Question

helpp solve (sin2x+sin4x)/(cos2x+cos4x) if ctan3x=1/5

Answer 1

Is that
\[\frac{\sin^2x+\sin^4x}{\cos^2x+\cos^4x}~~\text{give }\cot^3x=\frac{1}{5}~~?\]
Or
\[\frac{\sin2x+\sin4x}{\cos2x+\cos4x}~~\text{give }\cot3x=\frac{1}{5}~~?\]
Also, does "ctan" mean "cotangent?" I wasn't sure.

Answer 2

\[\frac{ \sin 2x+\sin 4x }{ \cos 2x+\cos 4x }\]

Answer 3

this

Answer 4

yes ctan is cotangent

Answer 5

So it's also \(\cot3x=\dfrac{1}{5}\), then?

Answer 6

yes

Answer 7

The way I see it, I think you would start off with the given information:
\[\cot3x=\frac{\cos3x}{\sin3x}=\frac{1}{5}\]
This tells you that the ratio of cosine to sine of the same angle is 1/5. For convenience, let's assume \(\cos3x=1\) and \(\sin3x=5\).

Now it looks like a matter of writing the \(\sin2x,\sin4x,\cos2x,\cos4x\) in terms of \(3x\).

Answer 8

yes and then?

Answer 9

To do that, you'll need some identities:
\[\cos(a+b)x=\cos ax\cos bx-\sin ax \sin bx\\
\cos(a-b)x=\cos ax\cos bx+\sin ax\sin bx\]
Notice that adding the two equations together gives you
\[\color{red}{2\cos ax\cos bx=\cos(a+b)x+\cos(a-b)x}\]
Some more identities:
\[\sin(a+b)x=\sin ax\cos bx+\sin bx\cos ax\\
\sin(a-b)x=\sin ax\cos bx-\sin bx\cos ax\]
Again, adding the two equations gives you something useful:
\[\color{red}{2\sin ax\cos bx=\sin(a+b)x+\sin(a-b)x}\]
Looking at the original fraction, you can see that if you let \(a=3\) and \(b=1\), you have exactly what you need:
\[\begin{align*}\frac{ \sin 2x+\sin 4x }{ \cos 2x+\cos 4x }&=\frac{\sin(3-1)x+\sin(3+1)x}{\cos(3-1)x+\cos(3+1)x}\\\\
&=\frac{2\sin 3x\cos x}{2\cos 3x\cos x}
\end{align*}\]

Answer 10

answer is 5 how?

Answer 11

Reduce that last fraction.
\[\frac{2\sin3x\cos x}{2\cos3x\cos x}=\frac{\sin 3x}{\cos 3x}=\frac{5}{1}=5\]

Answer 12

thankss