Question

Consider $△ACB,$ right-angled at C, in which $AB=29$ units, $BC=21$ units and $\angle ABC=\theta$. Determine the values of ${cos}^2\theta -{sin}^2\theta$.

Answer 1

Firstly we have to find the value of $AC$ with the help of Pythagoras theorem
According to Pythagoras theorem
${\left(Hypotenuse\right)}^2={\left(Base\right)}^2+{\left(Perpendicular\right)}^2$
$\Rightarrow {\left(AC\right)}^2+{\left(BC\right)}^2={\left(AB\right)}^2\Rightarrow {\left(AC\right)}^2+{\left(21\right)}^2={\left(29\right)}^2\Rightarrow {\left(AC\right)}^2={\left(29\right)}^2-{\left(21\right)}^2$
Using the identity $a^2-b^2=(a+b)(a-b)$
$\Rightarrow {\left(AC\right)}^2=(29-21)(29+21)\Rightarrow \left(8\right)\left(50\right)=400$
$\Rightarrow AC=\sqrt{400}=\pm 20$
But $AC$ can't be negative, so, $AC=20$ units
Now, we will find the $\sin \theta$ and $\cos \theta$
$\sin \theta =\frac{sideoppositetoangle\theta }{hypotenuse}$
In $△ACB$, side opposite to angle $\theta =AC=20$ and Hypotenuse $=AB=29$
So, $\sin \theta =\frac{AC}{AB}=\frac{20}{29}$
Now we know that
$\cos \theta =\frac{sideadjacenttoangle\theta }{hypotenuse}$
In $△ACB$, side adjacent to angle $\theta =BC=21$ and Hypotenuse $=AB=29$
So, $\cos \theta =\frac{BC}{AB}=\frac{21}{29}$

Putting values we get

${\cos }^2\theta -{\sin }^2\theta ={\left(\frac{21}{29}\right)}^2-{\left(\frac{20}{29}\right)}^2\Rightarrow \frac{441-400}{29\times 29}=\frac{41}{841}$