Order from least to greatest: 1/2 2/3 1/4 5/6

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rileu2661

05.12.2020

$\sin^2 x \cos^2 x = \dfrac{1 - \cos4x}{8}$ refresh if you see display errors such as \begin{aligned}, etc.focusing on the right side since it looks more complex: $\text{rhs} = \dfrac{1 - \cos4x}{8}$ if we rewrite $\cos 4x$ as $\cos(2 \cdot 2x)$ , then we can treat $2x$ as the angle that is being doubled. that will allow us to use the cosine double angle identity: $\begin{aligned} \cos 2x & = \cos^2 x - \sin^2 x \\ \cos(2 \cdot 2x) & = \cos^2 (2x) - \sin^2 (2x) \end{aligned}$ but notice how next to $\cos 4x$ , we have a 1. if we want to get rid of that 1, we can choose another version of the cosine double angle identity: $\begin{aligned} \cos 2x & = 1 - 2\sin^2 x \\ \cos(2 \cdot 2x) & = 1 - 2\sin^2 (2x) \end{aligned}$ so this leads us to $\begin{aligned} \text{rhs} & = \dfrac{1 - \cos4x}{8} \\ & = \dfrac{1 - \cos\big(2(2x)\big)}{8} \\ & = \dfrac{1 - \left(1 - 2\sin^2 (2x)\right)}{8} \\ & = \dfrac{1 - 1 + 2\sin^2 (2x)}{8} \\ & = \dfrac{2\sin^2 (2x)}{8} \end{aligned}$ we have a sine double angle identity: $\sin (2x) = 2 \sin (x) \cos (x)$ in order to use it, we need to get rid of that square.since $\sin^2 (2x)$ is the same as $\left(\sin (2x)\right)^2$ , which is the same as $\sin(2x) \cdot \sin(2x)$ , we have $\begin{aligned} \text{rhs} & = \dfrac{2\sin^2 (2x)}{8} \\ & = \dfrac{2\sin (2x) \cdot \sin (2x)}{8} \end{aligned}$ use the sine double angle identity twice. $\begin{aligned} \text{rhs} & = \dfrac{2 \big(2\sin(x)\cos(x) \big) \big(2\sin(x)\cos(x)\big)}{8} \end{aligned}$ multiplication order does not matter. multiply all the sines together. multiply all the cosines together. multiply all the 2s together to form 8, which will cancel with denominator. $\begin{aligned} \text{rhs} & = \dfrac{2 \big(2\sin(x)\cos(x) \big) \big(2\sin(x)\cos(x)\big)}{8} \\ & = \frac{(2)(2)(2) \cdot \sin^2 (x) \cos^2(x)}{8} \\ & = \frac{8 \cdot \sin^2 (x) \cos^2(x)}{8} \\ & = \sin^2 (x) \cos^2(x) \\ & = \text{lhs} \end{aligned}$

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