Question

Solve this question please
$\cos(a) + \cos(b) + \cos(c) = 1 + 4 \sin( \frac{a}{2} ) \sin( \frac{b}{2} ) \sin( \frac{c}{2} )$

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Answer 1

Appropriate Question :-

$\sf \: If \: a + b + c = \pi, \: prove \: that$

$\sf \: \cos(a) + \cos(b) + \cos(c) = 1 + 4sin\bigg[\dfrac{a}{2} \bigg]sin\bigg[\dfrac{b}{2} \bigg]sin\bigg[\dfrac{c}{2} \bigg]$

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

Consider,

$\rm :\longmapsto\:\cos(a) + \cos(b) + \cos(c)$

We know,

$\red{ \boxed{ \sf{ \:cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}}$

Using this, we get

$\rm \:  =  \: 2cos\bigg[\dfrac{a + b}{2} \bigg]cos\bigg[\dfrac{a - b}{2} \bigg] + cosc$

$\rm \:  =  \: 2cos\bigg[\dfrac{\pi - c}{2} \bigg]cos\bigg[\dfrac{a - b}{2} \bigg] + cosc$

$\rm \:  =  \: 2cos\bigg[\dfrac{\pi}{2} - \dfrac{c}{2} \bigg]cos\bigg[\dfrac{a - b}{2} \bigg] + cosc$

We know,

$\red{ \boxed{ \sf{ \:cos2x = 1 - 2 {sin}^{2}x}}}$

So, using this, we get

$\rm \:  =  \: 2sin\bigg[\dfrac{c}{2} \bigg]cos\bigg[\dfrac{a - b}{2} \bigg] + 1 - 2 {sin}^{2}\bigg[\dfrac{c}{2} \bigg]$

$\rm \:  = 1 +2sin\bigg[\dfrac{c}{2} \bigg]cos\bigg[\dfrac{a - b}{2} \bigg]- 2 {sin}^{2}\bigg[\dfrac{c}{2} \bigg]$

$\rm \:  = 1 +2sin\bigg[\dfrac{c}{2} \bigg]\bigg(cos\bigg[\dfrac{a - b}{2} \bigg]- sin\bigg[\dfrac{c}{2} \bigg]\bigg)$

$\rm \:  = 1 +2sin\bigg[\dfrac{c}{2} \bigg]\bigg(cos\bigg[\dfrac{a - b}{2} \bigg]- sin\bigg[\dfrac{\pi - (a + b)}{2} \bigg]\bigg)$

$\rm \:  = 1 +2sin\bigg[\dfrac{c}{2} \bigg]\bigg(cos\bigg[\dfrac{a - b}{2} \bigg]- sin\bigg[\dfrac{\pi}{2} - \bigg[\dfrac{a + b}{2} \bigg] \bigg]\bigg)$

$\rm \:  = 1 +2sin\bigg[\dfrac{c}{2} \bigg]\bigg(cos\bigg[\dfrac{a - b}{2} \bigg]- cos\bigg[\dfrac{a + b}{2} \bigg] \bigg)$

$\rm \:  = 1 +2sin\bigg[\dfrac{c}{2} \bigg]\bigg(cos\bigg[\dfrac{a}{2} - \dfrac{b}{2} \bigg]- cos\bigg[\dfrac{a}{2} + \dfrac{b}{2} \bigg] \bigg)$

We know,

$\red{ \boxed{ \sf{ \:cos(x - y) - cos(x + y) = 2sinxsiny}}}$

So, using this, we get

$\rm \:  =  \: 1 + 2sin\bigg[\dfrac{c}{2} \bigg] \times 2sin\bigg[\dfrac{a}{2} \bigg]sin\bigg[\dfrac{b}{2} \bigg]$

$\rm \:  =  \: 1 + 4sin\bigg[\dfrac{a}{2} \bigg]sin\bigg[\dfrac{b}{2} \bigg]sin\bigg[\dfrac{c}{2} \bigg]$

Hence, Proved

Additional Information :-

$\red{ \boxed{ \sf{ \:cosx - cosy = - \: 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}}$

$\red{ \boxed{ \sf{ \:sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}}$

$\red{ \boxed{ \sf{ \:sinx + siny = 2cos\bigg[\dfrac{x - y}{2} \bigg]sin\bigg[\dfrac{x + y}{2} \bigg]}}}$

$\red{ \boxed{ \sf{ \:2cosxcosy = cos(x + y) + cos(x - y)}}}$

$\red{ \boxed{ \sf{ \:2sinxcosy = sin(x + y) + sin(x - y)}}}$

Answer 2

This is actually an identity .

When a + b + c = π , this holds true.

If a + b + c = π , a+b = π - c; b+c = π - a and c+a = π - b

cos theta = 1-sin²(theta/2). This is a half angle identity

cos a + cos b + cos c

>> 2cos[(a+b)/2] . cos [(a-b)/2] + [ 1-sin²(c/2)]

>> 2cos[(π-c)/2] . cos[(a-b)/2] + [1-sin²(c/2)]

Let us seperate the 1 from here

>> 1 + { 2cos[(π-c)/2] . cos[(a-b)/2] - sin²(c/2) }

Now , this term 2 cos [(π-c)]/2 becomes 2 sin(c/2) .

As cos [ π/2 - x ] = sin x

So

>> 1 + { 2sin(c/2) . cos [(a-b)]/2 - sin²(c/2) }

We can take sin(c/2) common

>> 1 + 2sin(c/2){ cos(a-b)/2 - sin(c/2) }

Now , c = π-a-b ( Check the second line)

>> 1 + sin(c/2) { 2cos(a-b)/2 - sin(π-a-b)/2 }

Let us focus on the term , sin (π-a-b)/2

>> sin [ π/2 - (a+b)/2 ]

Now , sin [ π/2 - x] = cos x

So this term becomes cos(a+b)/2

>> 1 + 2 sin(c/2) { cos (a-b)/2 - cos(a+b)2 }

>> 1 + 2 sin(c/2) { 2 sin(a/2)(b/2) }. { Identity }

>> 1 + 2×2 [ sin(a/2) . sin (b/2) . sin (c/2) ]

>> 1 + 4sin(a/2)sin(b/2)sin(c/2)

Hence we have proved the identity .

In case you had any problems in understanding till here , feel free to drop a message in my ínbox. I will be glad to help :)

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