PROOF THE IDENTITY

[ugu]

QUESTION



ANSWER

 

[BigBeachBanana]

Check your signs in the denominator starting with the first equal sign.
The second line to last, please add parenthesis. Without them, it means something different.

 

[Dr.Peterson]

QUESTION
View attachment 30606


ANSWER
View attachment 30605

As written, your denominator is zero. Fix the error here:

 

[ugu]

i do not grab that, is it supposed to be - ?

As written, your denominator is zero. Fix the error here:
View attachment 30614

 

[ugu]

Check your signs in the denominator starting with the first equal sign.
The second line to last, please add parenthesis. Without them, it means something different.

i think i added parenthesis already? where else sir?

 

[Deleted member 4993]

i do not grab that, is it supposed to be - ?

Cos(A + B) = Cos(A) * Cos(B) - Sin(A) * Sin(B)

You wrote it incorrectly in the denominator. However, you wrote it correctly in the numerator.

 

[BigBeachBanana]

i do not grab that, is it supposed to be - ?

Yes. It's supposed to be "-". You have - for the same identity in the numerator.

This line is missing a parenthesis. What you wrote means [math]\frac{(2\cos A \cos B)(\sin A\sin B)}{2} \neq\frac{2\cos A \cos B}{2\sin A\sin B}[/math]

 

[ugu]

What about this method?

 

[BigBeachBanana]

The signs are still wrong but somehow got the right answer at the end.

 

[Dr.Peterson]

What about this method?

The work, as well as the answer, looks correct to me.

 

[BigBeachBanana]

The work, as well as the answer, looks correct to me.

 

[Dr.Peterson]

View attachment 30618

... which was corrected on the next line, so I didn't bother looking up. Such is the life of a proofreader.

 

[ugu]

View attachment 30618

appreciated

 

[ugu]

View attachment 30618

much thanks

Yes. It's supposed to be "-". You have - for the same identity in the numerator.
View attachment 30615
This line is missing a parenthesis. What you wrote means [math]\frac{(2\cos A \cos B)(\sin A\sin B)}{2} \neq\frac{2\cos A \cos B}{2\sin A\sin B}[/math]

so it should be enclosed as (2cosA.cosB)/(2sinA.sinB) ?

 

[BigBeachBanana]

much thanks


so it should be enclosed as (2cosA.cosB)/(2sinA.sinB) ?

Correct

 

[ugu]

The signs are still wrong but somehow got the right answer at the end.

But after collecting like terms, the sign has to change

 

[ugu]

View attachment 30618

It should not change after collecting like terms?

Is the sign at that point just the only error which does not affect other setting?

Just to change + to - and that is all?

 

[BigBeachBanana]

It should not change after collecting like terms?

Is the sign at that point just the only error which does not affect other setting?

Just to change + to - and that is all?

Signs do not change because of collecting like terms. You didn't distribute the negative sign properly in the picture I posted in #11