Question

Given $\Delta$ACB right angled at C in which AB = 29 units, BC = 21 units and $\angle$ABC = $\theta$. Determine the value of
$co{s}^2\theta -si{n}^2\theta$

Answer 1

In $\Delta$ ACB, we have
$A{B}^2=A{C}^2+B{C}^2$
$\Rightarrow AC=\sqrt{A{B}^2-B{C}^2}=\sqrt{{29}^2-{21}^2}=\sqrt{(29+21)(29-21)}=\sqrt{400}=20$ units
$thereforesin\theta =\frac{AC}{AB}=\frac{20}{29}andcos\theta =\frac{BC}{AB}=\frac{21}{29}$
(ii) We have,
$co{s}^2\theta -si{n}^2\theta ={\left(\frac{21}{29}\right)}^2-{\left(\frac{20}{29}\right)}^2=\frac{{21}^2-{20}^2}{{29}^2}=\frac{(21+20)(21-20)}{841}=\frac{41}{841}.$