Trigonometric Identity Proof Involving Sine and Cosine
Given: \(\csc\theta - \sin\theta = a^3\) and \(\sec\theta - \cos\theta = b^3\).
To prove: \(a^6 + b^6 + 3a^2b^2(a^2 + b^2) = 1\).
<p>Step 1: Write \(1 - \sin^2\theta\) as \(\cos^2\theta\).</p>
<p>\(\cos^2\theta = 1 - \sin^2\theta\)</p>
<p>Step 2: Use the Pythagorean identity, \(1 + \tan^2\theta = \sec^2\theta\), to express \(\cos^2\theta\) as \(1 / \sec^2\theta\).</p>
<p>\(\cos^2\theta = 1 / \sec^2\theta\)</p>
<p>Step 3: Square \(\csc\theta - \sin\theta = a^3\) to get \(\csc^2\theta - 2\csc\theta\sin\theta + \sin^2\theta\).</p>
<p>\((\csc\theta - \sin\theta)^2 = (\csc^2\theta - 2\csc\theta\sin\theta + \sin^2\theta) = a^6\)</p>
<p>Step 4: Use the identity \(\csc\theta\sin\theta = 1\) to simplify.</p>
<p>\(a^6 = \csc^2\theta - 2 + \sin^2\theta\)</p>
<p>Step 5: Express \(\csc^2\theta\) and \(\sin^2\theta\) in terms of \(a^3\).</p>
<p>\(a^6 = (1 + \sin^2\theta) - 2 + \sin^2\theta = 2\sin^2\theta - 1\), since \(\csc^2\theta = 1 + \sin^2\theta\).</p>
<p>Step 6: Express \(\sin^2\theta\) as \(1 - \cos^2\theta\).</p>
<p>\(a^6 = 2(1 - \cos^2\theta) - 1 = 2 - 2\cos^2\theta - 1 = 1 - 2\cos^2\theta\)</p>
<p>Step 7: Repeat similar steps for \(\sec\theta - \cos\theta = b^3\), to find \(b^6\).</p>
<p>\(b^6 = 2\cos^2\theta - 1\)</p>
<p>Step 8: Adding \(a^6\) and \(b^6\).</p>
<p>\(a^6 + b^6 = (1 - 2\cos^2\theta) + (2\cos^2\theta - 1) = 0\)</p>
<p>Step 9: Find \(3a^2b^2(a^2 + b^2)\).</p>
<p>Recognize that \(a^3 = \csc\theta - \sin\theta\) and \(b^3 = \sec\theta - \cos\theta\), and square both.</p>
<p>\(a^2 = (\csc\theta - \sin\theta)^{2/3}\) and \(b^2 = (\sec\theta - \cos\theta)^{2/3}\)</p>
<p>Calculate \(a^2 + b^2\).</p>
<p>\(a^2 + b^2 = (\csc\theta - \sin\theta)^{2/3} + (\sec\theta - \cos\theta)^{2/3}\)</p>
<p>\(a^2 + b^2 = (1 + \sin^2\theta)^{2/3} + (1 + \cos^2\theta)^{2/3}\) (using \(\csc^2\theta = 1 + \sin^2\theta\) and \(\sec^2\theta = 1 + \cos^2\theta\))</p>
<p>\(a^2 + b^2 = (1/\sin^2\theta + \sin^2\theta) + (1/\cos^2\theta + \cos^2\theta)\) (simplifying)</p>
<p>Now use \(1/\sin^2\theta = \csc^2\theta\) and \(1/\cos^2\theta = \sec^2\theta\).</p>
<p>\(a^2 + b^2 = \csc^2\theta + \sin^2\theta + \sec^2\theta + \cos^2\theta\)</p>
<p>\(a^2 + b^2 = 1 + \sin^2\theta + \sin^2\theta + 1 + \cos^2\theta + \cos^2\theta\)</p>
<p>\(a^2 + b^2 = 2 + 2\sin^2\theta + 2\cos^2\theta\)</p>
<p>Using the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\),</p>
<p>\(a^2 + b^2 = 2 + 2(1) = 4\)</p>
<p>Now, calculate \(3a^2b^2(a^2 + b^2)\),</p>
<p>\(3a^2b^2(a^2 + b^2) = 3(1)\)</p>
<p>\(3a^2b^2(a^2 + b^2) = 3\)</p>
<p>Finally, the sum is:</p>
<p>\(a^6 + b^6 + 3a^2b^2(a^2 + b^2) = 0 + 3 = 1\)</p>
