Answer:
To show that [tex]\sin^{2}\left(15^{\circ}\right) + \sin^{2}\left(45^{\circ}\right) + \sin^{2}\left(75^{\circ}\right) = (3/2)[/tex], make use of the following:
Step-by-step explanation:
Note that the sum of [tex]15^{\circ}[/tex] and [tex]75^{\circ}[/tex] is [tex]90^{\circ}[/tex]. Hence, using the fact that [tex]\sin(\theta) = \cos\left(90^{\circ} - \theta\right)[/tex] as long as [tex]0^{\circ} < \theta < 90^{\circ}[/tex]:
[tex]\displaystyle \sin\left(15^{\circ}\right) = \cos\left(90^{\circ} - 15^{\circ}\right) = \cos\left(75^{\circ}\right)[/tex].
Hence, the [tex]\displaystyle \sin\left(15^{\circ}\right)[/tex] in the original expression can be replaced with [tex]\cos\left(75^{\circ}\right)[/tex]. Additionally, since [tex]\sin\left(45^{\circ}\right) = \left(\sqrt{2} / 2\right)[/tex]:
[tex]\begin{aligned} & \sin^{2}\left(15^{\circ}\right) + \sin^{2}\left(45^{\circ}\right) + \sin^{2}\left(75^{\circ}\right)\\ =\; & \sin^{2}\left(15^{\circ}\right) + \left(\frac{\sqrt{2}}{2}\right)^{2} + \sin^{2}\left(75^{\circ}\right) \\ =\; &\cos^{2}\left(75^{\circ}\right) + \frac{1}{2} + \sin^{2}\left(75^{\circ}\right)\end{aligned}[/tex]
Since [tex]\cos^{2}(\theta) + \sin^{2}(\theta) = 1[/tex] for any real [tex]\theta[/tex] (the Pythagorean identity,) [tex]\cos^{2}\left(75^{\circ}\right) + \sin^{2}\left(75^{\circ}\right) = 1[/tex]. Hence:
[tex]\begin{aligned} & \sin^{2}\left(15^{\circ}\right) + \sin^{2}\left(45^{\circ}\right) + \sin^{2}\left(75^{\circ}\right) \\ =\; &\cos^{2}\left(75^{\circ}\right) + \frac{1}{2} + \sin^{2}\left(75^{\circ}\right) \\ =\; & \left(\cos^{2}\left(75^{\circ}\right) + \sin^{2}\left(75^{\circ}\right)\right) + \frac{1}{2} \\ =\; & 1 + \frac{1}{2} \\ =\; & \frac{3}{2}\end{aligned}[/tex].
