Question

2SIN A/2 COS A/2 IS EQUAL TO

Answer 1

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Answer:

$\displaystyle{\sf{2sin\frac{A}{2}cos\frac{A}{2}=sinA }}$

Proof :-

We know that,

sin(A + B) = sinAcosB + sinBcosA          ....(i)

Let us assume that ∠A = ∠B.

Then putting this assumption in eq.(i), we get :-

sin(A + A) = sinAcosA + sinAcosA

⇒sin2A = 2sinAcosA

If we divide the angles into their half, we get :-

⇒sinA = 2sinA/2 cosA/2

(Proved!)

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Regards,

CunningKing :)

Answer 2

Given Trigonometric Expression

2sin(A/2)cos(A/2)

Answer:

$\longrightarrow\rm\pink{2sin\dfrac{A}{2}cos\dfrac{A}{2}=sinA}$

Proof:

$\longrightarrow\rm{L.H.S.=2sin\dfrac{A}{2}cos\dfrac{A}{2}}$

$\longrightarrow\rm{L.H.S.=sin\bigg(\dfrac{A}{2}+\dfrac{A}{2}\bigg)+sin\bigg(\dfrac{A}{2}-\dfrac{A}{2}\bigg)}$

$\longrightarrow\rm{L.H.S.=sin\bigg(\dfrac{\cancel{2}A}{\cancel{2}}\bigg)+sin\bigg(0\bigg)}$

$\longrightarrow\rm{L.H.S.=sin(A)+0}$

$\longrightarrow\rm{L.H.S.=sin(A)}$

$\longrightarrow\rm{L.H.S.=R.H.S.}$

Hence Proved

Identities Used:

$\longrightarrow\rm{2sinAcosB=sin(A+B)+sin(A-B)}$

$\longrightarrow\rm{sin(0)=0}$

Some more Trigonometric identities:

$\longrightarrow\rm{2cosAsinB=sin(A+B)-sin(A-B)}$

$\longrightarrow\rm{2cosAcosB=cos(A+B)+cos(A-B)}$

$\longrightarrow\rm{2sinAsinB=cos(A-B)-cos(A+B)}$

$\longrightarrow\rm{sin(2A)=2sinAcosA}$

$\longrightarrow\sf{cos(2A)=2cos^{2}A-1=1-2sin^{2}A=cos^{2}A-sin^{2}A}$