Q13. Prove that, sin (A + B) = sin A.cos B + cos A.sin B
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Q13. Prove that, sin (A + B) = sin A.cos B + cos A.sin B
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Answer:
sin (A + B) = sin A.cos B + cos A.sin B
A = B = 45°
LHS:
sin (A + B) = sin(45°+45°)
= sin 90°
= 1
RHS:
sin A.cos B + cos A.sin B =
= sin45°. cos45°+cos 45°.sin 45°
= 1/√2•1/√2 + 1/√2•1/√2
= 1/2+1/2
= 1
LHS = RHS
sin (A + B) = sin A.cos B + cos A.sin B
Hence it proved.
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
We know,
➢ By Eulers form, we have
[tex]\rm :\longmapsto\: {e}^{ix} = cosx \: + \: i \: sinx[/tex]
So,
➢ Replace x by A + B.
[tex]\bf :\longmapsto\: {e}^{i(A + B)} = cos(A + B) \: + \: i \: sin(A + B) - - (1)[/tex]
Also,
[tex]\bf :\longmapsto\: {e}^{i(A + B)} [/tex]
[tex]\rm \: \: = \: {e}^{iA} \times {e}^{iB} [/tex]
[tex]\rm \: \: = \:(cosA + isinA)(cosB + isinB)[/tex]
[tex]\rm \: \: = \:cosAcosB + icosAsinB + isinAcosB + {i}^{2}sinAsinB[/tex]
[tex]\rm \: \: = \:cosAcosB + i(cosAsinB + sinAcosB) - sinAsinB[/tex]
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {l}^{2} = - \: 1 \bigg \}}[/tex]
[tex]\rm \: \: = \:(cosAcosB - sinAsinB) + i(cosAsinB + sinAcosB)[/tex]
[tex]\rm {e}^{i(A + B)} = \:(cosAcosB - sinAsinB) + i(cosAsinB + sinAcosB) - - (2)[/tex]
From equation (1) and (2), we concluded that
[tex]\rm\:cos(A + B)+isin(A + B) = (cosAcosBsinAsinB) + i(sinAcosB + sinBcosA)[/tex]
So, on comparing imaginary part, we get
[tex]\bf :\longmapsto\:sin(A + B) = sinAcosB + sinBcosA[/tex]
Hence, Proved
Additional Information :-
If we compare the real part, we get
[tex]\rm :\longmapsto\:cos(A + B) = cosAcosB - sinAsinB[/tex]