How to simplify my trigonometric fraction

Question

I have the following expression $$ \frac{4 \sin(a)}{\sin(a)^2 + \cos(a)^2 + r^4 \sin(a)^4\cos(a)^2 \sin(b)^2 \cos(b)^2 } $$ I would like Mathematica to recognize that this expression can be simplified to $$ \frac{4\sin(a)}{1+\frac{1}{4}r^4\sin(a)^4\cos(a)^2\sin(2b)^2} \ \ (1) $$ But although I can do the following

FullSimplify[Cos[a]^2 + Sin[a]^2]
=> 1

and

FullSimplify[Cos[b]^2 Sin[b]^2, ComplexityFunction -> Length]
=> 1/4 Sin[2 b]^2

when I try to use FullSimplify on the whole expression it will give me

FullSimplify[(4 Sin[a])/(Cos[a]^2 + Sin[a]^2 + r^4 Cos[a]^2 Cos[b]^2 Sin[a]^4 Sin[b]^2)]
=> (4 Sin[a])/(Sin[a]^2 + Cos[a]^2 (1 + r^4 Cos[b]^2 Sin[a]^4 Sin[b]^2))

What can I do to transform my fraction into the form $(1)$ ?

Answer 1

This is your expression:

expr = (4 Sin[a])/(Cos[a]^2+Sin[a]^2+r^4 Cos[a]^2 Cos[b]^2 Sin[a]^4 Sin[b]^2);

Why not to simply apply a rule:

    expr /. {Cos[a_]^2 + Sin[a_]^2 -> 1}

(*  (4 Sin[a])/(1 + r^4 Cos[a]^2 Cos[b]^2 Sin[a]^4 Sin[b]^2) *)

?? Have fun!

Answer 2

I was curious as to if the original workup was right because I was having problems doing this in my head/pencil/paper. Don't judge me.

Anyway, you can't just use MMA's built in commands to simplify things that involve fractions etc because the denominator may be 0 and MMA gives up, At least that's how I always thought of it.

Here's a work around.

As Alexei already pointed out

(4 Sin[a])/(Cos[a]^2 + Sin[a]^2 + r^4 Cos[a]^2 Cos[b]^2 Sin[a]^4 Sin[b]^2) /. {Cos[a]^2 + Sin[a]^2 -> 1}

gives

(4 Sin[a])/(1 + r^4 Cos[a]^2 Cos[b]^2 Sin[a]^4 Sin[b]^2)

---

In[20]:= expr = (4 Sin[a])/(Cos[a]^2+Sin[a]^2+r^4 Cos[a]^2 Cos[b]^2 Sin[a]^4 Sin[b]^2)/.{Cos[a]^2+Sin[a]^2-> 1}

Out[20]= (4 Sin[a])/(1+r^4 Cos[a]^2 Cos[b]^2 Sin[a]^4 Sin[b]^2)

In[21]:= expr/. {Sin[a]^4-> Sin[a]^4 (1+Cos[2b])/2*1/Cos[b]^2}

Out[21]= (4 Sin[a])/(1+1/2 r^4 Cos[a]^2 (1+Cos[2 b]) Sin[a]^4 Sin[b]^2)

In[22]:= %/.{Sin[b]^2-> Sin[b]^2 (1-Cos[2b])/(2 Sin[b]^2)}//Simplify

Out[22]= (16 Sin[a])/(4+r^4 Cos[a]^2 Sin[a]^4 Sin[2 b]^2)

Now Why MMA refuses to simplify the following I have no idea.

In[24]:= (16 Sin[a])/(4+r^4 Cos[a]^2 Sin[a]^4 Sin[2 b]^2) ((1/4)/(1/4))
Out[24]= (16 Sin[a])/(4+r^4 Cos[a]^2 Sin[a]^4 Sin[2 b]^2)

Doing it manually like this also fails.

In[25]:= ((1/4)16 Sin[a])/((1/4)(4+r^4 Cos[a]^2 Sin[a]^4 Sin[2 b]^2))
Out[25]= (16 Sin[a])/(4+r^4 Cos[a]^2 Sin[a]^4 Sin[2 b]^2)

But if you were to do this manually,

In[26]:= ((1/4)16 Sin[a])/((1/4)4+(1/4)r^4 Cos[a]^2 Sin[a]^4 Sin[2 b]^2)
Out[26]= (4 Sin[a])/(1+1/4 r^4 Cos[a]^2 Sin[a]^4 Sin[2 b]^2)

Give you the answer you seek.

I really need to refresh up on trig. I can't see why

Sin[b]^2 Cos[b]^2 == 1/4 Sin[2b]^2

Someone work that out by hand so I can see it.

---

To see how I came up with the replacement rules, see below.

In[19]:= Sin[a]^4== Sin[a]^4 (1+Cos[2b])/2*1/Cos[b]^2//Simplify
Out[19]= True

In[17]:= Sin[b]^2== Sin[b]^2 (1-Cos[2b])/(2 Sin[b]^2)//Simplify

Out[17]= True