\( 6.3 \quad \) If \( \sin 76^{\circ}=x \) and \( \cos 76^{\circ}=y \), show that \( x^{2}-y^{2}=\sin 62^{\circ} \)

To show that \( x^2 - y^2 = \sin 62^\circ \) given \( x = \sin 76^\circ \) and \( y = \cos 76^\circ \), we can start by using the identity for the difference of squares: \[ x^2 - y^2 = (\sin 76^\circ)^2 - (\cos 76^\circ)^2 \] We can use the Pythagorean identity, which states that: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Thus, we can express \( x^2 - y^2 \) in terms of sine and cosine: \[ x^2 - y^2 = \sin^2 76^\circ - \cos^2 76^\circ \] Using the identity for the difference of squares, we can rewrite this as: \[ x^2 - y^2 = -(\cos^2 76^\circ - \sin^2 76^\circ) = -\cos^2 76^\circ + \sin^2 76^\circ \] Now, we can use the co-function identity, which states that: \[ \sin(90^\circ - \theta) = \cos \theta \] Thus, we have: \[ \cos 76^\circ = \sin(90^\circ - 76^\circ) = \sin 14^\circ \] Now we can substitute this into our expression: \[ x^2 - y^2 = \sin^2 76^\circ - \sin^2 14^\circ \] Next, we can use the sine subtraction formula: \[ \sin^2 A - \sin^2 B = \sin(A + B) \sin(A - B) \] Let \( A = 76^\circ \) and \( B = 14^\circ \): \[ \sin^2 76^\circ - \sin^2 14^\circ = \sin(76^\circ + 14^\circ) \sin(76^\circ - 14^\circ) \] Calculating the angles: \[ 76^\circ + 14^\circ = 90^\circ \] \[ 76^\circ - 14^\circ = 62^\circ \] Thus, we have: \[ \sin^2 76^\circ - \sin^2 14^\circ = \sin 90^\circ \sin 62^\circ \] Since \( \sin 90^\circ = 1 \), we get: \[ \sin^2 76^\circ - \sin^2 14^\circ = 1 \cdot \sin 62^\circ = \sin 62^\circ \] Therefore, we conclude that: \[ x^2 - y^2 = \sin 62^\circ \] This completes the proof. Thus, we have shown that: \[ x^2 - y^2 = \sin 62^\circ \]