Consider $\triangle ACB$,right-angled at $C$,in which $AB = 29$ units,$BC = 21$ units and $\angle ABC = \theta$. Determine the values of:
$(i)$ $\cos^2 \theta + \sin^2 \theta$
$(ii)$ $\cos^2 \theta - \sin^2 \theta$
$(i)$ $\cos^2 \theta + \sin^2 \theta$
$(ii)$ $\cos^2 \theta - \sin^2 \theta$
(N/A) In $\triangle ACB$,we have:
$AC = \sqrt{AB^2 - BC^2} = \sqrt{(29)^2 - (21)^2}$
$= \sqrt{(29 - 21)(29 + 21)} = \sqrt{(8)(50)} = \sqrt{400} = 20$ units.
So,$\sin \theta = \frac{AC}{AB} = \frac{20}{29}$ and $\cos \theta = \frac{BC}{AB} = \frac{21}{29}$.
Now,
$(i)$ $\cos^2 \theta + \sin^2 \theta = \left(\frac{21}{29}\right)^2 + \left(\frac{20}{29}\right)^2 = \frac{441 + 400}{841} = \frac{841}{841} = 1$.
$(ii)$ $\cos^2 \theta - \sin^2 \theta = \left(\frac{21}{29}\right)^2 - \left(\frac{20}{29}\right)^2 = \frac{441 - 400}{841} = \frac{41}{841}$.
$AC = \sqrt{AB^2 - BC^2} = \sqrt{(29)^2 - (21)^2}$
$= \sqrt{(29 - 21)(29 + 21)} = \sqrt{(8)(50)} = \sqrt{400} = 20$ units.
So,$\sin \theta = \frac{AC}{AB} = \frac{20}{29}$ and $\cos \theta = \frac{BC}{AB} = \frac{21}{29}$.
Now,
$(i)$ $\cos^2 \theta + \sin^2 \theta = \left(\frac{21}{29}\right)^2 + \left(\frac{20}{29}\right)^2 = \frac{441 + 400}{841} = \frac{841}{841} = 1$.
$(ii)$ $\cos^2 \theta - \sin^2 \theta = \left(\frac{21}{29}\right)^2 - \left(\frac{20}{29}\right)^2 = \frac{441 - 400}{841} = \frac{41}{841}$.
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