The Sine of a Sum and Difference Identity

[mynamesmurph]

Hey there folks.

I'm struggling to make sense of the aforementioned proof, using trig identities and algebra. Here's a link of what I'm trying to comprehend.

http://facultypages.morris.umn.edu/...calculus/Lectures/SumDifferenceIdentities.pdf

I understand how to derive this formula, easy peasy:

cos(a ± b) = cos(a)cos(b) ∓ sin(a) sin(b)

But, I'm have having difficulty continuing to derive the sine formulas. So let's try to sin(a + b)

sin(a + b) = cos(Pi/2 - (a + b)) I understand this, this is a cofunction identity

sin(a + b) = cos (Pi/2 - a ) - b)) Distribute here and regroup



Then it gets a bit confusing.

cos (Pi/2 - a ) - b)

= cos(Pi/2 − a)cos(b) + sin(Pi/2 -a)sin(b)

I see this is a modification of the first formula, but where does the -b go? Why is the original formula brought in exactly?



From here I understand how

= cos(Pi/2 − a)cos(b) + sin(Pi/2 -a)sin(b)

turns into

= sin(a)cos(b) + cos(a)sin(b)


and in turn, I can do the difference formulas from there, just struggling with the middle part there.

Thanks!

 

[stapel]

I understand how to derive this formula, easy peasy:

cos(a ± b) = cos(a)cos(b) ∓ sin(a) sin(b)...

Then it gets a bit confusing.

cos (Pi/2 - a ) - b)

= cos(Pi/2 − a)cos(b) + sin(Pi/2 -a)sin(b)

I see this is a modification of the first formula, but where does the -b go? Why is the original formula brought in exactly?

All they've done is use "pi/2 - a" in place of "a" in the original identity. Then they just applied that identity.

There is no "-b"; there is only "subtracted by b".

 

[mynamesmurph]

I understand that they are subtracting by B and I understand that Pi/2 - a is replacing a, but I suppose there is a disconnect as to why it's being replaced. I'm probably not explaining myself very well.

Why is cos(a) replaced by cos(pi/2 -a), when it's sin(a + b) = cos((pi/2 - a) - b) ?

Maybe it's so obvious my question doesn't make sense...

 

[Ishuda]

I understand that they are subtracting by B and I understand that Pi/2 - a is replacing a, but I suppose there is a disconnect as to why it's being replaced. I'm probably not explaining myself very well.

Why is sin(a) replaced by cos(pi/2 -a), when it's sin(a + b) = cos((pi/2 - a) - b) ?

Maybe it's so obvious my question doesn't make sense...

First, I think you made a typo [see red part of quote where I replace cos by sin].

Next, If I'm understanding you correctly (and that's no guarantee), the answer is because it works. Mathematics is (sometimes) just learning how to re-write a problem so that it is easier to solve [see the tag line of Suubhotosh Khan which is "“... mathematics is only the art of saying the same thing in different words” - B. Russell"]. So, for this particular problem, if we re-write sin(a + b) as cos(Pi/2 - (a + b)) the stated problem is easier to solve if we regroup the terms and use the sum and difference laws for cosines.

 

[Deleted member 4993]

I understand that they are subtracting by B and I understand that Pi/2 - a is replacing a, but I suppose there is a disconnect as to why it's being replaced. I'm probably not explaining myself very well.

Why is cos(a) replaced by cos(pi/2 -a), when it's sin(a + b) = cos((pi/2 - a) - b) ?

Maybe it's so obvious my question doesn't make sense...

You know

sin(Θ) = cos (π/2 - Θ)

Now replace

Θ = a+b

sin(a+b) = cos[π/2 - (a + b)] = cos[π/2 - a - b] = cos[(π/2 - a) - b] = cos(π/2-a) * cos(b) + sin(π/2-a) * sin(b) = sin(a)*cos(b) + cos(a)*sin(b)

 

[mynamesmurph]

You know

sin(Θ) = cos (π/2 - Θ)

Now replace

Θ = a+b

sin(a+b) = cos[π/2 - (a + b)] = cos[π/2 - a - b] = cos[(π/2 - a) - b] = cos(π/2-a) * cos(b) + sin(π/2-a) * sin(b) = sin(a)*cos(b) + cos(a)*sin(b)

I understand up to the bold, and after the bold, ordinarily I understand these things pretty easily, but I think I'm missing something quite obvious.

cos[(π/2 - a) - b] = cos(π/2-a) * cos(b) + sin(π/2-a) * sin(b)


Why doescos[(π/2 - a) - b] equal cos(π/2-a) * cos(b) + sin(π/2-a) * sin(b) ​?Really appreciate the help folks.

 

[Ishuda]

...Why doescos[(π/2 - a) - b] equal cos(π/2-a) * cos(b) + sin(π/2-a) * sin(b) ​?Really appreciate the help folks.

That is just the law of the cosine of sum and differences of angles. Let
c = (π/2 - a)
then what we have is cos(c - b) and, according to the general formula you gave at the beginning
cos(a ± b) = cos(a)cos(b) ∓ sin(a) sin(b),

cos(c - b) = cos(c)cos(b) + sin(c) sin(b)

That is c has just replaced a in the general formula. Now replace c by its value π/2-a. [Notice, we are talking about two different a's here]

 

[Deleted member 4993]

I understand up to the bold, and after the bold, ordinarily I understand these things pretty easily, but I think I'm missing something quite obvious.

cos[(π/2 - a) - b] = cos(π/2-a) * cos(b) + sin(π/2-a) * sin(b)


Why doescos[(π/2 - a) - b] equal cos(π/2-a) * cos(b) + sin(π/2-a) * sin(b) ​?Really appreciate the help folks.

In your first post, you wrote,

cos(a ± b) = cos(a)cos(b) ∓ sin(a) sin(b)

Let's rewrite that:

cos(x - y) = cos(x) * cos(y) + sin(x) * sin(y) .... (1) .......... You understand this

Now replace:

x = π/2 - a

y = b

then we get:

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

cos[(π/2 - a) - b] = cos(π/2 - a) * cos(b) + sin(π/2 - a) * sin(b)

It is just a matter of re-writing the "known" equation ... that's it!

 

[mynamesmurph]

So we could set x and y to equal anything then right? It's just that we choose to set x to (pi/2 - a) which is an identity and can thus change cos(pi/2 -a) to sin(a) and sin(pi/2 - a) to cos(a), right?

 

[Ishuda]

So we could set x and y to equal anything then right? It's just that we choose to set x to (pi/2 - a) which is an identity and can thus change cos(pi/2 -a) to sin(a) and sin(pi/2 - a) to cos(a), right?

Re-writing the law of cosines as
cos(x ± y) = cos(x)cos(y) ∓ sin(x) sin(y)
then the answer is yes, we could set x and y to equal anything

 

[mynamesmurph]

Ok, well that seems to have been the hangup, I was somehow using the trig identity one step early and the whole thing became circular. It seems to be implied in the proof, and I wasn't picking up on it. Thank you all.