Question

How do you simplify $$\cos 4\theta$$ to a trigonometric function of a unit $$\theta ?$$

Answer 1

Hint: We are given a ratio as $$\cos 4\theta$$ and we are asked to simplify it to the ratio in which the angle $$\theta$$ is in the unit, that is just $$\theta .$$ To do so we will need the relation type of $$\cos n\theta$$ and the other ratio. We will be using the identity given the relation between $$\cos 2\theta$$ and $${\cos }^2\theta$$ and $${\sin }^2\theta$$ as $$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta .$$ We will also use that $${\cos }^2\theta +{\sin }^2\theta =1.$$

Complete step by step answer:
We are given a trigonometric ratio of $$\cos 4\theta$$ and we have to change it into a ratio with unit $$\theta$$ means to change it into a ratio where the $$\theta$$ is in unit form. To do so we will see what the ways to change $$\cos n\theta$$ into unit $$\theta$$ form are. We know that $$\cos 2\theta$$ is given as $${\cos }^2\theta -{\sin }^2\theta .$$ So we will use it continuously till n = 4.
So, first, we have $$\cos 4\theta$$ and we can write $$4\theta =2\times 2\theta ,$$ so we can write $$2\theta$$ as y. So, we get,
$$4\theta =2y$$
Hence,
$$\cos 4\theta =\cos 2y$$
Now, we will apply the identity $$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$$ on cos 2y. So, we will get that
$$\cos 4\theta =\cos 2y$$
$$\Rightarrow \cos 4\theta ={\cos }^{2}y-{\sin }^{2}y[\text{As }\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta ]$$
Now we also know that
$${\cos }^2\theta +{\sin }^2\theta =1$$
So,
$${\sin }^2\theta =1-{\cos }^2\theta$$
Hence,
$$\Rightarrow {\sin }^{2}y=1-{\cos }^{2}y$$
So, using this above, we get,
$$\Rightarrow \cos 4\theta ={\cos }^{2}y-(1-{\cos }^{2}y)$$
$$\Rightarrow \cos 4\theta ={\cos }^{2}y+1{\cos }^{2}y-1$$
$$\Rightarrow \cos 4\theta =2{\cos }^{2}y-1$$
Now putting back $$y=2\theta$$ in the above equation we will get
$$\Rightarrow \cos 4\theta =2{\cos }^{2}2\theta -1$$
Again, using $$\cos 2\theta$$ as $${\cos }^2\theta -{\sin }^2\theta$$ we get,
$$\Rightarrow \cos 4\theta =2{({\cos }^2\theta -{\sin }^2\theta )}^2-1$$
As $${\sin }^2\theta =1-{\cos }^2\theta ,$$ so we get,
$$\Rightarrow \cos 4\theta =2{({\cos }^2\theta -(1-{\cos }^2\theta ))}^2-1$$
On simplifying, we get,
$$\Rightarrow \cos 4\theta =2{({\cos }^2\theta +{\cos }^2\theta -1)}^2-1$$
$$\Rightarrow \cos 4\theta =2{(2{\cos }^2\theta -1)}^2-1$$
Now using $${(a-b)}^2=a^2+b^2-2ab$$ we get by letting $$a=2{\cos }^2\theta$$ and b = 1, we get,
$$\Rightarrow \cos 4\theta =2(4{\cos }^4\theta -4{\cos }^2\theta +1)-1$$
So, opening the brackets, we get,
$$\Rightarrow \cos 4\theta =8{\cos }^4\theta -8{\cos }^2\theta +1$$
Hence, we get $$\cos 4\theta$$ is written as $$8{\cos }^4\theta -8{\cos }^2\theta +1$$ in unit $$\theta .$$

Note: While solving this we need to be very careful with sign and bracket as one wrong opening of brackets or wrong solving of the sign will lead us to the wrong answer.
$$-(1-\sin \theta )\ne -1-\sin \theta$$ as – 1 in from will be multiplied by both terms and will give as $$-1+\sin \theta .$$ So, we need to use proper step by step solution.