Question

\( \sin(60^\circ + \theta) - \cos(30^\circ - \theta) \) is equal to :

• A. \(2 \cos\theta\)
• B. \(2 \sin\theta\)
• C. 0
• D. 1

Hint:

Use the complementary angle identity: \(\cos A = \sin(90^\circ - A)\) or \(\sin B = \cos(90^\circ - B)\). Let's change \(\cos(30^\circ - \theta)\). \(\cos(30^\circ - \theta) = \sin(90^\circ - [30^\circ - \theta])\) \( = \sin(90^\circ - 30^\circ + \theta) \) \( = \sin(60^\circ + \theta) \). So the expression becomes: \( \sin(60^\circ + \theta) - \sin(60^\circ + \theta) = 0 \).

Answer 1

The Correct Option is C

Concept: This problem uses the complementary angle identity \(\cos x = \sin(90^\circ - x)\). Step 1: Apply the complementary angle identity to one of the terms We need to evaluate \( \sin(60^\circ + \theta) - \cos(30^\circ - \theta) \). Let's convert the cosine term to a sine term. We know that \(\cos x = \sin(90^\circ - x)\). Let \(x = 30^\circ - \theta\). Then, \(\cos(30^\circ - \theta) = \sin(90^\circ - (30^\circ - \theta))\). Simplify the angle inside the sine function: \(90^\circ - (30^\circ - \theta) = 90^\circ - 30^\circ + \theta = 60^\circ + \theta\). So, \(\cos(30^\circ - \theta) = \sin(60^\circ + \theta)\). Step 2: Substitute this back into the original expression The original expression is \( \sin(60^\circ + \theta) - \cos(30^\circ - \theta) \). Substitute \(\cos(30^\circ - \theta) = \sin(60^\circ + \theta)\): \[ \sin(60^\circ + \theta) - \sin(60^\circ + \theta) \] Step 3: Simplify \[ \sin(60^\circ + \theta) - \sin(60^\circ + \theta) = 0 \] The expression is equal to 0. Alternative Method: Convert sine to cosine We know \(\sin y = \cos(90^\circ - y)\). Let \(y = 60^\circ + \theta\). Then \(\sin(60^\circ + \theta) = \cos(90^\circ - (60^\circ + \theta))\). \(90^\circ - (60^\circ + \theta) = 90^\circ - 60^\circ - \theta = 30^\circ - \theta\). So, \(\sin(60^\circ + \theta) = \cos(30^\circ - \theta)\). Substitute this into the original expression: \[ \cos(30^\circ - \theta) - \cos(30^\circ - \theta) = 0 \] Both methods yield 0.