Question

Express \[\cos 4x\] as powers of \[\cos x\]?

Answer 1

Hint:We need to express \[\cos 4x\] as powers of \[\cos x\]. For that, we need to use the formula \[\cos 2t = 2{\cos ^2}t - 1\], We need to use this formula twice by first considering \[4x = 2t\], which implies \[2x = t\]. Then we will again apply this formula in the term \[2{\cos ^2}2x\]. After that, we need to apply the formula \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]. Using these formulas, we will obtain \[\cos 4x\] as the power of \[\cos x\].

Complete step by step answer:
We need to express \[\cos 4x\] as powers of \[\cos x\].
Using the formula \[\cos 2t = 2{\cos ^2}t - 1\], we get
Comparing \[4x = 2t\], we get \[2x = t\]
Using this in the formula \[\cos 2t = 2{\cos ^2}t - 1\], we get
\[ \Rightarrow \cos 4x = 2{\cos ^2}\left( {2x} \right) - 1\]
\[ \Rightarrow \cos 4x = 2{\left( {\cos \left( {2x} \right)} \right)^2} - 1\]

Now, again using the formula \[\cos 2t = 2{\cos ^2}t - 1\], we get
\[ \Rightarrow \cos 4x = 2{\left( {2{{\cos }^2}x - 1} \right)^2} - 1\]
Now, using the formula \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], we get
\[ \Rightarrow \cos 4x = 2\left( {{{\left( {2{{\cos }^2}x} \right)}^2} + {{\left( 1 \right)}^2} - 2\left( {2{{\cos }^2}x} \right)\left( 1 \right)} \right) - 1\]
Now, solving the brackets, we get
\[ \Rightarrow \cos 4x = 2\left( {4{{\cos }^4}x + 1 - 4{{\cos }^2}x} \right) - 1\]

Now, opening the brackets, we get
\[ \Rightarrow \cos 4x = 8{\cos ^4}x + 2 - 8{\cos ^2}x - 1\]
Now, combining like terms, we get
\[ \Rightarrow \cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + \left( {2 - 1} \right)\]
Now, solving the brackets, we get
\[ \Rightarrow \cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + 1\]
Now, taking out \[8{\cos ^2}x\] common form first two terms, we get
\[ \therefore \cos 4x = 8{\cos ^2}x\left( {{{\cos }^2}x - 1} \right) + 1\]

Hence, \[\cos 4x\], in terms of powers of \[\cos x\], can be written as \[\cos 4x = 8{\cos ^2}x\left( {{{\cos }^2}x - 1} \right) + 1\].

Note:We have many formulas for \[\cos 2x\], in terms of \[\sin x\] and \[\cos x\], only \[\cos x\], only \[\sin x\], only \[\tan x\] but since we have to write in powers of \[\cos x\] only, we will use the formula \[\cos 2x = 2{\cos ^2}x - 1\]. We need to compare the angles and then apply the formulas. Not able to identify the angles properly, we will make mistakes and then it will result in the wrong answer. Applying once, we see that we get the angle to be \[2x\] and so we need to again apply the formula for \[\cos 2x\] so that we can get it in terms of \[\cos x\].