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

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,

\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.