# Angle formula of trigonometric functions

• Aug 22nd 2013, 12:21 PM
reindeer7
Angle formula of trigonometric functions
We know cos (2x) =2cos2(x) -1. Now my exercise book gives an example, asking to find cos (0.5x) if cos (x) =0.3.It says that you must replace x with 0.5x so that your new angle formula will be cos (x) =2cos2(0.5x) -1.I want to know how the domain of the cos function changed on either side of the equality sign (=).What is the principle involved in changing the domain, the operations done.It seems that you just multiplied all the xs in the whole expression with the coefficient of the x which is in the domain of the function of which you are finding the range.This sort of thing I have seen in simple algebra but how does this work in trig functions? An explanation in which you substitute the 2x or 0.5x etc in the domain with some other variable like u would be greatly appreciated.
• Aug 22nd 2013, 06:12 PM
chiro
Re: Angle formula of trigonometric functions
Hey reindeer7.

The domain or sine and cosine functions is always the whole real number line. The range is always [-1,1] for these functions.

All you are doing is using a relationship between cos(2x) and cos(x) which is valid for all values of x that are real numbers (and also most likely for complex numbers as well).

The general proof of the identity can be done through Eulers formula which says e^(iax) = [e^(ix)]^a = cos(ax) + isin(ax) = (cos(x) + isin(x))^a

Once you use this, the only thing left is to realize the quadrants when cos(x) and/or sin(x) is positive or negative. Cos(x) is negative in between branches of (pi/2,3pi/2] and sine is positive in [0,2pi). This is needed since x^2 has both positive and negative solutions and you need to choose the right solution based on the quadrant.
• Aug 23rd 2013, 12:09 AM
Prove It
Re: Angle formula of trigonometric functions
Quote:

Originally Posted by reindeer7
We know cos (2x) =2cos2(x) -1. Now my exercise book gives an example, asking to find cos (0.5x) if cos (x) =0.3.It says that you must replace x with 0.5x so that your new angle formula will be cos (x) =2cos2(0.5x) -1.I want to know how the domain of the cos function changed on either side of the equality sign (=).What is the principle involved in changing the domain, the operations done.It seems that you just multiplied all the xs in the whole expression with the coefficient of the x which is in the domain of the function of which you are finding the range.This sort of thing I have seen in simple algebra but how does this work in trig functions? An explanation in which you substitute the 2x or 0.5x etc in the domain with some other variable like u would be greatly appreciated.

Here if \displaystyle \displaystyle \begin{align*} x = \frac{1}{2}\theta \end{align*}, you have

\displaystyle \displaystyle \begin{align*} \cos{(2x)} &= 2\cos^2{(x)} - 1 \\ \cos{ \left( 2\cdot \frac{1}{2} \theta \right) } &= 2\cos^2{ \left( \frac{1}{2} \theta \right) } - 1 \\ \cos{(\theta)} &= 2\cos^2{ \left( \frac{1}{2} \theta \right) } - 1 \\ 0.3 &= 2\cos^2{ \left( \frac{1}{2} \theta \right) } - 1 \end{align*}

Can you go from here?