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Simplify-write as single function?
- Thread starter Pingu
- Start date
G'day, Pingu.
\(\displaystyle f(x) = \cos{(2x)} \cdot\ cos{x} \, + \, \sin{(2x)} \cdot \sin{x}\) passes as a function, so it may better to ask to write the expression as a single trigonometric function.
By that I mean to simplify it down to just \(\displaystyle cos(x)\) or just \(\displaystyle \sin^2{(2x)}\), for examples, as opposed to having both sin and cos in there.
A lot of this work is a bit of trial and error, playing with identities and hoping you get a nice result out. It can be quite difficult to see the end of the path from the beginning; you can just make your way until you can see it. With practise, your foresight improves.
One option to start with would be to put \(\displaystyle \cos{(2x)}\) and \(\displaystyle \sin{(2x)}\) in terms of \(\displaystyle \cos{x}\) and \(\displaystyle \sin{x}\), respectively:
You should be familiar with the following identities
. . \(\displaystyle \L \cos{2x} = 2\cos^2{x} - 1\)
. . \(\displaystyle \L \sin{2x} = 2\sin{x}\cos{x}\)
So
. . \(\displaystyle \L \cos{2x} \cdot \cos{x} + \sin{2x} \cdot \sin{x}\)
becomes
. . \(\displaystyle \L (2\cos^2{x} - 1)\cos{x} \, + \, (2\sin{x}\cos{x}) \cdot \sin{x}\)
Simplify:
. . \(\displaystyle \L 2\cos^3{x} \, - \, \cos{x} \, + \, 2\sin^2{x}\cos{x}\)
See if you click onto the next step. (Give it a few minutes.)
\(\displaystyle f(x) = \cos{(2x)} \cdot\ cos{x} \, + \, \sin{(2x)} \cdot \sin{x}\) passes as a function, so it may better to ask to write the expression as a single trigonometric function.
By that I mean to simplify it down to just \(\displaystyle cos(x)\) or just \(\displaystyle \sin^2{(2x)}\), for examples, as opposed to having both sin and cos in there.
A lot of this work is a bit of trial and error, playing with identities and hoping you get a nice result out. It can be quite difficult to see the end of the path from the beginning; you can just make your way until you can see it. With practise, your foresight improves.
One option to start with would be to put \(\displaystyle \cos{(2x)}\) and \(\displaystyle \sin{(2x)}\) in terms of \(\displaystyle \cos{x}\) and \(\displaystyle \sin{x}\), respectively:
You should be familiar with the following identities
. . \(\displaystyle \L \cos{2x} = 2\cos^2{x} - 1\)
. . \(\displaystyle \L \sin{2x} = 2\sin{x}\cos{x}\)
So
. . \(\displaystyle \L \cos{2x} \cdot \cos{x} + \sin{2x} \cdot \sin{x}\)
becomes
. . \(\displaystyle \L (2\cos^2{x} - 1)\cos{x} \, + \, (2\sin{x}\cos{x}) \cdot \sin{x}\)
Simplify:
. . \(\displaystyle \L 2\cos^3{x} \, - \, \cos{x} \, + \, 2\sin^2{x}\cos{x}\)
See if you click onto the next step. (Give it a few minutes.)
There are several identities you can use.
\(\displaystyle cos2xcosx+sin2xsinx\)
\(\displaystyle cos2x=cos^{2}x-sin^{2}x\)
\(\displaystyle (cos^{2}x-sin^{2}x)cosx+2sinxcosxsinx\)
\(\displaystyle cos^{3}x-sin^{2}xcosx+2sin^{2}xcosx\)
\(\displaystyle cosx(cos^{2}x-sin^{2}x+2sin^{2}x)\)
\(\displaystyle cosx(cos^{2}x+sin^{2}x)\)
Now you should see something familiar in the parentheses.
\(\displaystyle cos2xcosx+sin2xsinx\)
\(\displaystyle cos2x=cos^{2}x-sin^{2}x\)
\(\displaystyle (cos^{2}x-sin^{2}x)cosx+2sinxcosxsinx\)
\(\displaystyle cos^{3}x-sin^{2}xcosx+2sin^{2}xcosx\)
\(\displaystyle cosx(cos^{2}x-sin^{2}x+2sin^{2}x)\)
\(\displaystyle cosx(cos^{2}x+sin^{2}x)\)
Now you should see something familiar in the parentheses.