Question

solve this question anyone​

Answer 1

------------[To prove]- - - - -

$\sf cos^2(\frac{A}{2})+cos^2(\frac{B}{2})-cos^2(\frac{c}{2})$

$=\sf 2cos(\frac{A}{2})\times cos(\frac{B}{2})\times sin(\frac{c}{2})$

--------------[proof]- - - - - - - -

LHS = $\sf cos^2(\frac{A}{2})+cos^2(\frac{B}{2})-cos^2(\frac{c}{2})$

[ As we know, 1+cos2Φ=2cos²Φ

so, 1+cosΦ=2cos²(Φ/2)

Angle in terms of A ,

1+cosA=2cos²(A/2)

cos²(A/2)=[{1+cosA}/2]

similarly,

cos²(B/2)=[{1+cosB}/2]

As we know the, trigonometric identity , sin²Φ+cos²Φ=1

So, cos²(c/2)={1-sin²(c/2)}]

therefore ,

$=\sf (\frac{1+cosA}{2})+(\frac{1+cosB}{2})-[{1-sin^2(\frac{c}{2})}]$

$=\sf (\frac{1}{2})+(\frac{1}{2}cosA)+(\frac{1}{2})+(\frac{1}{2}cosB)-1+(sin^2(\frac{c}{2}))$

$=\sf \frac{1}{2}(cosA+cosB)+1-1+(sin^2(\frac{c}{2}))$

$=\sf \frac{1}{2}[2cos(\frac{A+B}{2}).cos(\frac{A-B}{2})]+(sin^2(\frac{c}{2}))$

we know that, the angles of trigonometric functions in a triangle,

let, the angles are,{ A+B+C=π...(angle sum property) }

taking (1/2)on both the sides,

So, 1/2[(A)+(B)+(c)]=1/2(π)

=(A/2)+(B/2)=(π/2)-(C/2)

taking cos on both the sides,

∴ cos[(A/2)+(B/2)]=cos[(π/2)-(C/2)]

therefore ,

$=\sf cos(\frac{\pi}{2}-\frac{C}{2}).cos(\frac{A-B}{2})+sin^2(\frac{C}{2})$

$=\sf sin(\frac{C}{2}).cos(\frac{A-B}{2})+sin^2(\frac{C}{2})$

$=\sf sin(\frac{C}{2})[cos(\frac{A-B}{2})+sin(\frac{C}{2})]$

similarly in a triangle it also applicable for sin and other trigonometric ratios i.e

{sin(c/2)=[sin(π/2)-{(A+B)/2}]}

therefore,

$=\sf sin(\frac{c}{2})[cos(\frac{A-B}{2})+sin(\frac{\pi}{2}-(\frac{A+B}{2}))]$

$=\sf sin(\frac{C}{2})[cos(\frac{A-B}{2})+cos(\frac{A+B}{2})]$

-----{since sin(π/2-Φ)=cosΦ}

Now use the formula,

cosA+cosB=2cos[(A+B)/2].cos[(A-B)/2]

So,

$=\sf sin(\frac{C}{2}).2cos(\frac{\frac{A+B}{2}+\frac{A-B}{2}}{2}).cos(\frac{\frac{A+B}{2}-\frac{A-B}{2}}{2})$

$=\sf sin(\frac{C}{2}).2cos(\frac{\frac{2A}{2}}{2}).cos(\frac{\frac{2B}{2}}{2})$

$=\sf sin(\frac{C}{2}).2cos(\frac{A}{2}).cos(\frac{B}{2})$

Rearrange the terms

$=\sf 2cos(\frac{A}{2}).cos(\frac{B}{2}).sin(\frac{C}{2})$

= RHS, HENCE PROVED