Question

How do you use the power reducing formulas to rewrite the expression \[{\cos ^4}x\] in terms of the first power of cosine?

Answer 1

Hint:We can solve this using the cosine double angle formula. That is we know the formula \[\cos (2x) = 2{\cos ^2}x - 1\]. We also use the formula \[{(a + b)^2} = {a^2} + {b^2} + 2ab\].

We know that \[{\cos ^4}x = {(\cos x)^4}\]. It can be written as \[{(\cos x)^4} = {({\cos ^2}x)^2}\].

Now we need to express this in terms of the first power of cosine.

Complete step by step solution:
Now, we have \[{\cos ^4}x = {(\cos x)^4}\]

It can be rewritten as \[{(\cos x)^4} = {\left( {{{(\cos x)}^2}} \right)^2} = {\left( {{{\cos }^2}x}
\right)^2}{\text{ - - - - - (a)}}\].

We need \[{\cos ^2}x\] value,

We know the cosine double angle formula \[\cos (2x) = 2{\cos ^2}(x) - 1\].

Rearranging we have,
\[ \Rightarrow 2{\cos ^2}(x) - 1 = \cos (2x)\]

Adding 1 on both side we get,
\[ \Rightarrow 2{\cos ^2}(x) = 1 + \cos (2x)\]

Since we need \[{\cos ^2}x\] we divide the whole equation by 2 we get,
\[{\cos ^2}(x) = \dfrac{1}{2}\left( {1 + \cos (2x)} \right){\text{ }} - - - - - (1)\]

Now substituting in the equation (a) we have,

That is \[{\cos ^4}x = {({\cos ^2}x)^2}\]

\[ = {\left( {\dfrac{1}{2}\left( {1 + \cos (2x)} \right)} \right)^2}\]
\[ = \dfrac{1}{4}{\left( {1 + \cos (2x)} \right)^2}\]

Now using the formula \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] we will get,
\[ = \dfrac{1}{4}\left( {{1^2} + {{\cos }^2}(2x) + 2\cos (2x)} \right)\]

Since we can see that in the above simplified equation we have \[{\cos ^2}(2x)\]. So we need to convert this into the first power of cosine.

Now from equation (1) we can write that \[{\cos ^2}(2x) = \dfrac{1}{2}\left( {1 + \cos (4x)} \right)\], that is multiply 2 with the angels.

Substituting in above equation we get,
\[ = \dfrac{1}{4}\left( {1 + \left( {\dfrac{1}{2}\left( {1 + \cos (4x)} \right)} \right) + 2\cos (2x)}
\right)\]

Multiplying \[\dfrac{1}{4}\] inside the brackets we have,

\[ = \left( {\dfrac{1}{4} + \left( {\dfrac{1}{8}\left( {1 + \cos (4x)} \right)} \right) + 2\dfrac{1}{4}\cos
(2x)} \right)\]

Taking \[\dfrac{1}{8}\]as common we will get,
\[ = \dfrac{1}{8}\left( {2 + \left( {1 + \cos (4x)} \right) + 4\cos (2x)} \right)\] is the required answer.

That is \[{\cos ^4}x = \dfrac{1}{8}\left( {3 + \cos (4x) + 4\cos (2x)} \right)\]

Note: As we can see that we have a large calculation part, so we need to be careful. Remember all the cosine double angle formula that is \[\cos (2x) = 2{\cos ^2}x - 1\], \[\cos (2x) = {\cos ^2}x - {\sin^2}x\] and \[\cos (2x) = 1 - 2{\sin ^2}x\]. We simplify the given equation until we only have first power of cosine in the simplified equation.