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Math  /  Trigonometry

QuestionEvaluate sec⁡−1(−2)\sec ^{-1}(-\sqrt{2}). Enter an exact answer in radians.
Provide your answer below: sec⁡−1(−2)=\sec ^{-1}(-\sqrt{2})= □\square rad

Studdy Solution

STEP 1

1. The secant inverse function sec⁡−1(x)\sec^{-1}(x) gives an angle θ\theta such that sec⁡(θ)=x\sec(\theta) = x.
2. The secant function sec⁡(θ)\sec(\theta) is the reciprocal of the cosine function, i.e., sec⁡(θ)=1cos⁡(θ)\sec(\theta) = \frac{1}{\cos(\theta)}.
3. The principal range of the secant inverse function sec⁡−1(x)\sec^{-1}(x) is [0,π]∖{π2}[0, \pi] \setminus \left\{\frac{\pi}{2}\right\}.

STEP 2

1. Express the problem in terms of the secant function.
2. Determine the angle θ\theta such that sec⁡(θ)=−2\sec(\theta) = -\sqrt{2}.
3. Verify that the obtained angle is within the principal range of the secant inverse function.
4. Provide the exact value of θ\theta in radians.

STEP 3

Express the problem sec⁡−1(−2)\sec^{-1}(-\sqrt{2}) using the definition of the secant function.
We need to find θ\theta such that sec⁡(θ)=−2\sec(\theta) = -\sqrt{2}.

STEP 4

Recall that sec⁡(θ)=1cos⁡(θ)\sec(\theta) = \frac{1}{\cos(\theta)}. Therefore, we need 1cos⁡(θ)=−2\frac{1}{\cos(\theta)} = -\sqrt{2}.
This implies: cos⁡(θ)=−12=−22 \cos(\theta) = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}

STEP 5

Identify the angle θ\theta for which cos⁡(θ)=−22\cos(\theta) = -\frac{\sqrt{2}}{2}.
The cosine function equals −22-\frac{\sqrt{2}}{2} at θ=3π4\theta = \frac{3\pi}{4} and θ=5π4\theta = \frac{5\pi}{4}. However, only one of these is in the principal range of sec⁡−1(x)\sec^{-1}(x).

STEP 6

Verify that the angle θ=3π4\theta = \frac{3\pi}{4} is within the principal range [0,π]∖{π2}[0, \pi] \setminus \left\{\frac{\pi}{2}\right\}.
Since 3π4\frac{3\pi}{4} is within this range, it is the valid solution.

STEP 7

The exact value of sec⁡−1(−2)\sec^{-1}(-\sqrt{2}) in radians is: sec⁡−1(−2)=3π4 \sec^{-1}(-\sqrt{2}) = \frac{3\pi}{4}
Provide the answer in the desired format: sec⁡−1(−2)=3π4 rad\sec ^{-1}(-\sqrt{2})=\frac{3\pi}{4} \text{ rad}

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