How Can You Transform Sin(t)*Sin(x) into f(x+t)+g(x-t) Using Trig Identities?

[Bassoonmac]

[SOLVED] Trigonometric Transformation

This is a calculus 3 problem, but this part involves only trig identities:
Make the function f(x,t) = sin(t)*sin(x) into the form: f(x+t)+g(x-t).
I'm not sure whether to use half angle formulas, or what?

 

[cristo]

I'm not sure whether to use half angle formulas...

Yup, I'd try that!

 

[Architeuthis Dux]

I would approach this problem by first identifying the given function f(x,t) = sin(t)*sin(x) and its desired form f(x+t)+g(x-t). Then, I would use trigonometric identities to manipulate the given function and transform it into the desired form.

One way to approach this problem is by using the double angle formula for sine: sin(2x) = 2sin(x)cos(x). We can rewrite the given function as f(x,t) = (1/2)sin(2t)*sin(2x), which is in the form of the double angle formula.

Next, we can use the sum and difference formulas for sine to rewrite the double angle formula as f(x,t) = (1/2)[sin(x+t)sin(x-t) + cos(x+t)cos(x-t)]. This is close to the desired form, but we still have the extra cosine term.

To eliminate the cosine term, we can use the Pythagorean identity: cos^2(x) + sin^2(x) = 1. Rearranging this equation, we get cos^2(x) = 1 - sin^2(x). Substituting this into our function, we get f(x,t) = (1/2)[sin(x+t)sin(x-t) + (1-sin^2(x+t))(1-sin^2(x-t))].

Finally, we can use the double angle formula again to rewrite the sine terms, giving us the desired form of f(x,t) = sin(x+t) + g(x-t), where g(x-t) = (1/2)(1-sin^2(x-t))(1-sin^2(x+t)).

In conclusion, by using trigonometric identities such as the double angle formula and the Pythagorean identity, we can transform the given function into the desired form of f(x+t)+g(x-t). This process is an important application of trigonometric transformations in calculus and can be used in various mathematical and scientific contexts.