How to make this expression independent from B? thank you!

(sin(B- A) + cos A + sin(B + A))/((2 tan B + sec B) *cos Bsin(B−A)+cosA+sin(B+A)(2tanB+secB)⋅cosB

I don't whether that's the right transltion. But the goal of this exercise is to make sure you have a B in your final answer anymore using Simpsons formula's, double-angle formula's, sum and difference formula's... It's for a preparation for an exam, but I can't seem to figure it out. Thank you!

2 Answers
Ratnaker Mehta
Dec 8, 2017

Please refer to the Discussion given in the Explanation.

Explanation:

We will use : sinC+sinD=2sin((C+D)/2)cos((C-D)/2).sinC+sinD=2sin(C+D2)cos(C−D2).

"The Expression="{sin(B-A)+cosA+sin(B+A)}/{(2tanB+secB)cosB},The Expression=sin(B−A)+cosA+sin(B+A)(2tanB+secB)cosB,

=[{sin(B+A)+sin(B-A)}+cosA]/{(2tanB+secB)cosB},={sin(B+A)+sin(B−A)}+cosA(2tanB+secB)cosB,

=[{2sinBcosA}+cosA]/{(2tanB+secB)cosB},={2sinBcosA}+cosA(2tanB+secB)cosB,

=[{2sinBcosA}+cosA]/{2tanBcosB+secBcosB},={2sinBcosA}+cosA2tanBcosB+secBcosB,

=[{2sinBcosA}+cosA]/{(2sinB)/cosB*cosB+1/cosB*cosB},={2sinBcosA}+cosA2sinBcosB⋅cosB+1cosB⋅cosB,

=[{2sinBcosA}+cosA]/(2sinB+1),={2sinBcosA}+cosA2sinB+1,

=[cancel((2sinB+1))cosA}/cancel((2sinB+1)),

=cosA, which is free from B.

Q.E.D.

Cesareo R.
Dec 8, 2017

See below.

Explanation:

We have

g(a,b) = u(a)v(b) then

{(g(0,b)=u(0)v(b)),(g(a,0)=u(a)v(0)),(g(0,0)=u(0)v(0)):}

and then

g(a,b) = (g(0,b)g(a,0))/g(0,0)

but now with a = A and b = B

g(0,B) =(SecB (1 + 2 SinB))/(SecB + 2 TanB)
g(A,0) = cosA
g(0,0) = 1

and with those considerations we conclude:

(Sin(B-A) + CosA + Sin(B-A))/((2 TanB + SecB) CosB) = v(B)cos(A)

but v(B) = 1

then finally

(Sin(B-A) + CosA + Sin(B-A))/((2 TanB + SecB) CosB) = cos(A)

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