Consider the following for the three (03) items that follow: Let p...

Given p = \sin 35^{}^{\circ}, q = \sin 25^{}^{\circ}, and r = \sin(-95^{}^{\circ}).

Simplifying $r$:

r = -\sin(90^{}^{\circ} + 5^{}^{\circ}) = -\cos 5^{}^{\circ}

We need to find the value of $p^2 + q^2 + r^2$:

p^2 + q^2 + r^2 = \sin^2 35^{}^{\circ} + \sin^2 25^{}^{\circ} + \cos^2 5^{}^{\circ}

Using the identities $\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2} and \cos^2 \theta = \dfrac{1 + \cos 2\theta}{2}$:

p^2 + q^2 + r^2 = \dfrac{1 - \cos 70^{}^{\circ}}{2} + \dfrac{1 - \cos 50^{}^{\circ}}{2} + \dfrac{1 + \cos 10^{}^{\circ}}{2}

= \dfrac{3}{2} - \dfrac{1}{2}(\cos 70^{}^{\circ} + \cos 50^{}^{\circ} - \cos 10^{}^{\circ})

Using the sum-to-product formula \cos C + \cos D = 2 \cos \left(\dfrac{C+D}{2}\right) \cos\left(\dfrac{C-D}{2}\right):<br><br>\cos 70^{}^{\circ} + \cos 50^{}^{\circ} = 2 \cos \left(\dfrac{70^{}^{\circ} + 50^{}^{\circ}}{2}\right) \cos\left(\dfrac{70^{}^{\circ} - 50^{}^{\circ}}{2}\right)<br><br>= 2 \cos 60^{}^{\circ} \cos 10^{}^{\circ}$<br><br>$= 2 \left(\dfrac{1}{2}\right) \cos10^{}^{\circ} = \cos 10^{}^{\circ}

Substituting this back into the expression:

$p^2 + q^2 + r^2 =$\dfrac{3}{2} - \dfrac{1}{2}(\cos10^{}^{\circ} - \cos 10^{}^{\circ})

=\dfrac{3}{2} -$0 =$\dfrac{3}{2}

Answer: \dfrac{3}{2}