Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value of θ when these two points coincide. (0 ≤ θ < 2π)
Show your work, thanks!​

Answer:

[tex] \rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ} [/tex]

[tex] \rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1, \frac{\sqrt{3}}{2} - 1[/tex]

Step-by-step explanation:

we are given two coincident points

[tex] \displaystyle P( \sin(θ)+2, \tan(θ)-2) \: \text{and } \\ \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)[/tex]

since they are coincident points

[tex] \rm \displaystyle P( \sin(θ)+2, \tan(θ)-2) = \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)[/tex]

By order pair we obtain:

[tex] \begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) = \sin( \theta) + 2 \\ \\ \displaystyle 3 \sin( \theta) - 2 \cos( \theta) + a = \tan( \theta) - 2\end{cases}[/tex]

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using substitution method

to do so, make a the subject of the equation.therefore from the second equation we acquire:

[tex] \begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sinθ \cos(θ)+2a \cos(θ )= \sin( \theta) + 2 \\ \\ \boxed{\displaystyle a = \tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) } \end{cases}[/tex]

now substitute:

[tex] \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) \}= \sin( \theta) + 2 [/tex]

distribute:

[tex]\rm\displaystyle \displaystyle 4 \sin ^{2}( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) - 6 \sin( \theta) \cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2 [/tex]

collect like terms:

[tex]\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2 [/tex]

rearrange:

[tex]\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) + 4 \cos ^{2} ( \theta) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + = \sin( \theta) + 2 [/tex]

by Pythagorean theorem we obtain:

[tex]\rm\displaystyle \displaystyle 4 - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) + 2 [/tex]

cancel 4 from both sides:

[tex]\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) - 2[/tex]

move right hand side expression to left hand side and change its sign:

[tex]\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2 = 0[/tex]

factor out sin:

[tex]\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2 = 0[/tex]

factor out 2:

[tex]\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) + 2(- 2\cos( \theta) + 1 ) = 0[/tex]

group:

[tex]\rm\displaystyle \displaystyle ( \sin (θ) + 2)(- 2 \cos(θ)+1) = 0[/tex]

factor out -1:

[tex]\rm\displaystyle \displaystyle - ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0[/tex]

divide both sides by -1:

[tex]\rm\displaystyle \displaystyle ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0[/tex]

by Zero product property we acquire:

[tex] \begin{cases}\rm\displaystyle \displaystyle \sin (θ) + 2 = 0 \\ \displaystyle2 \cos(θ) - 1= 0 \end{cases}[/tex]

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

[tex] \begin{cases}\rm\displaystyle \displaystyle \sin (θ) \neq - 2 \\ \displaystyle2 \cos(θ) = 1\end{cases}[/tex]

divide both sides by 2:

[tex] \rm\displaystyle \displaystyle \displaystyle \cos(θ) = \frac{1}{2}[/tex]

by unit circle we get:

[tex] \rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ} [/tex]

so when θ is 60° a is:

[tex] \rm \displaystyle a = \tan( {60}^{ \circ} ) - 2 - 3 \sin( {60}^{ \circ} ) + 2 \cos( {60}^{ \circ} ) [/tex]

recall unit circle:

[tex] \rm \displaystyle a = \sqrt{3} - 2 - \frac{ 3\sqrt{3} }{2} + 2 \cdot \frac{1}{2} [/tex]

simplify which yields:

[tex] \rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1[/tex]

when θ is 300°

[tex] \rm \displaystyle a = \tan( {300}^{ \circ} ) - 2 - 3 \sin( {300}^{ \circ} ) + 2 \cos( {300}^{ \circ} ) [/tex]

remember unit circle:

[tex] \rm \displaystyle a = - \sqrt{3} - 2 + \frac{3\sqrt{ 3} }{2} + 2 \cdot \frac{1}{2} [/tex]

simplify which yields:

[tex] \rm \displaystyle a = \frac{ \sqrt{3} }{2} - 1[/tex]

and we are done!

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