Question

Prove that
sin+a+cos? (a+B)+2 sina sinß cos(a+B)
is independent of a​

Answer 1

Answer:

Explanation:

        Proof of the product and sum formulas

Products as sums

a)  sin alpha cos β  =   ½[sin (alpha + β) + sin (alpha − β)]

 b)  cos alpha sin β  =   ½[sin (alpha + β) − sin (alpha − β)]

 c)  cos alpha cos β  =   ½[cos (alpha + β) + cos (alpha − β)]

 d)  sin alpha sin β  =   −½[cos (alpha + β) − cos (alpha − β)]

Proof

These formulas are also derived from the sum and difference formulas.  To derive (a), write

sin (alpha + β)   =   sin alpha cos β + cos alpha sin β

 

sin (alpha − β)   =   sin alpha cos β − cos alpha sin β

and add vertically.  The last terms in each line will cancel:

sin (alpha + β) + sin (alpha − β) = 2 sin alpha cos β.

Therefore, on exchanging sides,

2 sin alpha cos β = sin (alpha + β) + sin (alpha − β),

so that

sin alpha cos β = ½[sin (alpha + β) + sin (alpha − β)].  

This is the identity (a)).

Formula (b) is derived in exactly the same manner, only instead of adding, subtract sin (alpha − β) from sin (alpha + β).  

Formulas (c) and (d) are derived similarly.  To derive (c), write

cos (alpha + β) = cos alpha cos β − sin alpha sin β,

cos (alpha − β) = cos alpha cos β + sin alpha sin β,

and add.  To derive (d), subtract.

Let us derive (d).  On subtracting, the first terms on the right will cancel.  We will have

cos (alpha + β) − cos (alpha − β) = −2 sin alpha sin β.

Therefore, on solving for sin alpha sin β,

sin alpha sin β = −½[cos (alpha + β) − cos (alpha − β)].

Sums as products

e)  sin A + sin B   =   2 sin ½ (A + B) cos ½ (A − B)

 

f)  sin A − sin B   =   2 sin ½ (A − B) cos ½ (A + B)

 

g)  cos A + cos B   =   2 cos ½ (A + B) cos ½ (A − B)

 

h)  cos A − cos B   =   −2 sin ½ (A + B) sin ½ (A − B)

Proof

The formulas (e), (f), (g), (h) are derived from (a), (b), (c), (d) respectively; that is, (e) comes from (a), (f) comes from (b), and so on.  

To derive (e), exchange sides in (a):

½[sin (alpha + β) + sin (alpha − β)] = sin alpha cos β,

so that

sin (alpha + β) + sin (alpha − β) = 2 sin alpha cos β.  .  .  .  .  . (1)

Now put

 alpha + β = A

 and

 alpha − β = B.  .   .  .  .  .  .  .  .  .  .  .  .  .  .  .(2)

The left-hand side of line (1) then becomes

sin A + sin B.

This is now the left-hand side of (e), which is what we are trying to prove.

To complete the right−hand side of line (1), solve those simultaneous equations (2) for alpha and β.  

On adding them,  2alpha = A + B,

so that

alpha = ½(A + B).

On subtracting those two equations,  2β = A − B,

so that

β = ½(A − B).

On the right−hand side of line (1), substitute those expressions for alpha and β.  Line (1) then becomes

sin A + sin B = 2 sin ½(A + B) cos ½(A − B).

This is the identity (e).

Read it as follows:

"sin A + sin B equals twice the sine of half their sum

times the cosine of half their difference."

Identities (f), (g), and (h) are derived in exactly the same manner from (b), (c), and (d) respectively.

Trigonometric identities