# How to simplify my trigonometric fraction

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 $(1)$ ?

• How about '1/(TrigExpand[1/expr] // FullSimplify)' Mar 13, 2017 at 8:21

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!

• Thanks for the hint. I was able to do what I wanted with expr //. { Sin[a]^2 + Cos[a]^2 -> 1, Cos[b]^2*Sin[b]^2 -> 1/4 Sin[2 b]^2} Mar 14, 2017 at 23:03

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.

(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

• Regarding the trig identities: $\sin(2b)=2 \sin(b) \cos(b)$ is a well-known identity. It can be derived from the angle addition identity, $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$, which I believe can be derived from simpler principles (e.g., 2-dimensional rotation matrices). Mar 14, 2017 at 22:51
• I don't remember it! Where did you find it? Show work. =P Mar 14, 2017 at 22:59
• Identities and proofs from Wikipedia, but a lot of people just have them memorized. Mar 14, 2017 at 23:07
• Also, if you want to force Mathematica to multiply the numerator and denominator by a common factor, you can use something like mapNumDenom[ff_, expr_] := ff@Numerator@expr/ff@Denominator@expr and then mapNumDenom[ Distribute[#/4]&, expr ]. I also frequently use similarly defined mapNum and mapDenom functions that apply only to the numerator or denominator. Feel free to add this to your answer if you would like. Mar 14, 2017 at 23:14
• I would if I could understand that code... Mar 14, 2017 at 23:40