Example Question - proving identity
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Proving a Trigonometric Identity
The given equation is a trigonometric identity that we need to prove. Let's start by working with the left-hand side (LHS) of the equation and try to transform it into the right-hand side (RHS). The LHS of the equation is: \[ \frac{2 \tan \theta}{1 + \tan^2 \theta} \] We know that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), and we can use this to rewrite the LHS: \[ \frac{2 \frac{\sin \theta}{\cos \theta}}{1 + \left(\frac{\sin \theta}{\cos \theta}\right)^2} \] Simplify it further by combining the terms: \[ \frac{2\sin \theta}{\cos \theta} \times \frac{\cos^2 \theta}{\cos^2 \theta + \sin^2 \theta} \] Since \(\cos^2 \theta + \sin^2 \theta = 1\) (which is another trigonometric identity), we can simplify: \[ \frac{2\sin \theta}{\cos \theta} \times \frac{\cos^2 \theta}{1} \] \[ 2\sin \theta \times \cos \theta \] Which is the RHS of the equation: \[ 2\sin \theta \cos \theta \] Therefore, we have proved that the LHS is equal to the RHS: \[ \frac{2 \tan \theta}{1 + \tan^2 \theta} = 2 \sin \theta \cos \theta \] And this confirms that the initial statement is a valid trigonometric identity.
Proving a Trigonometric Identity
To prove the identity \[ \frac{\sin^3x - \cos^3x}{\sin x - \cos x} = \frac{1 + \sin 2x}{2} \] let's start by factoring the numerator using the difference of cubes formula: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] where \( a = \sin x \) and \( b = \cos x \). Thus, \[ \sin^3x - \cos^3x = (\sin x - \cos x)(\sin^2x + \sin x\cos x + \cos^2x) \] We know that \(\sin^2x + \cos^2x = 1\), so we can simplify as follows: \[ \sin^3x - \cos^3x = (\sin x - \cos x)(1 + \sin x\cos x) \] Now, divide both sides by \(\sin x - \cos x\): \[ \frac{\sin^3x - \cos^3x}{\sin x - \cos x} = \frac{(\sin x - \cos x)(1 + \sin x\cos x)}{\sin x - \cos x} = 1 + \sin x\cos x \] We also know the double-angle identity for sine: \[ \sin 2x = 2\sin x\cos x \] So, rewriting \( \sin x\cos x \) in terms of the double-angle identity: \[ \frac{1 + \sin 2x}{2} \] This means: \[ 1 + \sin x\cos x = \frac{1 + \sin 2x}{2} \] Hence, we've shown that: \[ \frac{\sin^3x - \cos^3x}{\sin x - \cos x} = \frac{1 + \sin 2x}{2} \] And the identity is proved.
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