# Thread: Use a double angle to rewrite each trigonometric ratio

1. ## Use a double angle to rewrite each trigonometric ratio

cos 3x

The answer at the back of the book says

2 sin² (1.5x) - 1

but how do you get that?
Isn't the "Double Angle Formula" for Cosine supposed to follow
cos 2Θ = 2 cos²Θ - 1 ??? Why does the answer say "sin"? And how do they know to use 1.5x? (the book is known to have many typos, if that helps...)

2. Originally Posted by lanvin
cos 3x

The answer at the back of the book says

2 sin² (1.5x) - 1

but how do you get that?
Isn't the "Double Angle Formula" for Cosine supposed to follow
cos 2Θ = 2 cos²Θ - 1 ??? Why does the answer say "sin"? And how do they know to use 1.5x? (the book is known to have many typos, if that helps...)
We know that $\cos(nx)=\cos^2\left(\frac{n}{2}\right)-\sin^2\left(\frac{n}{2}x\right)={\color{red}2\cos^ 2\left(\frac{n}{2}x\right)}=1-2\sin^2\left(\frac{n}{2}x\right)$

So we may do the following

\begin{aligned}\cos(3x)&=\cos\left(2\cdot\frac{3}{ 2}x\right)\\
&=\cos^2\left(\frac{3}{2}x\right)-\sin^2\left(\frac{3}{2}x\right)\\
&={\color{red}2\cos^2\left(\frac{3}{2}x\right)-1}\\
&=1-2\sin^2\left(\frac{3}{2}x\right)\end{aligned}

It just goes to show the fallibility of books.

3. thanx a lot for your help,
but where did you get (n/2)x from?

the Double Angle Formulas for Cosine we were given say
cos 2Θ = cos²Θ - sin²Θ = 2cos²Θ -1 = 1- 2sin²Θ

is it just accepted that (n/2)x = Θ?

4. Originally Posted by lanvin
thanx a lot for your help,
but where did you get (n/2)x from?

the Double Angle Formulas for Cosine we were given say
cos 2Θ = cos²Θ - sin²Θ = 2cos²Θ -1 = 1- 2sin²Θ

is it just accepted that (n/2)x = Θ?
Yes

5. Originally Posted by Mathstud28
Yes
Ok... that helps. Thank you