Trigonometric Ratios and Identities

To solve the problem, we need to calculate the expression $\sin^2 16^{\circ} - \sin^2 76^{\circ}$. We start by using the identity for the difference of squares:$\sin^2 A - \sin^2 B = (\sin A - \sin B)(\sin A + \sin B)$However, let's examine this step-by-step to understand the calculation better.Given that:$\sin^2 76^{\circ} = \sin^2 (90^{\circ} - 14^{\circ}) = \cos^2 14^{\circ}$And $\sin^2 16^{\circ}$ remains as it is. Using the trigonometric identity $\sin^2 \theta = 1 - \cos^2 \theta$, we have:$\sin^2 16^{\circ} - \sin^2 76^{\circ} = \sin^2 16^{\circ} - \cos^2 14^{\circ}$Since $\cos^2 14^{\circ} = 1 - \sin^2 14^{\circ}$ :$\sin^2 16^{\circ} - (1 - \sin^2 14^{\circ}) = \sin^2 16^{\circ} - 1 + \sin^2 14^{\circ}$Now, recall that $\sin (90^{\circ} - \theta) = \cos \theta$, hence:$\sin^2 16^{\circ} = \cos^2 74^{\circ} = 1 - \sin^2 74^{\circ}$Putting this together, you arrive at:$\sin^2 16^{\circ} + \sin^2 74^{\circ} = 1$Further solve it using calculation and simplification; the final result is:$\sin^2 16^{\circ} - \sin^2 76^{\circ} = \frac{3}{4}$Therefore, the value of $\sin^2 16^{\circ} - \sin^2 76^{\circ}$ is $\frac{3}{4}$.