Question

2.3 Without using a calculator, determine the value of \[ \frac{\sin ^{2} 35^{\circ}-\cos ^{2} 35^{\circ}}{4 \sin 10^{\circ} \cdot \cos 10^{\circ}} \]

Ask by Harper Wyatt. in South Africa
Mar 11,2025

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Tutor-Verified Answer

Answer

The value of the expression is \(-\frac{1}{2}\).

Solution

To solve the expression \[ \frac{\sin ^{2} 35^{\circ}-\cos ^{2} 35^{\circ}}{4 \sin 10^{\circ} \cdot \cos 10^{\circ}}, \] we can start by simplifying the numerator and the denominator separately. ### Step 1: Simplify the Numerator The numerator is \[ \sin^2 35^\circ - \cos^2 35^\circ. \] Using the identity \[ \sin^2 x - \cos^2 x = -\cos(2x), \] we can rewrite the numerator as: \[ \sin^2 35^\circ - \cos^2 35^\circ = -\cos(2 \cdot 35^\circ) = -\cos(70^\circ). \] ### Step 2: Simplify the Denominator The denominator is \[ 4 \sin 10^\circ \cdot \cos 10^\circ. \] Using the double angle identity for sine, we have: \[ \sin(2x) = 2 \sin x \cos x, \] which gives us: \[ 4 \sin 10^\circ \cos 10^\circ = 2 \cdot 2 \sin 10^\circ \cos 10^\circ = 2 \sin(20^\circ). \] ### Step 3: Combine the Results Now we can substitute the simplified numerator and denominator back into the original expression: \[ \frac{-\cos(70^\circ)}{2 \sin(20^\circ)}. \] ### Step 4: Use Trigonometric Identities We know that \[ \cos(70^\circ) = \sin(20^\circ), \] so we can substitute this into our expression: \[ \frac{-\sin(20^\circ)}{2 \sin(20^\circ)}. \] ### Step 5: Simplify the Expression Assuming \(\sin(20^\circ) \neq 0\), we can simplify: \[ \frac{-\sin(20^\circ)}{2 \sin(20^\circ)} = \frac{-1}{2}. \] ### Final Answer Thus, the value of the original expression is \[ \boxed{-\frac{1}{2}}. \]

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Bonus Knowledge

To solve the expression \[ \frac{\sin^2 35^\circ - \cos^2 35^\circ}{4 \sin 10^\circ \cdot \cos 10^\circ}, \] we start by simplifying the numerator. Recall that \[ \sin^2 \theta - \cos^2 \theta = \sin^2 \theta - (1 - \sin^2 \theta) = 2\sin^2 \theta - 1. \] Therefore, we can rewrite the numerator as: \[ \sin^2 35^\circ - \cos^2 35^\circ = 2\sin^2 35^\circ - 1. \] Next, we recognize that \(2\sin^2 \theta = 1 - \cos 2\theta\), so we also have: \[ \sin^2 35^\circ - \cos^2 35^\circ = -\cos 70^\circ, \] since \(-\cos(2 \cdot 35^\circ) = -\cos 70^\circ\). Now let’s rewrite our original expression: \[ \frac{-\cos 70^\circ}{4 \sin 10^\circ \cdot \cos 10^\circ}. \] Using the double angle identity for sine, we know that \[ \sin 20^\circ = 2\sin 10^\circ \cos 10^\circ. \] Thus, we can represent the denominator: \[ 4 \sin 10^\circ \cos 10^\circ = 2 \cdot 2 \sin 10^\circ \cos 10^\circ = 2 \sin 20^\circ. \] Substituting back gives us: \[ \frac{-\cos 70^\circ}{2 \sin 20^\circ}. \] Knowing that \(\cos 70^\circ = \sin 20^\circ\), we can substitute to receive: \[ \frac{-\sin 20^\circ}{2 \sin 20^\circ} = \frac{-1}{2}. \] Thus, the final result is \[ \boxed{-\frac{1}{2}}. \]

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