Question

Can anyone explain me how the equations comes which is underlined or what formulas is used here?
Don't post irrelevant answers to gain points.​

Answer 1

Explanation:

This probably looks like a sum on friction.

The mathematical part actually falls under the multiple and sub-multiple angles of trigonometric ratios.

Please find attached the explanation.

Hope your doubt is solved.

Answer 2

Explanation :

mg sinθ ≥ μ(mg - mgcosθ)

sinθ can be written as :

$\sin \theta = \sin 2\bigg(\dfrac{\theta}{2}\bigg) [ \because \tt 2 \ cancels \ 2]$

⇒ sin 2(θ/2) is of the form sin2A

sin2A :

sin 2A = sin (A + A)

we know, sin (C + D) = sin C cos D + cos C sin D

put C = D = A,

sin 2A

= sin (A + A)

= sin A cos A + cos A sin A

= sin A cos A (1 + 1)

= sin A cos A (2)

= 2 sinA cosA

sin2(θ/2) :

$\sf \sin 2 \bigg(\dfrac{\theta}{2} \bigg) = 2\sin \bigg(\dfrac{\theta}{2}\bigg) \cos \bigg(\dfrac{\theta}{2} \bigg)$

Now, taking the given expression :

mg sinθ ≥ μ(mg - mgcosθ)

$\sf mgsin2\bigg( \dfrac{\theta}{2} \bigg) \geq \mu (mg-mg \cos \theta) \\\\ \sf mg \times 2\sin \bigg( \dfrac{\theta}{2} \bigg) \cos \bigg( \dfrac{\theta}{2} \bigg) \geq \mu mg (1- \cos \theta) \\\\ \sf 2\sin \bigg( \dfrac{\theta}{2} \bigg) \cos \bigg( \dfrac{\theta}{2} \bigg) \geq \mu (1- \cos \theta)$

cosθ :

cosθ can be written as cos 2(θ/2)

it is of the form cos2A

we know,

cos2A = cos (A+A)

cos2A = cos A cos A - sinA sinA

cos2A = cos²A - sin²A

cos2A = (1 - sin²A) - sin²A

cos2A = 1 - sin²A - sin²A

cos2A = 1 - 2sin²A

1 - cos2A = 1 - [1 - 2sin²A]

1 - cos2A = 1 - 1 + 2sin²A

1 - cos2A = 2sin²A

Put A = θ/2 :

$\sf 1-\cos \theta \\\\ =1-\cos 2\bigg( \dfrac{\theta}{2} \bigg) \\\\ =2 \sin ^2 \bigg( \dfrac{\theta}{2} \bigg)$

Now,

$\sf 2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} \geq \mu (1- \cos \theta) \\\\ \sf 2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} \geq \mu \times 2 \sin^2 \dfrac{\theta}{2} \\\\ \sf 2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} \geq 2\mu sin^2 \dfrac{\theta}{2}$