Additional Solved Examples
1
Find cot θ, if sin θ = $\dfrac{5}{13}$
and $\dfrac{π}{2}$ < θ < $π$
Solution
We know that $sin^2 θ$ + $cos^2 θ$ = 1
$\begin{align*}
∴ cos θ & = - \sqrt{1 - sin^2 θ} \\
& = - \sqrt{1 - \dfrac{25}{169}} \\
& = - \dfrac{12}{13}
\end{align*}
$
tan θ = $\dfrac{sin θ}{cos θ}$ = $\dfrac{\dfrac{5}{13}}{\dfrac{-12}{13}}$
= $\dfrac{-5}{12}$
cot θ = $\dfrac{1}{tan θ} = -\dfrac{1}{\dfrac{5}{12}} = -\dfrac{12}{5}$
2
Calculate sin 65° sin 25°
Solution
We know that $sin^2 θ$ + $cos^2 θ$ = 1
$\begin{align*}
∴ cos θ & = - \sqrt{1 - sin^2 θ} \\
& = - \sqrt{1 - \dfrac{25}{169}} \\
& = - \dfrac{12}{13}
\end{align*}
$
tan θ = $\dfrac{sin θ}{cos θ}$
= $- \dfrac{\dfrac{5}{13}}{\dfrac{12}{13}} $
= $- \dfrac{5}{12}$
$cot θ$ =
$\dfrac{1}{tan θ}$
= $-\dfrac{1}{\dfrac{5}{12}} = -\dfrac{12}{5}$