Question

Can someone help me with this equation?

Answer 1

cosXcosY = [cos(X+Y) + cos(X-Y)]/2

Answer 2

I love when the parentheses are implied. This sub normally has a few opinions about what is grouped with what.

Answer 3

Start by simplifying both sides with a trig identity:  cos(a)cos(b) = 1/2( cos(a+b) + cos(a-b) ) should do the trick.

If you do that to both sides you'll have a cos(5x) term on both sides (because 2+3 = 4+1 = 5) which will cancel.

After you also eliminate the 1/2 from both sides, you should be left with cos(3x) = cos(x).  This is true for x=0, and also for x equal to some fractions of pi. There are an infinite number of solutions, so once you find the pattern you have your answer.

If you graph y=cos(3x) and y=cos(x) on the same graph it'll be easier to see the pattern.

Answer 4

Um can you add parenthesis? Without them I can’t really help

Answer 5

Using chatGPT:


Solve Cosine Equation



How do you solve:. CosXcos4X = cos2Xcos3X

To solve the equation, you can start by using the identity cos(a)cos(b) = (cos(a+b) + cos(a-b))/2.

So, cosXcos4X = (cos(X+4X) + cos(X-4X))/2 = (cos5X + cos(-3X))/2

and cos2Xcos3X = (cos(2X+3X) + cos(2X-3X))/2 = (cos5X + cos(-X))/2

then you get (cos5X + cos(-3X))/2 = (cos5X + cos(-X))/2

then you can combine the cos5X terms cos(-3X) = cos(-X)

so you can now say that -3X = -X + 2nPi

then you can solve for X, where n is any integer X = (2nPi)/4 = n*Pi/2

So, the solutions of the equation are X = n*Pi/2, where n is any integer.

Answer 6

Oh this is too funny. I read the problem completely differently. I was just going to say "uh, they're not equal" LOL