Question

The most general values of $\text{ }\\!\\!\theta\\!\\!\text{ }$ for which $\text{sin }\\!\\!\theta\\!\\!\text{ -cos }\\!\\!\theta\\!\\!\text{ }\,\text{=}\,\underset{\text{a }\\!\\!\hat{\mathrm{I}}\\!\\!\text{ }\,\text{R}}{\mathop{\text{min}}}\,\,\text{(1,}\,{{\text{a}}^{\text{2}}}\text{-6a}\,\text{+}\,\text{10)}$ are given by:
A.$\text{n }\\!\\!\pi\\!\\!\text{ +}\,{{\text{(-1)}}^{\text{n}}}\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{4}}\,\text{-}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{4}}$
B.$\text{n }\\!\\!\pi\\!\\!\text{ +}\,{{\text{(-1)}}^{\text{n}}}\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{4}}\,\text{+}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{4}}$
C.$\text{2n }\\!\\!\pi\\!\\!\text{ }\,\text{+}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{4}}$
D.None of these

Answer 1

Try to find out the minimum value of ${{\text{a}}^{\text{2}}}\text{- 6a+10}$ and then form the trigonometric equation.

Complete step-by-step answer:
Let us rewrite ${{\text{a}}^{\text{2}}}\text{- 6a+10}$ as \text{(}{{\text{a}}^{\text{2}}}\text{- 6a+9)}\,\text{+}\,\text{1}$$$=\,{{(\text{a- 3)}}^{2}}\text{+1}$$ Clearly, since $${{\text{(a-3)}}^{\text{2}}}\,\ge \,\text{0}$$, the minimum value {{\text{a}}^{\text{2}}}\text{- 6a+10}$can take is 1, when$\text{a}=,3$Hence$,\underset{\text{a}\in ,\text{R}}{\mathop{\min }},,(1,,{{\text{a}}^{2}}-6\text{a},\text{+},\text{10)}$comes out to be 1 since at its lowest value,$,\underset{\text{a}\in ,\text{R}}{\mathop{\min }},,(1,,1\text{)}$is still 1. Now, construct the trigonometric function equation. That is,$\begin{aligned}
& \text{sin }\!\!\theta\!\!\text{ -},\text{cos }\!\!\theta\!\!\text{ },\text{=},\underset{\text{a }\!\!\hat{\mathrm{I}}\!\!\text{ },\text{R}}{\mathop{\text{min}}},,\text{(1,},{{\text{a}}^{\text{2}}}\text{-6a},\text{+},\text{10)} \\
& \Rightarrow ,\text{sin }\!\!\theta\!\!\text{ },\text{-},\text{cos }\!\!\theta\!\!\text{ },\text{=},\text{1} \\
& \Rightarrow ,\text{sin }\!\!\theta\!\!\text{ },\text{=},\text{1+},\text{cos }\!\!\theta\!\!\text{ },
\end{aligned}$
Using the trigonometric identities

& \text{a)}\,\text{1+cos }\\!\\!\theta\\!\\!\text{ }\,\text{=}\,\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}} \\\ & \text{b)}\,\text{sin }\\!\\!\theta\\!\\!\text{ }\,\text{=}\,\text{2sin }\\!\\!\theta\\!\\!\text{ }\,\text{cos }\\!\\!\theta\\!\\!\text{ } \\\ \end{aligned}$$ $$\begin{aligned} & \text{2sin}\,\dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}}\text{cos}\dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}}\,\text{=}\,\text{2}\,\text{cos}\dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}} \\\ & \Rightarrow \text{2sin}\,\dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}}\left[ \text{cos}\dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}}\,\text{=}\,\text{2}\,\text{cos}\dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}} \right]\,\text{=}\,\text{0} \\\ \end{aligned}$$ This equation has two sets of solutions. $$\begin{aligned} & \text{cos}\,\dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}}\,\text{=}\,\text{0} \\\ & \Rightarrow \,\dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}}\,\text{=}\,\text{(2n}\,\text{+}\,\text{1)}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{2}} \\\ & \text{so}\,\text{ }\\!\\!\theta\\!\\!\text{ }\,\text{=}\,\text{(2n+1) }\\!\\!\pi\\!\\!\text{ } \end{aligned}$$ Or $$\text{sin}\,\dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}}\,\text{=}\,\text{cos}\,\dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}}\,\Rightarrow \,\text{tan}\,\dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}}\,\text{=}\,\text{1}$$ Which means $$\begin{aligned} & \dfrac{\text{ }\\!\\!\theta\\!\\!\text{ }}{\text{2}}\,\text{=}\,\text{n }\\!\\!\pi\\!\\!\text{ }\,\text{+}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{4}} \\\ & \text{ }\\!\\!\theta\\!\\!\text{ }\,\text{=}\,\text{2n }\\!\\!\pi\\!\\!\text{ }\,\text{+}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{2}} \end{aligned}$$ Combining the above two solutions, the most general solution we get $$\text{ }\\!\\!\theta\\!\\!\text{ }\,\text{=}\,\text{n }\\!\\!\pi\\!\\!\text{ }\,\text{+}\,{{\text{(-1)}}^{\text{n}}}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{4}}\text{+}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{4}}$$ Which reduces to $$\left\\{ \begin{aligned} & \text{ }\\!\\!\theta\\!\\!\text{ =}\,\text{n }\\!\\!\pi\\!\\!\text{ }\,\text{+}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{2}}\,\text{for}\,\text{even}\,\text{n} \\\ & \text{and} \\\ & \text{ }\\!\\!\theta\\!\\!\text{ }\,\text{=}\,\text{n }\\!\\!\pi\\!\\!\text{ }\,\text{for}\,\text{odd}\,\text{n} \\\ \end{aligned} \right\\}$$ consistent with above solutions **Hence, option B is correct.** **Note:** Another way to solar the trig equation is the following: $$\begin{aligned} & \text{sin}\,\text{ }\\!\\!\theta\\!\\!\text{ }\,\text{=}\,\text{cos}\,\text{ }\\!\\!\theta\\!\\!\text{ }\,\text{=}\,\text{1} \\\ & \text{divide}\,\text{by}\,\sqrt{\text{2}} \\\ & \dfrac{\text{1}}{\sqrt{\text{2}}}\,\text{sin}\,\text{ }\\!\\!\theta\\!\\!\text{ }\,\text{-}\,\dfrac{\text{1}}{\sqrt{\text{2}}}\,\text{cos}\,\text{ }\\!\\!\theta\\!\\!\text{ }\,\text{=}\,\dfrac{\text{1}}{\sqrt{\text{2}}} \\\ & \Rightarrow \,\text{sin}\,\left( \text{ }\\!\\!\theta\\!\\!\text{ }\,\text{-}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{4}} \right)\,\text{=}\,\dfrac{\text{1}}{\sqrt{\text{2}}}\,\left[ \text{using}\,\text{sin}\,\text{(A-}\,\text{B)}\,\text{=}\,\text{sinA}\,\text{cosB-cosAsinB} \right] \\\ & \Rightarrow \,\text{ }\\!\\!\theta\\!\\!\text{ }\,\text{-}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{4}}\,=\,\text{2n }\\!\\!\pi\\!\\!\text{ }\,\text{+}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{4}} \\\ & \\\ \end{aligned}$$ Or $$\left\\{ \begin{aligned} & \text{ }\\!\\!\theta\\!\\!\text{ =}\,\text{2n }\\!\\!\pi\\!\\!\text{ }\,\text{+}\,\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{2}}\text{)}\, \\\ & \text{or} \\\ & \text{ }\\!\\!\theta\\!\\!\text{ }\,\text{=}\,\text{(2n + 1)}\, \\\ \end{aligned} \right\\}$$which is the same set of answers we got before.