Avg. rate of change of trig functions...

[botley111222]

f(x)=sinx
Interval [x, x+h]
Avg. rate of change = (f(x+h) - f(x))/((x+h) - (x))
therefore = (sin(x+h)-sinx)/h. Right?

I need to get the answer
sinx * (cosh-1)/(h) + cosx * (sinh)/(h)

 

[BigGlenntheHeavy]

\(\displaystyle Hint: \ sin(x+h) \ = \ sin(x)cos(h)+cos(x)sin(h)\)

 

[botley111222]

Thanks. I figured that out, and I figured out the problem f(x)=cosx.

Now how about f(x)=tanx?

(tan(x+h)-tan(x)) / h

I use tan addition and get

[(tanx+tanh)/(1-tanxtanh)]/h

Please help me get the answer (tanh/h)*(sec^2(x)/(1-tanxtanh).

 

[Deleted member 4993]

Thanks. I figured that out, and I figured out the problem f(x)=cosx.

Now how about f(x)=tanx?

(tan(x+h)-tan(x)) / h

I use tan addition and get

[(tanx+tanh)/(1-tanxtanh)]/h

Please help me get the answer (tanh/h)*(sec^2(x)/(1-tanxtanh).

\(\displaystyle \frac{tan(h)+tan(x)}{1-tan(x)\cdot tan(h)} \ - \ tan(x)\)

Now do the subtraction carefully

\(\displaystyle = \ \frac{tan(h) \ + \ tan(x) \ - \ tan(x) \ + \ tan^2(x)\cdot tan(h)}{1-tan(x)\cdot tan(h)} \\)

\(\displaystyle = \ \frac{tan(h) \cdot [1 \ + \ tan^2(x)]}{1-tan(x)\cdot tan(h)} \\)

Now continue - this is simple algebra, just be careful....