Prove sin(A+B)=sinAcosB+cosAsinB

In this article, we will show that sin(A+B)=sinAcosB+cosAsinB.

Proof Geometrically

Let a triangle right angle triangle \Delta ABC such that the line AB extend upto a point D. AC is the hypotenuse of \Delta ABC. Let \angle CAB = \alpha +\beta.

right angle triangle

Draw a line from AF from A that insect the line CB at E such that \angle EAB = \alpha \; and \; \angle CAE = \beta

Proof of sin(A+B)=sinAcosB+cosAsinB Geometrically

Draw a line from C perpendicular to AF at point G. After that, draw two lines from G, one perpendicular to CB, and another perpendicular to AD.

proof the given trigonometry identify

We can easily show that \Delta AEB is similar to \Delta GEH and \Delta GEH is similar to \Delta CGH. Therefore, we can say \angle GCH = \angle EGH = \angle EAB = \alpha.

show sin(A+B)=sinAcosB+cosAsinB geometrical method

Since, \Delta ABC is right angled. Using sin formula, we get

\\*\sin(\alpha +\beta ) = \dfrac{CB}{AC}\newline\\*\sin(\alpha +\beta ) = \dfrac{CH + HB}{AC}\newline\\*\sin(\alpha +\beta ) = \dfrac{CH}{AC} + \dfrac{HB}{AC}\newline\\*\sin(\alpha +\beta ) = \dfrac{CH}{AC}*\dfrac{CG}{CG} + \dfrac{HB}{AC}*\dfrac{AG}{AG}\newline\\*\sin(\alpha +\beta ) = \dfrac{CH}{CG}*\dfrac{CG}{AC} + \dfrac{HB}{AG}*\dfrac{AG}{AC}

BHGJ is a rectangle. Therefore, HB = JG . Substituting in above equation,

\\*\sin(\alpha +\beta ) = \dfrac{CH}{CG}*\dfrac{CG}{AC} + \dfrac{JG}{AG}*\dfrac{AG}{AC}\;\;\;\;...1

From \Delta AGJ,

\\*\sin(\alpha) = \dfrac{JG}{AG}\;\;\;\;...2

From \Delta AGC,

\\*\cos(\beta) = \dfrac{AG}{AC}\;\;\;\;...3

\\*\sin(\beta) = \dfrac{CG}{AC}\;\;\;\;...4

From \Delta CHG,

\\*\cos(\alpha) = \dfrac{CH}{CG}\;\;\;\;...5

Substituting equation 2, 3, 4, 5 in equation 1.

\\*\sin(\alpha +\beta ) = \sin(\alpha)cos(\beta) + \cos(\alpha)sin(\beta)

Hence Proved.

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