\( \def\dfrac#1#2{\displaystyle\frac{#1}{#2}} \def\solve#1{\begin{array}{rcl}#1\end{array} } \)

Home / 08 Trigonometric Functions / 60 Double Angle for Power Reduction

Example: Express \(\cos^4(x)\) using only first powers of \(\cos(x)\).


Solution

This one is a bit different than the last few where we needed to make some drawings. In this example, the process will be purely algebraic by looking for opportunities to apply the double angle formula for Cosine: \[ \solve{ \cos(2\theta) &=& 2\cos^2\theta-1\\ \cos(2\theta)+1 &=& 2\cos^2\theta \\ \dfrac{\cos(2\theta)+1}{ 2 } &=& \cos^2\theta } \]With this Power Reduction formula in hand, we can tackle the original problem:

\[ \solve{ \cos^4(x) &=& (\cos^2 x)^2 \\ &=& \left(\frac{\cos(2x)+1}{ 2 }\right)^2\\ &=&\dfrac{\cos^2(2x)+2\cos(2x)+1}{{4}}\\ &=&\dfrac{\left(\frac{\cos(4x)+1}{{2}}\right) + 2\cos(2x)+1}{{4}}\\ &=&\dfrac{\left(\frac{\cos(4x)+1}{{2}}\right) + 2\cos(2x)+1}{{4}}\times {\color{ red } \dfrac{ 2 }{ 2} }\\ &=&\dfrac{{\color{ red }2 }\left(\frac{\cos(4x)+1}{{2}}\right)+{\color{ red }2 }\cdot2\cos(2x)+{\color{ red }2 }\cdot 1}{ {\color{ red }2 } \cdot 4}\\ &=&\dfrac{\cos(4x)+1+4\cos(2x)+2}{ 8 } \\&=&\dfrac{\cos(4x)+4\cos(2x)+3}{ 8 } } \]

So, that is a lot of algebra, and I want to make sure that I didn't make any mistakes. What can I do? One way to double check your work in a situation like this, is to graph the original expression and the new expression and see if their graphs line up:

aligned graphs

As you can see (feel free to toggle the graphs on/off and pan around) the graphs are indeed identical, so we can be very confident that our solution is correct.