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,
$${\cos }^2(x)={\displaystyle \frac{1}{2}}\left(1+\cos (2x)\right)-----(1)$$

Now substituting in the equation (a) we have,

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

$$={\left({\displaystyle \frac{1}{2}}\left(1+\cos (2x)\right)\right)}^2$$
$$={\displaystyle \frac{1}{4}}{\left(1+\cos (2x)\right)}^2$$

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

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 $${\cos }^2(2x)={\displaystyle \frac{1}{2}}\left(1+\cos (4x)\right)$$, that is multiply 2 with the angels.

Substituting in above equation we get,
$$={\displaystyle \frac{1}{4}}\left(1+\left({\displaystyle \frac{1}{2}}\left(1+\cos (4x)\right)\right)+2\cos (2x)\right)$$

Multiplying $${\displaystyle \frac{1}{4}}$$ inside the brackets we have,

$$=\left({\displaystyle \frac{1}{4}}+\left({\displaystyle \frac{1}{8}}\left(1+\cos (4x)\right)\right)+2{\displaystyle \frac{1}{4}}\cos (2x)\right)$$

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

That is $${\cos }^{4}x={\displaystyle \frac{1}{8}}\left(3+\cos (4x)+4\cos (2x)\right)$$

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.