁡θ\sin(2\theta) = \sqrt{2} \cos\theta true. Watch out! Remember your double-angle formulas, and don't divide by cos⁡θ\cos\theta without considering what happens if cos⁡θ=0\cos\theta = 0!

STEP 2

1. Rewrite using the double-angle formula
2. Factor
3. Solve for each factor

STEP 3

We know the double-angle formula for sine is sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta.
Let's **substitute** that into our equation: 2sin⁡θcos⁡θ=2cos⁡θ2\sin\theta\cos\theta = \sqrt{2}\cos\theta

STEP 4

To solve this, we want to set the equation equal to **zero**.
We can do this by subtracting 2cos⁡θ\sqrt{2}\cos\theta from both sides: 2sin⁡θcos⁡θ−2cos⁡θ=02\sin\theta\cos\theta - \sqrt{2}\cos\theta = 0

STEP 5

Notice that both terms have a common factor of cos⁡θ\cos\theta.
Let's **factor** that out! cos⁡θ(2sin⁡θ−2)=0\cos\theta(2\sin\theta - \sqrt{2}) = 0

STEP 6

Now, we have two factors multiplied together equaling zero.
This means either one or both of the factors must be zero.
So, let's set the first factor equal to zero: cos⁡θ=0\cos\theta = 0 This happens when θ=π2+nπ\theta = \frac{\pi}{2} + n\pi, where nn is any integer.

STEP 7

Now, let's set the second factor equal to zero: 2sin⁡θ−2=02\sin\theta - \sqrt{2} = 0

STEP 8

**Add** 2\sqrt{2} to both sides: 2sin⁡θ=22\sin\theta = \sqrt{2}

STEP 9

**Divide** both sides by 22 to isolate sin⁡θ\sin\theta: sin⁡θ=22\sin\theta = \frac{\sqrt{2}}{2} This happens when θ=π4+2nπ\theta = \frac{\pi}{4} + 2n\pi and θ=3π4+2nπ\theta = \frac{3\pi}{4} + 2n\pi, where nn is any integer.

STEP 10

The solutions to the equation sin⁡(2θ)=2cos⁡θ\sin(2\theta) = \sqrt{2}\cos\theta are θ=π2+nπ\theta = \frac{\pi}{2} + n\pi, θ=π4+2nπ\theta = \frac{\pi}{4} + 2n\pi, and θ=3π4+2nπ\theta = \frac{3\pi}{4} + 2n\pi, where nn is any integer.

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