Question

If $\sin A+\cos A=\dfrac{7}{5}$ and $\sin A\cos A=\dfrac{12}{25}$, find the values of $\sin A$ and $\cos A$.

Answer 1

Hint:Use the relation: $\sin A\cos A=\dfrac{12}{25}$ and write $\cos A$ in terms of $\sin A$. Substitute the value of \[\cos A\] in the relation: $\sin A+\cos A=\dfrac{7}{5}$ to obtain a quadratic equation in $\sin A$. Solve this quadratic equation by the help of middle term split and get the values of $\sin A$. Corresponding to these values of $\sin A$, find the values of $\cos A$ using first relation.

Complete step-by-step answer:
We have been provided with two equations:
$\sin A+\cos A=\dfrac{7}{5}.................(i)$
$\sin A\cos A=\dfrac{12}{25}...................(ii)$
Now, from equation (ii), we have,
$\begin{align}
  & \sin A\cos A=\dfrac{12}{25} \\
 & \Rightarrow \cos A=\dfrac{12}{25\sin A} \\
\end{align}$
Substituting this value of $\cos A$ in equation (i), we have,
$\sin A+\dfrac{12}{25\sin A}=\dfrac{7}{5}$
Taking L.C.M we get,
$\begin{align}
  & \dfrac{25{{\sin }^{2}}A+12}{25\sin A}=\dfrac{7}{5} \\
 & \Rightarrow \dfrac{25{{\sin }^{2}}A+12}{5\sin A}=7 \\
\end{align}$
By cross-multiplication, we get,
$\begin{align}
  & 25{{\sin }^{2}}A+12=35\sin A \\
 & \Rightarrow 25{{\sin }^{2}}A-35\sin A+12=0 \\
\end{align}$
This is a quadratic equation with its variable as: $\sin A$. Now, splitting the middle term, we get,
$\begin{align}
  & 25{{\sin }^{2}}A-15\sin A-20\sin A+12=0 \\
 & \Rightarrow 5\sin A\left( 5\sin A-3 \right)-4\left( 5\sin A-3 \right)=0 \\
 & \Rightarrow \left( 5\sin A-3 \right)\left( 5\sin A-4 \right)=0 \\
\end{align}$
Substituting each term equal to 0, we get,
$\begin{align}
  & \left( 5\sin A-3 \right)=0\text{ or }\left( 5\sin A-4 \right)=0 \\
 & \Rightarrow 5\sin A=3\text{ or }5\sin A=4 \\
 & \Rightarrow \sin A=\dfrac{3}{5}\text{ or }\sin A=\dfrac{4}{5} \\
\end{align}$
We have obtained two values of $\sin A$ and we have to find the values of $\cos A$ corresponding to these. Using equation (i): $\sin A+\cos A=\dfrac{7}{5}$, we have,
Case (i): When $\sin A=\dfrac{3}{5}$.
\[\begin{align}
  & \Rightarrow \dfrac{3}{5}+\cos A=\dfrac{7}{5} \\
 & \Rightarrow \cos A=\dfrac{7}{5}-\dfrac{3}{5} \\
 & \Rightarrow \cos A=\dfrac{7-3}{5} \\
 & \Rightarrow \cos A=\dfrac{4}{5} \\
\end{align}\]
Case (ii): When $\sin A=\dfrac{4}{5}$.
\[\begin{align}
  & \Rightarrow \dfrac{4}{5}+\cos A=\dfrac{7}{5} \\
 & \Rightarrow \cos A=\dfrac{7}{5}-\dfrac{4}{5} \\
 & \Rightarrow \cos A=\dfrac{7-4}{5} \\
 & \Rightarrow \cos A=\dfrac{3}{5} \\
\end{align}\]
Hence, the values of $\sin A$ are: $\dfrac{3}{5}$ and $\dfrac{4}{5}$ and the corresponding values of $\cos A$ are: $\dfrac{4}{5}$ and $\dfrac{3}{5}$.

Note: One may note that, we can also substitute the value of $\sin A$ from relation (ii), in relation (i). The answer will be the same. Here, we have to consider both the values of $\sin A$ and $\cos A$ because they are both satisfying the equations and are in the range of these functions. There is nothing on the basis of which one value can be rejected.