Let f:R→R be a twice-differentiable function such that
(sinxcosy)[f(2x+2y)−f(2x−2y)]=(cosxsiny)[f(2x+2y)+f(2x−2y)]
for all x,y∈R. If f′(0)=21, then the value of 24f′′(35π) is:
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Step-by-Step Solution
Step 1: Rearrange the given equation
We begin by rearranging the given functional equation to isolate the terms involving f. Dividing both sides by (sinxcosy)[f(2x+2y)+f(2x−2y)] allows us to express the ratio of the difference and sum of f terms in terms of trigonometric functions.
Step 2: Apply Componendo and Dividendo
Applying the componendo and dividendo rule, which states that if ba=dc, then a−ba+b=c−dc+d, we can simplify the expression. In our case, we apply it as (a+b)−(a−b)(a+b)+(a−b)=c−dc+d, which simplifies to 2b2a=c−dc+d.
Step 3: Simplify the expression
We simplify the right-hand side of the equation by expressing cotx and tany in terms of sin and cos. Then, we use the trigonometric identities for sin(A+B) and sin(A−B) to further simplify the expression.
Step 4: Introduce new variables and differentiate
Let u=2x+2y and v=2x−2y. Then x=4u+v and y=4u−v. Substituting these into the simplified equation, we get a general form for f(u)/f(v). Differentiating with respect to u while treating v as a constant, we get f′(u)=Ccos(2u) for some constant C.
Step 5: Find the constant C and the function f(x)
Given f′(0)=21, we substitute x=0 into f′(x)=Ccos(2x). This gives Ccos(0)=C=21. However, upon re-evaluating the differentiation, a simpler approach is to recognize that the form f(v)f(u)=sin((u−v)/2)sin((u+v)/2) implies f(x)=ksin(x/2) for some constant k. Differentiating this, f′(x)=k21cos(x/2). Using f′(0)=21, we get k21cos(0)=21⟹k21=21⟹k=1. Thus, f(x)=sin(x/2) and f′(x)=21cos(x/2).
Step 6: Calculate the second derivative
Now we differentiate f′(x)=21cos(2x) with respect to x to find the second derivative, f''(x). The derivative of cos(ax) is −asin(ax).
Step 7: Evaluate 24f′′(35π)
Finally, we substitute x=35π into the expression for f''(x) and multiply by 24. We know that sin(65π)=sin(π−6π)=sin(6π)=21.