Sum and differences identities equations

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The discussion revolves around solving the equation cos(∏/2 + x) - cos(∏/2 - x) and determining its equivalent form from given options. Participants emphasize using the sum and difference formulas for cosine, specifically cos(A - B) = cosAcosB + sinAsinB. Clarification is provided on how to assign values to A and B for each cosine term separately. The conversation highlights confusion about applying the formula correctly and reassures that each cosine expression can be treated independently. Ultimately, the focus is on correctly applying trigonometric identities to find the solution.
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Homework Statement


Which is equivalent to: cos(∏/2 + x) - cos(∏/2 - x)?
A) -2cos(x)
B) -2
C) 0
D)-2sin(x)


Homework Equations


Cos (A-B)


The Attempt at a Solution


I am totally stuck :( please help!
 
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keishaap said:

Homework Statement


Which is equivalent to: cos(∏/2 + x) - cos(∏/2 - x)?
A) -2cos(x)
B) -2
C) 0
D)-2sin(x)


Homework Equations


Cos (A-B)


The Attempt at a Solution


I am totally stuck :( please help!

You mentioned the sum and difference formulas. I think you should try to use them.
 
Dick said:
You mentioned the sum and difference formulas. I think you should try to use them.

I have tried to use them but i totally get stuck like i don't know which is a or b and we are only given cos (a-b) = cosAcosB + sinAsinB
 
keishaap said:
I have tried to use them but i totally get stuck like i don't know which is a or b and we are only given cos (a-b) = cosAcosB + sinAsinB

In cos(pi/2-x) the 'a' is pi/2 and the 'b' is x. So what does that turn into? What are cos(pi/2) and sin(pi/2)?
 
Cos (pi/2) =0
Sin(pi/2)= 1
 
keishaap said:
Cos (pi/2) =0
Sin(pi/2)= 1

Ok, do go on. So what is cos(pi/2-x)?
 
How come the b is not a -x ?
 
So cos( 0-x)?
 
keishaap said:
I have tried to use them but i totally get stuck like i don't know which is a or b and we are only given cos (a-b) = cosAcosB + sinAsinB

It doesn't matter. Pick one angle and make it A. The other then must be B.
 
  • #10
keishaap said:
How come the b is not a -x ?

Why would you think that?? If you have a formula for cos(a-b) and you want to apply it to cos(pi/2-x) then you should put a=pi/2 and b=x. Your formula already has the '-' in it. Just use the formula and stop trying to double think it.
 
  • #11
Dick said:
Why would you think that?? If you have a formula for cos(a-b) and you want to apply it to cos(pi/2-x) then you should put a=pi/2 and b=x. Your formula already has the '-' in it. Just use the formula and stop trying to double think it.

But the formula also has pi/2 + x there's 2 A's and 2 B's
 
  • #12
keishaap said:
But the formula also has pi/2 + x there's 2 A's and 2 B's

Use the formula SEPARATELY for each one. You can choose them differently for cos(pi/2-x) and cos(pi/2+x). Finish cos(pi/2-x) first, then worry about the other one.
 
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  • #13
Dick said:
Use the formula SEPARATELY for each one. You can choose them differently for cos(pi/2-x) and cos(pi/2+x). Finish cos(pi/2-x) first, then worry about the other one.

Okay i thought they had to be the same because the teacher didn't show us any other way thanks!
 
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