Distance between the points[tex] \large{\red{\mathfrak{dont\:spam.}}}

1.	Distance between the points[tex] \large{\red{\mathfrak{dont\:spam.}}}
Answer» $\large\underline{\sf{Solution-}}$ Consider, $\rm :\longmapsto\:cos\bigg(\theta + \dfrac{<klux>2</klux>\pi}{3} \bigg) - cos\bigg( \theta + \dfrac{\pi}{3}\bigg)$ We know, $\boxed{ \sf \: cosx - cosy = - \: 2sin\bigg( \dfrac{<klux>X</klux> + y}{2} \bigg)<klux>SIN</klux>\bigg( \dfrac{x - y}{2} \bigg)}$ So, using this identity, we get $\: \: = \sf \: - \: 2 \: sin\bigg( \dfrac{\theta + \dfrac{2\pi}{3} + \theta + \dfrac{\pi}{3}}{2} \bigg)sin\bigg( \dfrac{\theta + \dfrac{2\pi}{3} - \theta - \dfrac{\pi}{3}}{2} \bigg)$ $\: \: = \sf \: - \: 2 \: sin\bigg( \dfrac{\pi}{2} + \theta \bigg)sin\bigg( \dfrac{\pi}{6} \bigg)$ $\: \: \sf \: = \: - \: 2 \: cos\theta \: \times \dfrac{1}{2} \: \: \: \: \: \: \: \{ \because \:sin\bigg( \dfrac{\pi}{2} + \theta\bigg) = cos\theta \}$ $\: \: = \sf \: - cos\theta$ $\bf\implies \: \boxed{ \sf \: cos\bigg(\theta + \dfrac{2\pi}{3} \bigg) - cos\bigg( \theta + \dfrac{\pi}{3}\bigg) = - cos\theta} - - (1)$ Now, Consider, $\rm :\longmapsto\:sin\bigg(\theta + \dfrac{2\pi}{3} \bigg) - sin\bigg( \theta + \dfrac{\pi}{3}\bigg)$ We know that $\boxed{ \sf \: sinx - siny = 2 \: cos\bigg( \dfrac{x + y}{2} \bigg)sin\bigg( \dfrac{x - y}{2} \bigg)}$ So, using this identity, we get $\: \: \sf \: = \:2 \: cos\bigg( \dfrac{\theta + \dfrac{2\pi}{3} + \theta + \dfrac{\pi}{3}}{2} \bigg)sin\bigg( \dfrac{\theta + \dfrac{2\pi}{3} - \theta - \dfrac{\pi}{3}}{2} \bigg)$ $\: \: = \sf \: \: 2 \: cos\bigg( \dfrac{\pi}{2} + \theta \bigg)sin\bigg( \dfrac{\pi}{6} \bigg)$ $\: \: = \sf \: - \: 2 \: sin\theta \times \dfrac{1}{2}$ $\: \: = \sf \: - \: sin\theta$ $\rm :\implies\: \boxed{ \sf \: sin\bigg(\theta + \dfrac{2\pi}{3} \bigg) - sin\bigg( \theta + \dfrac{\pi}{3}\bigg) = - sin\theta } - - (2)$ Now, We know, Distance between two points is given by Distance Formula. $\: \: \: \boxed{\sf\ \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }}$ $\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: (x_2,y_2)$ Here, we have to find Distance between $\rm :\longmapsto\: \sf \:A \: \bigg[a \: cos\bigg( \dfrac{2\pi}{3} + \theta\bigg), \: a \: sin\bigg(\dfrac{2\pi}{3} + \theta \bigg)\bigg]$ and $\rm :\longmapsto\: \sf \: B \: \bigg[a \: cos\bigg(\dfrac{\pi}{3} + \theta\bigg), \: sin\bigg(\dfrac{\pi}{3} + \theta\bigg)\bigg]$ So, Distance between A and B is given by $\sf \: AB = \sqrt{ { {a}^{2} \bigg( cos\bigg(\theta + \dfrac{2\pi}{3} \bigg) - cos\bigg( \theta + \dfrac{\pi}{3}\bigg) \bigg) }^{2} + {a}^{2} {\bigg( sin\bigg(\theta + \dfrac{2\pi}{3} \bigg) - sin\bigg( \theta + \dfrac{\pi}{3}\bigg) \bigg) }^{2}}$ $\: \: = \sf \: \sqrt{ {a}^{2} {( - cos\theta)}^{2} + {a}^{2} {( - sin\theta)}^{2}}$ $\: \: = \sf \: \sqrt{ {a}^{2} {cos}^{2}\theta + {a}^{2} {sin}^{2}\theta }$ $\: \: = \sf \: \sqrt{ {a}^{2} \bigg( {sin}^{2}\theta + {cos}^{2}\theta\bigg) }$ $\: \: = \sf \: a$ ADDITIONAL Information : - Trigonometry Formulas sin(−θ) = −sin θ cos(−θ) = cos θ TAN(−θ) = −tan θ cosec(−θ) = −cosecθ sec(−θ) = sec θ cot(−θ) = −cot θ Product to Sum Formulas sin x sin y = 1/2 [cos(x–y) − cos(x+y)] cos x cos y = 1/2[cos(x–y) + cos(x+y)] sin x cos y = 1/2[sin(x+y) + sin(x−y)] cos x sin y = 1/2[sin(x+y) – sin(x−y)] Sum to Product Formula sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2] sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2] cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2] cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2] Sum or Difference of angles cos (A + B) = cos A cos B – sin A sin B cos (A – B) = cos A cos B + sin A sin B sin (A+B) = sin A cos B + cos A sin B sin (A -B) = sin A cos B – cos A sin B tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)] tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)] cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)] cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)] cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A Multiple and Submultiple angles sin2A = 2sinA cosA = [2tan A /(1+tan²A)] cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)] tan 2A = (2 tan A)/(1-tan²A)

Answer»

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\:cos\bigg(\theta + \dfrac{<klux>2</klux>\pi}{3} \bigg) - cos\bigg( \theta + \dfrac{\pi}{3}\bigg)$

We know,

$\boxed{ \sf \: cosx - cosy = - \: 2sin\bigg( \dfrac{<klux>X</klux> + y}{2} \bigg)<klux>SIN</klux>\bigg( \dfrac{x - y}{2} \bigg)}$

So, using this identity, we get

$\: \: = \sf \: - \: 2 \: sin\bigg( \dfrac{\theta + \dfrac{2\pi}{3} + \theta + \dfrac{\pi}{3}}{2} \bigg)sin\bigg( \dfrac{\theta + \dfrac{2\pi}{3} - \theta - \dfrac{\pi}{3}}{2} \bigg)$

$\: \: = \sf \: - \: 2 \: sin\bigg( \dfrac{\pi}{2} + \theta \bigg)sin\bigg( \dfrac{\pi}{6} \bigg)$

$\: \: \sf \: = \: - \: 2 \: cos\theta \: \times \dfrac{1}{2} \: \: \: \: \: \: \: \{ \because \:sin\bigg( \dfrac{\pi}{2} + \theta\bigg) = cos\theta \}$

$\: \: = \sf \: - cos\theta$

$\bf\implies \: \boxed{ \sf \: cos\bigg(\theta + \dfrac{2\pi}{3} \bigg) - cos\bigg( \theta + \dfrac{\pi}{3}\bigg) = - cos\theta} - - (1)$

Now, Consider,

$\rm :\longmapsto\:sin\bigg(\theta + \dfrac{2\pi}{3} \bigg) - sin\bigg( \theta + \dfrac{\pi}{3}\bigg)$

We know that

$\boxed{ \sf \: sinx - siny = 2 \: cos\bigg( \dfrac{x + y}{2} \bigg)sin\bigg( \dfrac{x - y}{2} \bigg)}$

So, using this identity, we get

$\: \: \sf \: = \:2 \: cos\bigg( \dfrac{\theta + \dfrac{2\pi}{3} + \theta + \dfrac{\pi}{3}}{2} \bigg)sin\bigg( \dfrac{\theta + \dfrac{2\pi}{3} - \theta - \dfrac{\pi}{3}}{2} \bigg)$

$\: \: = \sf \: \: 2 \: cos\bigg( \dfrac{\pi}{2} + \theta \bigg)sin\bigg( \dfrac{\pi}{6} \bigg)$

$\: \: = \sf \: - \: 2 \: sin\theta \times \dfrac{1}{2}$

$\: \: = \sf \: - \: sin\theta$

$\rm :\implies\: \boxed{ \sf \: sin\bigg(\theta + \dfrac{2\pi}{3} \bigg) - sin\bigg( \theta + \dfrac{\pi}{3}\bigg) = - sin\theta } - - (2)$

Now,

We know,

Distance between two points is given by Distance Formula.

$\: \: \: \boxed{\sf\ \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }}$

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: (x_2,y_2)$

Here,

we have to find Distance between

$\rm :\longmapsto\: \sf \:A \: \bigg[a \: cos\bigg( \dfrac{2\pi}{3} + \theta\bigg), \: a \: sin\bigg(\dfrac{2\pi}{3} + \theta \bigg)\bigg]$

and

$\rm :\longmapsto\: \sf \: B \: \bigg[a \: cos\bigg(\dfrac{\pi}{3} + \theta\bigg), \: sin\bigg(\dfrac{\pi}{3} + \theta\bigg)\bigg]$

So, Distance between A and B is given by

$\sf \: AB = \sqrt{ { {a}^{2} \bigg( cos\bigg(\theta + \dfrac{2\pi}{3} \bigg) - cos\bigg( \theta + \dfrac{\pi}{3}\bigg) \bigg) }^{2} + {a}^{2} {\bigg( sin\bigg(\theta + \dfrac{2\pi}{3} \bigg) - sin\bigg( \theta + \dfrac{\pi}{3}\bigg) \bigg) }^{2}}$

$\: \: = \sf \: \sqrt{ {a}^{2} {( - cos\theta)}^{2} + {a}^{2} {( - sin\theta)}^{2}}$

$\: \: = \sf \: \sqrt{ {a}^{2} {cos}^{2}\theta + {a}^{2} {sin}^{2}\theta }$

$\: \: = \sf \: \sqrt{ {a}^{2} \bigg( {sin}^{2}\theta + {cos}^{2}\theta\bigg) }$

$\: \: = \sf \: a$

ADDITIONAL Information : -

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

TAN(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formula

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)