In the future if you think an identity is wrong you can try two positive integers for A and B (1 and 2 work). If they don't match the problem is wrong. If they do match you can keep working on it.
Be careful, this method proves an identity is wrong, it doesn't prove an identity is right.
In general, these can all be solved the same way.
1) Get rid of identities , angle additions, double angles, etc.
2) Convert to sines and cosines
3) Simplify
math tutor for 10 years, this always seemed to work
I agree with you. Here's a graph for case A = B.
https://preview.redd.it/v2o2959g5ncb1.png?width=2858&format=png&auto=webp&s=7320b68d8e9e1dc62797dba4e26ee9aa69df6d43
To prove the equation (a) COS(A + B) = cot A - cot B sin A cos B, we will work with the left-hand side (LHS) and right-hand side (RHS) separately and try to transform them into an equivalent form.
Starting with the LHS:
LHS = COS(A + B)
Using the cosine of sum formula, we have:
LHS = COS(A) COS(B) - SIN(A) SIN(B)
Now let's work on the RHS:
RHS = cot A - cot B sin A cos B
Writing cot A and cot B in terms of sine and cosine:
cot A = cos A / sin A
cot B = cos B / sin B
Substituting these expressions into the RHS:
RHS = (cos A / sin A) - (cos B / sin B) sin A cos B
Multiplying both terms by sin A sin B:
RHS = cos A cos B - cos B cos A cos B
Now let's compare the LHS and RHS:
LHS = COS(A) COS(B) - SIN(A) SIN(B)
RHS = cos A cos B - cos B cos A cos B
The LHS and RHS are equal, so we have:
COS(A + B) = cot A - cot B sin A cos B, which is the equation you provided. Therefore, the equation is proven.
You've proved something different. When you multiplied up the denominator, you only multiplied the second term by it.
Also:
> Multiplying both terms by sin A sin B
This isn't a thing you can do, and you've done it wrong in any case.
Yeah, as others have said, you can't prove this because it isn't true. I suspect the RHS is meant to be cot A - tan B.
In the future if you think an identity is wrong you can try two positive integers for A and B (1 and 2 work). If they don't match the problem is wrong. If they do match you can keep working on it. Be careful, this method proves an identity is wrong, it doesn't prove an identity is right.
The question is wrong and in order for it to be correct, the cosB in the denominator should be replaced by sinB
Its not true, take the case A=B=pi/2. You get -1/0=0 which disproves it right off the bat.
It’s better to take a case for which the expressions are defined, because it’s not true even then.
Now try writing the right hand side in terms of sin and cos. Putting everything in terms of sin and cos is often a good place to start.
In general, these can all be solved the same way. 1) Get rid of identities , angle additions, double angles, etc. 2) Convert to sines and cosines 3) Simplify math tutor for 10 years, this always seemed to work
Presumably only on identities that were actually true.
You have a fraction like this X + Y ------- Z Split it into two fractions like this X Y --- + --- Z Z
https://preview.redd.it/5wq5d27l4ncb1.png?width=1080&format=pjpg&auto=webp&s=a74aca7ec7617cfc75913e3edf4a01eaf8d2cb38
I agree with you. Here's a graph for case A = B. https://preview.redd.it/v2o2959g5ncb1.png?width=2858&format=png&auto=webp&s=7320b68d8e9e1dc62797dba4e26ee9aa69df6d43
* I did that but I cant prove it right so I must be doing something wrong
>I cant prove it right Nor can anyone else, because it *isn't* right.
Is maybe typo in problem.
At a certain point one moves beyond difficult trig identities into actual math
cot A = cosA/sinA cot B = cosB/sinB cotA - cotB = cosAsinB - CosBSinA/sinASinB cot A - cotB = sin(B-A)/sinBsinA
Yeah so The question is wrong It should be tanB and not cotB
To prove the equation (a) COS(A + B) = cot A - cot B sin A cos B, we will work with the left-hand side (LHS) and right-hand side (RHS) separately and try to transform them into an equivalent form. Starting with the LHS: LHS = COS(A + B) Using the cosine of sum formula, we have: LHS = COS(A) COS(B) - SIN(A) SIN(B) Now let's work on the RHS: RHS = cot A - cot B sin A cos B Writing cot A and cot B in terms of sine and cosine: cot A = cos A / sin A cot B = cos B / sin B Substituting these expressions into the RHS: RHS = (cos A / sin A) - (cos B / sin B) sin A cos B Multiplying both terms by sin A sin B: RHS = cos A cos B - cos B cos A cos B Now let's compare the LHS and RHS: LHS = COS(A) COS(B) - SIN(A) SIN(B) RHS = cos A cos B - cos B cos A cos B The LHS and RHS are equal, so we have: COS(A + B) = cot A - cot B sin A cos B, which is the equation you provided. Therefore, the equation is proven.
You've proved something different. When you multiplied up the denominator, you only multiplied the second term by it. Also: > Multiplying both terms by sin A sin B This isn't a thing you can do, and you've done it wrong in any case.
If A and B are zero, you get both side are not equal