Question

What is the formula of \[\cos 4x\]?

Answer 1

Hint: Here in this question, we have to find the formula of given trigonometric function. To solve this, First rewrite the given angle in the form of addition or difference, then the standard trigonometric formula cosine sum i.e., \[cos\,(A + B)\]or cosine difference i.e., \[cos\,(A - B)\] identity defined as \[cos\,A.cos\,B - sin\,A.sin\,B\] and \[cos\,A.cos\,B + sin\,A.sin\,B\]using one of these we get required value.

Complete step-by-step solution:
To evaluate the given question by using a formula of cosine addition defined as the cosine addition formula calculates the cosine of an angle that is either the sum or difference of two other angles. It arises from the law of cosines and the distance formula. By using the cosine addition formula, the cosine of both the sum and difference of two angles can be found with the two angles' sines and cosines.
Consider the given function
\[ \Rightarrow \,\,\,\cos 4x\]-------(1)
It can be rewritten as
\[ \Rightarrow \,\,\,\cos \left( {2x + 2x} \right)\]
Now, by using a cosine sum identity: \[cos\,(A + B) = cos\,A.cos\,B - sin\,A.sin\,B\]
Here, \[A = 2x\] and \[B = 2x\] on substituting, we have
 \[ \Rightarrow \,\,\,cos\,(2x + 2x) = cos\,2x.cos\,2x - sin\,2x.sin\,2x\]
\[ \Rightarrow \,\,\,co{s^2}2x - si{n^2}2x\]
As by the standard trigonometric identity: \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] \[ \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta \], then on simplification we have
\[ \Rightarrow \,\,\,co{s^2}2x - \left( {1 - {{\cos }^2}2x} \right)\]
\[ \Rightarrow \,\,\,co{s^2}2x - 1 + {\cos ^2}2x\]
On simplification, we have
\[ \Rightarrow \,\,\,2co{s^2}2x - 1\]
Or it can be written as
\[ \Rightarrow \,\,\,2{\left( {cos\left( {x + x} \right)} \right)^2} - 1\]
Again, by the cosine sum identity we have
\[ \Rightarrow \,\,\,2{\left( {\cos x \cdot \cos x - \sin x \cdot \sin x} \right)^2} - 1\]
\[ \Rightarrow \,\,\,2{\left( {{{\cos }^2}x - {{\sin }^2}x} \right)^2} - 1\]
Again, by the trigonometric identity \[{\sin ^2}\theta = 1 - {\cos ^2}\theta \], we have
\[ \Rightarrow \,\,\,2{\left( {{{\cos }^2}x - \left( {1 - {{\cos }^2}x} \right)} \right)^2} - 1\]
\[ \Rightarrow \,\,\,2{\left( {2{{\cos }^2}x - 1} \right)^2} - 1\]
Now using a algebraic identity \[ \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
Here \[a = 2{\cos ^2}x\] and \[b = 1\], then
\[ \Rightarrow \,\,\,2\left( {{{\left( {2{{\cos }^2}x} \right)}^2} + {1^2} - 2\left( {2{{\cos }^2}x} \right)\left( 1 \right)} \right) - 1\]
\[ \Rightarrow \,\,\,2\left( {4{{\cos }^4}x + 1 - 4{{\cos }^2}x} \right) - 1\]
On simplification, we have
\[ \Rightarrow \,\,\,8{\cos ^4}x + 2 - 8{\cos ^2}x - 1\]
\[ \Rightarrow \,\,\,8{\cos ^4}x - 8{\cos ^2}x + 1\]

Hence, the formula \[\cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + 1\].

Note: This question also can be solved by using a double angle formula \[\cos 2x = 2{\cos ^2}x - 1\] and further simplification will be the same in the above method. Since they have mentioned to solve the given function by using the cosine sum or difference identity, for this we have standard formula and also known the trigonometric identities, double and half angle formulas.