Complex Numbers
Trigonometric Functions of Complex Numbers
Grade 11
Question:
<p>If <span>cos</span>(<span>1</span> - <span>i</span>) = <span>a</span> + <span>ib</span>, where <span>a</span>, <span>b</span> ∈ ℝ and <span>i</span> = <span>√-1</span>, then</p>
<p>(a) <span>a</span> = <span>\frac{1}{2}(e^{-1} - e) \cos 1</span>, <span>b</span> = <span>\frac{1}{2}(e + e^{-1}) \sin 1</span></p>
<p>(b) <span>a</span> = <span>\frac{1}{2}(e + e^{-1}) \cos 1</span>, <span>b</span> = <span>\frac{1}{2}(e - e^{-1}) \sin 1</span></p>
<p>(c) <span>a</span> = <span>\frac{1}{2}e^{-1} + \frac{1}{2}e \cos 1</span>, <span>b</span> = <span>\frac{1}{2}e + \frac{1}{2}e^{-1} \sin 1</span></p>
<p>(d) <span>a</span> = <span>\frac{1}{2}(e - e^{-1}) \cos 1</span>, <span>b</span> = <span>\frac{1}{2}(e - e^{-1}) \sin 1</span></p>
Step-by-Step Solution
Key Concept: Use Euler's formula e^(ix) = cos(x) + i·sin(x) to extend cosine to complex arguments. For cos(z) with z = u + iv, apply the formula cos(u + iv) = cos(u)cosh(v) - i·sin(u)sinh(v).
<p><strong>Step 1: Use the extension of cosine to complex numbers</strong></p><p>For a complex number z = u + iv, we have:</p><p>cos(u + iv) = cos(u)cosh(v) - i·sin(u)sinh(v)</p><p>Here u = 1 and v = -1, so:</p><p>cos(1 - i) = cos(1)cosh(-1) - i·sin(1)sinh(-1)</p></p><p><strong>Step 2: Evaluate hyperbolic functions at -1</strong></p><p>Since cosh(-1) = cosh(1) = (e + e^(-1))/2</p><p>And sinh(-1) = -sinh(1) = -(e - e^(-1))/2</p></p><p><strong>Step 3: Substitute values</strong></p><p>cos(1 - i) = cos(1)·(e + e^(-1))/2 - i·sin(1)·(-(e - e^(-1))/2)</p><p>cos(1 - i) = (e + e^(-1))/2·cos(1) + i·(e - e^(-1))/2·sin(1)</p></p><p><strong>Step 4: Identify real and imaginary parts</strong></p><p>Comparing with a + ib:</p><p>a = (1/2)(e + e^(-1))cos(1)</p><p>b = (1/2)(e - e^(-1))sin(1)</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B