Verify that sin(A+B) + sin(A-B) = 2sinA sinB ?

Question

Asher Friedman · Accepted Answer

#"see explanation"# #"using the "color(blue)"addition formulae for sin"# #•color(white)(x)sin(A+-B)=sinAcosB+-cosAsinB# #rArrsin(A+B)=sinAcosB+cosAsinB# #rArrsin(A-B)=sinAcosB-cosAsinB# #rArrsin(A+B)+sin(A-B)=2sinAcosB# #!=2sinAsinBlarr"check your question"#

Josiah Anderson · Answer

It is not an identity.

It is not an identity. #A = 90° , B = 0° # LS: #sin(A+B) + sin(A-B) = sin (90°+0°) + sin ( 90°-0°) = 2# RS: # 2sinA sinB = 2 sin 90° sin 0° = 2 xx1xx0 = 0# #2!=0# # = 2sinA sinB # #sin(A+B) + sin(A-B) = 2sinA sinB # #LHS: sin(A+B) + sin(A-B) # #sinAcosB + cosAsinB + sinAcosB - cosAsinB = # #sinAcosB + sinAcosB = 2sinAcosB #

Anna Andrews · Answer

undefined To verify the identity sin(A+B) + sin(A-B) = 2sinA sinB, we'll start by expanding sin(A+B) and sin(A-B) using the sum and difference identities for sine:

sin(A+B) = sinA cosB + cosA sinB
sin(A-B) = sinA cosB - cosA sinB

Now, let's add sin(A+B) and sin(A-B) together:

sin(A+B) + sin(A-B) = (sinA cosB + cosA sinB) + (sinA cosB - cosA sinB)

Grouping like terms:

= sinA cosB + sinA cosB + cosA sinB - cosA sinB

Simplify by combining the like terms:

= 2sinA cosB

Using the identity sin(2θ) = 2sinθ cosθ, where θ = A, we can rewrite 2sinA cosB as sin(2A):

= sin(2A)

So, sin(A+B) + sin(A-B) = sin(2A).

Therefore, the identity sin(A+B) + sin(A-B) = 2sinA sinB is not valid. The correct identity is sin(A+B) + sin(A-B) = 2sinA cosB.