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

If sin⁡A+cos⁡A=75\sin A+\cos A=\dfrac{7}{5} and sin⁡Acos⁡A=1225\sin A\cos A=\dfrac{12}{25}, find the values of sin⁡A\sin A and cos⁡A\cos A.

Explanation

Solution

Hint:Use the relation: sin⁡Acos⁡A=1225\sin A\cos A=\dfrac{12}{25} and write cos⁡A\cos A in terms of sin⁡A\sin A. Substitute the value of cos⁡A\cos A in the relation: sin⁡A+cos⁡A=75\sin A+\cos A=\dfrac{7}{5} to obtain a quadratic equation in sin⁡A\sin A. Solve this quadratic equation by the help of middle term split and get the values of sin⁡A\sin A. Corresponding to these values of sin⁡A\sin A, find the values of cos⁡A\cos A using first relation.

Complete step-by-step answer:
We have been provided with two equations:
sin⁡A+cos⁡A=75.................(i)\sin A+\cos A=\dfrac{7}{5}.................(i)
sin⁡Acos⁡A=1225...................(ii)\sin A\cos A=\dfrac{12}{25}...................(ii)
Now, from equation (ii), we have,
sin⁡Acos⁡A=1225 ⇒cos⁡A=1225sin⁡A \begin{aligned} & \sin A\cos A=\dfrac{12}{25} \\\ & \Rightarrow \cos A=\dfrac{12}{25\sin A} \\\ \end{aligned}
Substituting this value of cos⁡A\cos A in equation (i), we have,
sin⁡A+1225sin⁡A=75\sin A+\dfrac{12}{25\sin A}=\dfrac{7}{5}
Taking L.C.M we get,
25sin⁡2A+1225sin⁡A=75 ⇒25sin⁡2A+125sin⁡A=7 \begin{aligned} & \dfrac{25{{\sin }^{2}}A+12}{25\sin A}=\dfrac{7}{5} \\\ & \Rightarrow \dfrac{25{{\sin }^{2}}A+12}{5\sin A}=7 \\\ \end{aligned}
By cross-multiplication, we get,
25sin⁡2A+12=35sin⁡A ⇒25sin⁡2A−35sin⁡A+12=0 \begin{aligned} & 25{{\sin }^{2}}A+12=35\sin A \\\ & \Rightarrow 25{{\sin }^{2}}A-35\sin A+12=0 \\\ \end{aligned}
This is a quadratic equation with its variable as: sin⁡A\sin A. Now, splitting the middle term, we get,
25sin⁡2A−15sin⁡A−20sin⁡A+12=0 ⇒5sin⁡A(5sin⁡A−3)−4(5sin⁡A−3)=0 ⇒(5sin⁡A−3)(5sin⁡A−4)=0 \begin{aligned} & 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{aligned}
Substituting each term equal to 0, we get,
(5sin⁡A−3)=0 or (5sin⁡A−4)=0 ⇒5sin⁡A=3 or 5sin⁡A=4 ⇒sin⁡A=35 or sin⁡A=45 \begin{aligned} & \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{aligned}
We have obtained two values of sin⁡A\sin A and we have to find the values of cos⁡A\cos A corresponding to these. Using equation (i): sin⁡A+cos⁡A=75\sin A+\cos A=\dfrac{7}{5}, we have,
Case (i): When sin⁡A=35\sin A=\dfrac{3}{5}.

& \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{aligned}$$ Case (ii): When $\sin A=\dfrac{4}{5}$. $$\begin{aligned} & \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{aligned}$$ 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.