I need to prove sin A cos B = 1/2 [ sin (A + B) + sin (A - B) ]?

Question

Thank you in advance.

Answer 1

Use the addition and subtraction theorem:

sin (A + B) = sin A + cos B
sin (A - B) = sin A - cos B

From there it's just a matter of simplification.

sin A cos B = 1/2 [ sin (A + B) + sin (A - B) ]
sin A cos B = 1/2 [ sin A cos B + cos A sin B + sin (A - B) ]
sin A cos B = 1/2 [ sin A cos B + cos A sin B + sin A cos B - cos A sin B ]
sin A cos B = 1/2 [ 2 sin A cos B ]
sin A cos B = sin A cos B

Answer 2

Sin A Cos B

Answer 3

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RE:
I need to prove sin A cos B = 1/2 [ sin (A + B) + sin (A - B) ]?
Thank you in advance.

Answer 4

Start with:

sin (A + B) + sin (A - B)
= sinAcosB + cosAsinB + sinAcosB - cosAsinB
eliminate ( cosAsinB - cosAsinB)
= 2 sinAcosB + 0
So, we got: 2 sin A cos B = [ sin (A + B) + sin (A - B) ]

Multiply both sides by 1/2, you get the original equation.

Answer 5

Sin A Sin B

Answer 6

sin (A + B) = sin A cos B + cos A sin B
sin (A - B) = sin A cos B - cos A sin B
sum = 2 sin A cos B
(1/2) sum = sin A cos B as required

Answer 7

There are three formulae you need to apply. The first is the sum of angles formula:

sin(x + y) = sin(x)cos(y) + sin(y)cos(x)

The other two are the even and odd properties of cos and sin:

cos(-x) = cos(x)
sin(-x) = -sin(x)

We can start with the RHS of the identity. Applying the sum rule we get:

1/2[sin(A+B) + sin(A + (-B))]
= 1/2[sin(A)cos(B) + sin(B)cos(A) + sin(A)cos(-B) + sin(-B)cos(A)]

Applying the even property of cos and the odd property of sin to move the negatives from the functions:

=1/2[sin(A)cos(B) + sin(B)cos(A) + sin(A)cos(B) - sin(B)cos(A)]

Here, the second and fourth terms cancel out, while the first and third terms are the same, so we get:

=1/2[2sin(A)cos(B)]
= sin(A)cos(B)

which is the required result.

Answer 8

sin(A+B) = sinAcosB + cosAsinB
sin(A-B) = sinAcosB - cosAsinB

sin(A+B) + sin(A-B) = 2sinAcosB
sinAcosB = (1/2)[sin(A+B) + sin(A-B)]

Answer 9

sin(A+B)=sinAcosB+cosAsinB
sin(A-B)=sinAcosB-cosAsinB
add this two equations
sin(A+B)+sin(A-B)=2sinAcosB
sinAcosB=1/2[sin(A+B)+sin(A-B)]

Answer 10

cos(a+b)=cosa cosb-sina sinb cos(a-b)=cosa cosb+sina sinb now subtract them and you will get: cos(a-b) - cos(a+b)=2sina sinb