SOLUTION: Prove the following is an identity: cos5y+cos3y/cos5y-cos3y=-cot4ycoty

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Question 815656: Prove the following is an identity: cos5y+cos3y/cos5y-cos3y=-cot4ycoty
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Please
  • Put multiple-term numerators and denominators in parentheses. What you posted meant:
    cos5y%2Bcos3y%2Fcos5y-cos3y=-cot4ycoty
    when I suspect you meant:
    %28cos5y%2Bcos3y%29%2F%28cos5y-cos3y%29=-cot4ycoty
  • Put arguments of functions in parentheses. cos(5y) is clearer than cos5y (which could mean (cos(5))*y)
%28cos%285y%29%2Bcos%283y%29%29%2F%28cos%285y%29-cos%283y%29%29=-cot%284y%29cot%28y%29

One of the keys to proving identities is to get the arguments to match. On the left side of our equation we have arguments of 5y and 3y. On the right side we have arguments of 4y and y. One way to match these arguments is to rewrite the arguments on the left side in terms of the arguments on the right. With a little effort we should be able to find that 5y = 4y + y and 3y = 4y - y. Rewriting the left side using these arguments we get:

Now we can use the cos(A+B) and cos(A-B) formulas on the left side:

In the numerator the terms with sin in them cancel each other out. In the denominator it is the terms with cos that cancel out. (Be careful with the "-" in front of the parentheses in the denominator!). So this simplifies to:
%282cos%284y%29cos%28y%29%29+%2F+%28-2sin%284y%29sin%28y%29%29+=+-cot%284y%29cot%28y%29
The two's cancel:
%28cos%284y%29cos%28y%29%29+%2F+%28-sin%284y%29sin%28y%29%29+=+-cot%284y%29cot%28y%29
The path to the end should be clear now. Moving the minus out in front and "un-multiplying" fractions we get:
-%28cos%284y%29%2Fsin%284y%29%29%28cos%28y%29%2Fsin%28y%29%29+=+-cot%284y%29cot%28y%29
Since cot = cos/sin we get:
-cot%284y%29cot%28y%29+=+-cot%284y%29cot%28y%29