91Ó°ÊÓ

We can use the angle addition identities to rewrite the terms in the expression. The angle addition identities are as follows: $$ \cos(a + b) = \cos a \cos b - \sin a \sin b \\ \cos(a - b) = \cos a \cos b + \sin a \sin b $$ We can rewrite the terms as follow: $$ \cos^{2} 10^{\circ} = \frac{1}{2}(\cos 20^{\circ} + 1) \\ \cos^{2} 50^{\circ} = \frac{1}{2}(\cos 100^{\circ} + 1) \\ \cos^{2} 70^{\circ} = \frac{1}{2}(\cos 140^{\circ} + 1) $$

Now we are going to use the Sum-to-Product formula, which states: $$ \cos x + \cos y = 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} $$ By applying the formula to each term, we get: $$ \cos 20^{\circ} + \cos 140^{\circ} = 2\cos{80^{\circ}}\cos{-60^{\circ}} \\ \cos 100^{\circ} + \cos 40^{\circ} = 2\cos{70^{\circ}}\cos{-30^{\circ}} \\ \cos 20^{\circ} + \cos 100^{\circ} = 2\cos{60^{\circ}}\cos{-40^{\circ}} $$

Now we are going to sum the expressions we got in Steps 1 and 2: $$ \frac{1}{2}(\cos 20^{\circ} + 1) + \frac{1}{2}(\cos 100^{\circ} + 1) + \frac{1}{2}(\cos 140^{\circ} + 1) \\ = \frac{1}{2}(3 + \cos{80^{\circ}}\cos{-60^{\circ}} + \cos{70^{\circ}}\cos{-30^{\circ}} + \cos{60^{\circ}}\cos{-40^{\circ}}) $$

Now, we need to simplify the expression. The main thing to notice is that the multiplication of cosines in each term can be written as the cosines of their sum or difference using the product-to-sum formula. So we get: $$ \frac{1}{2}(3 + \cos(80^{\circ} - (-60^{\circ})) + \cos(80^{\circ} + (-60^{\circ})) + \cos(100^{\circ} - (-40^{\circ})) \\ = \frac{1}{2}(3 + \cos 140^{\circ} + \cos 20^{\circ} + \cos 140^{\circ}) $$ Since cosine is an even function, we can eliminate the negative values from the cosines: $$ = \frac{1}{2}(3 + \cos 140^{\circ} + \cos 20^{\circ} + \cos 140^{\circ}) $$ And finally, by summing these values, we obtain: $$ \frac{1}{2}(3 + 2\cos 140^{\circ} + \cos 20^{\circ}) = \frac{1}{2}(3 + 0) = \frac{3}{2} $$ The given expression: $$ \cos ^{2} 10^{\circ}+\cos ^{2} 50^{\circ}+\cos ^{2} 70^{\circ} = \frac{3}{2} $$ So, the proof is complete.