\( \left. \begin{array} { l } { 5 \quad \sin 4 x \sin 3 x + \cos 4 x \cos 3 x } \\ { 6 \quad \sin ( 90 ^ { \circ } - x ) \cos x + \cos ( 90 ^ { \circ } + x ) \sin x } \end{array} \right. \)

Let's solve the two expressions step by step. ### Expression 1: \[ 5 \quad \sin 4x \sin 3x + \cos 4x \cos 3x \] This expression can be simplified using the product-to-sum identities. The identity states: \[ \sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)] \] \[ \cos A \cos B = \frac{1}{2} [\cos (A - B) + \cos (A + B)] \] Applying these identities, we have: 1. For \( \sin 4x \sin 3x \): \[ \sin 4x \sin 3x = \frac{1}{2} [\cos (4x - 3x) - \cos (4x + 3x)] = \frac{1}{2} [\cos x - \cos 7x] \] 2. For \( \cos 4x \cos 3x \): \[ \cos 4x \cos 3x = \frac{1}{2} [\cos (4x - 3x) + \cos (4x + 3x)] = \frac{1}{2} [\cos x + \cos 7x] \] Now substituting these back into the expression: \[ 5 \left( \frac{1}{2} [\cos x - \cos 7x] \right) + \frac{1}{2} [\cos x + \cos 7x] \] Combining these: \[ = \frac{5}{2} (\cos x - \cos 7x) + \frac{1}{2} (\cos x + \cos 7x) \] \[ = \frac{5}{2} \cos x - \frac{5}{2} \cos 7x + \frac{1}{2} \cos x + \frac{1}{2} \cos 7x \] \[ = \left( \frac{5}{2} + \frac{1}{2} \right) \cos x + \left( -\frac{5}{2} + \frac{1}{2} \right) \cos 7x \] \[ = 3 \cos x - 2 \cos 7x \] ### Expression 2: \[ 6 \quad \sin (90^{\circ} - x) \cos x + \cos (90^{\circ} + x) \sin x \] Using the co-function identities: 1. \( \sin (90^{\circ} - x) = \cos x \) 2. \( \cos (90^{\circ} + x) = -\sin x \) Substituting these into the expression: \[ 6 \cos x \cos x + (-\sin x) \sin x \] \[ = 6 \cos^2 x - \sin^2 x \] Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can express \( \sin^2 x \) as \( 1 - \cos^2 x \): \[ = 6 \cos^2 x - (1 - \cos^2 x) \] \[ = 6 \cos^2 x - 1 + \cos^2 x \] \[ = 7 \cos^2 x - 1 \] ### Final Results: 1. The first expression simplifies to: \[ 3 \cos x - 2 \cos 7x \] 2. The second expression simplifies to: \[ 7 \cos^2 x - 1 \]