pre calculus

[frez$]

Use sum and difference identities of
sin(a+b) and sin(a-b) to derive
sin(a)+sin(b)

 

[blamocur]

Use sum and difference identities of
sin(a+b) and sin(a-b) to derive
sin(a)+sin(b)

From Read Before Posting: Show your beginning work, or ask a specific question about the exercise, or explain why you're stuck.

 

[frez$]

I don't understand what I'm supposed to derive, or how deriving works

 

[blamocur]

I don't understand what I'm supposed to derive, or how deriving works

Assume [imath]a=u+v[/imath] and [imath]b=u-v[/imath] and get the sum in terms of [imath]u,v[/imath]. Then express [imath]u,v[/imath] through [imath]a,b[/imath].

 

[Deleted member 4993]

Use sum and difference identities of
sin(a+b) and sin(a-b) to derive
sin(a)+sin(b)

The problem as stated above does not make sense to me. Can you post the exact problem - verbatim? May be you want to ask somebody to translate the problem correctly.

 

[frez$]

The problem as stated above does not make sense to me. Can you post the exact problem - verbatim? May be you want to ask somebody to translate the problem correctly.

i think what the question is asking is using sin(a+b) which is sin(a)cos(b)+cos(a)sin(b) and sin(a-b) which is sin(a)cos(b) - cos(a)sin(b)
to derive sin(a) + sin(b) which is 2sin(a+b/2) cos (a-b/2)

 

[blamocur]

i think what the question is asking is using sin(a+b) which is sin(a)cos(b)+cos(a)sin(b) and sin(a-b) which is sin(a)cos(b) - cos(a)sin(b)
to derive sin(a) + sin(b) which is 2sin(a+b/2) cos (a-b/2)

Actually the answer is 2sin((a+b)/2)cos((a-b)/2). I.e., the answer you wrote is missing some important parentheses.

 

[frez$]

oh really? why do we add those parentheses, and how can I get the final answer

 

[skeeter]

oh really? why do we add those parentheses, and how can I get the final answer

because sin(a + b/2) can be interpreted as [imath]\sin\left(a + \dfrac{b}{2}\right)[/imath]

 

[BigBeachBanana]

Restate the question:
Show that
[math]\sin(a) + \sin(b) = 2\sin \left(\frac{a+b}{2}\right)\cos \left(\frac{a-b}{2}\right)[/math]Using [math]\sin(a+b)=\sin(a) \cos(b)+\cos(a) \sin(b)\\ \sin(a-b)= \sin(a)\cos(b) - \cos(a) \sin(b)[/math]

why do we add those parentheses?

Without the parenthesis: [math]2\sin\left(a+\frac{b}{2}\right) \cos \left(a-\frac{b}{2}\right)[/math]Notice the difference

 

[frez$]

Restate the question:
Show that
[math]\sin(a) + \sin(b) = 2\sin \left(\frac{a+b}{2}\right)\cos \left(\frac{a-b}{2}\right)[/math]Using [math]\sin(a+b)=\sin(a) \cos(b)+\cos(a) \sin(b)\\ \sin(a-b)= \sin(a)\cos(b) - \cos(a) \sin(b)[/math]
Without the parenthesis: [math]2\sin\left(a+\frac{b}{2}\right) \cos \left(a-\frac{b}{2}\right)[/math]Notice the difference

oh I see, but how could i get the final answer?

 

[BigBeachBanana]

oh I see, but how could i get the final answer?

First let a= x+y and b=x-y. Then solve for x, and y in term of a and b. We get [imath]x= \frac{a+b}{2}[/imath] and [imath]y= \frac{a-b}{2}[/imath]
Next,subtracting identity 2 from identity 1:
[imath]\sin(x+y)=\sin(x) \cos(y)+\cos(x) \sin(y)\dots(1)[/imath]
[imath]\sin(x-y)= \sin(x)\cos(y) - \cos(x) \sin(y)\dots (2)[/imath]
We have:
[imath]\sin(x+y)-sin(x-y)=2\sin(y)\cos(x)\dots(3)[/imath]
Then, subtitute back equation 3 in term a and b
Can you continue?