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Math  /  Trigonometry

Question4) Determine the values of sin⁡2θ,cos⁡2θ\sin 2 \theta, \cos 2 \theta, and tan⁡2θ\tan 2 \theta, given cos⁡θ=35\cos \theta=\frac{3}{5} and 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2}

Studdy Solution

STEP 1

What is this asking? Given that cos⁡θ\cos \theta is 35\frac{3}{5} and θ\theta is between 00 and π2\frac{\pi}{2}, we need to find sin⁡2θ\sin 2\theta, cos⁡2θ\cos 2\theta, and tan⁡2θ\tan 2\theta. Watch out! Remember the different formulas for cos⁡2θ\cos 2\theta!
Also, make sure you're using the correct quadrant for θ\theta to find sin⁡θ\sin \theta.

STEP 2

1. Find sin⁡θ\sin \theta
2. Calculate sin⁡2θ\sin 2\theta
3. Calculate cos⁡2θ\cos 2\theta
4. Calculate tan⁡2θ\tan 2\theta

STEP 3

We know cos⁡θ=35\cos \theta = \frac{3}{5} and 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2}.
This tells us θ\theta is in the **first quadrant**, so both sin⁡θ\sin \theta and cos⁡θ\cos \theta will be **positive**!

STEP 4

Let's use the Pythagorean identity: sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1.
We're doing this because it connects sine and cosine, and we already know cos⁡θ\cos \theta.

STEP 5

Substitute the value of cos⁡θ\cos \theta into the identity: sin⁡2θ+(35)2=1\sin^2 \theta + \left(\frac{3}{5}\right)^2 = 1 sin⁡2θ+925=1\sin^2 \theta + \frac{9}{25} = 1

STEP 6

To isolate sin⁡2θ\sin^2 \theta, we subtract 925\frac{9}{25} from both sides of the equation.
This gives us: sin⁡2θ=1−925\sin^2 \theta = 1 - \frac{9}{25} sin⁡2θ=2525−925\sin^2 \theta = \frac{25}{25} - \frac{9}{25}sin⁡2θ=1625\sin^2 \theta = \frac{16}{25}

STEP 7

Now, take the square root of both sides to find sin⁡θ\sin \theta: sin⁡θ=±1625=±45\sin \theta = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5} Since θ\theta is in the first quadrant, sin⁡θ\sin \theta is **positive**.
So, sin⁡θ=45\sin \theta = \frac{4}{5}.
Awesome!

STEP 8

We'll use the double-angle formula: sin⁡2θ=2⋅sin⁡θ⋅cos⁡θ\sin 2\theta = 2 \cdot \sin \theta \cdot \cos \theta.
We're using this because we now know both sin⁡θ\sin \theta and cos⁡θ\cos \theta.

STEP 9

Substitute the values we found: sin⁡2θ=2⋅45⋅35\sin 2\theta = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} sin⁡2θ=2425\sin 2\theta = \frac{24}{25}

STEP 10

We can use the double-angle formula: cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = \cos^2 \theta - \sin^2 \theta.
We're using this formula because we already know sin⁡θ\sin \theta and cos⁡θ\cos \theta.

STEP 11

Substitute the values: cos⁡2θ=(35)2−(45)2\cos 2\theta = \left(\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2 cos⁡2θ=925−1625\cos 2\theta = \frac{9}{25} - \frac{16}{25}cos⁡2θ=−725\cos 2\theta = -\frac{7}{25}

STEP 12

We can use the formula: tan⁡2θ=sin⁡2θcos⁡2θ\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta}.
We're using this because we just calculated sin⁡2θ\sin 2\theta and cos⁡2θ\cos 2\theta.

STEP 13

Substitute the values: tan⁡2θ=2425−725\tan 2\theta = \frac{\frac{24}{25}}{-\frac{7}{25}} tan⁡2θ=2425⋅(−257)\tan 2\theta = \frac{24}{25} \cdot \left(-\frac{25}{7}\right)tan⁡2θ=−247\tan 2\theta = -\frac{24}{7}

STEP 14

We found sin⁡2θ=2425\sin 2\theta = \frac{24}{25}, cos⁡2θ=−725\cos 2\theta = -\frac{7}{25}, and tan⁡2θ=−247\tan 2\theta = -\frac{24}{7}.

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