# Thread: This method of double angle formula's acceptable? (For test)

1. ## This method of double angle formula's acceptable? (For test)

apologies in advance, i'm quite new to these forums and don't know how to illustrate my question the proper way. i'll state my steps clearly and what i did in each step in brackets next to the equation.

$Cos pi/8 =$

(Using double angle formula $cos(2x) =2cos^2(x) -1$)

$Cos\frac{Pi}{4} = 2cos^2\frac{Pi}{8} -1$

$sqrt(2)/2 = 2cos^2(Pi/8) -1$ (figured out the value of $cos pi/4$)

$sqrt(2)/2 +1 = 2cos^2(Pi/8)$ (Moved the -1 to the left, making +1)

$sqrt(2)/4 + 1/2 = cos^2(Pi/8)$ (Divided both sides by 2)

$sqrt(sqrt(2)/4 + 1/2) = cos(Pi/8)$ (Square root of both sides)

$cos(pi/8) = Sqrt(Sqrt(2) + 2)/2$

seem correct? havn't got the answers in the back of my book. Seems right to me but i have to be sure i can use this method in my test. Would i be able to use this method for Tan, Sin, Cos and their recipricols?

2. Originally Posted by SirNostalgic

$Cos pi/8 =$

(Using double angle formula $cos(2x) =2cos^2(x) -1$)

$cos(pi/8) = Sqrt(Sqrt(2) + 2)/2$

seem correct? havn't got the answers in the back of my book. Seems right to me but i have to be sure i can use this method in my test. Would i be able to use this method for Tan, Sin, Cos and their recipricols?
Indeed, your solution is correct. I would suspect you are free to use the trig identities you mentioned as long as you are sure you never divide by zero in your computations.

to show $\pi$ type
Code:
\pi
to show the radical $\sqrt{\pi} \sqrt[n]{\pi}$ type
Code:
\sqrt{\pi} \sqrt[n]{\pi}

3. thank you, makes me feel better now. I'll try a few with Tan and Cotan as those 2 always get the better of me