how to incorporate the mathematical practice standards and constructivism into an elementary classroom

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RealGibbon857

23.04.2023

Mathematics

Useful identities: $\begin{aligned} \cos\left(\tfrac{\pi}{2} - \theta\right) & = \sin\theta & & \text{cofunction identity} \\ \tan(-\theta) & = -\tan\theta & & \text{odd tangent identity} \\ \tan\theta & = \dfrac{\sin\theta}{\cos\theta} \\ \sin^2\theta + \cos^2\theta & = 1 & & \text{pythagorean identity} \end{aligned}$ ======answer: $\sec\theta$ ====== $\dfrac{\cos \theta}{1 - \cos\left(\tfrac{\pi}{2} - \theta\right) } + \tan(-\theta)$ use cofunction identity to convert into sine $\begin{aligned} & = \dfrac{\cos \theta}{1 - \sin\theta } + \tan(-\theta) \end{aligned}$ use odd tangent identity to move the negative out from tangent: $\begin{aligned} & = \dfrac{\cos \theta}{1 - \sin\theta } + \big(- \tan(\theta) \big) \\ & =\dfrac{\cos \theta}{1 - \sin\theta } - \tan\theta \end{aligned}$ use the tangent identity $\tan\theta = \sin \theta / \cos\theta$ $\begin{aligned} & =\dfrac{\cos \theta}{1 - \sin\theta } - \dfrac{\sin \theta}{\cos\theta} \end{aligned}$ we have two fractions with unlike denominators. let us get them to have the same denominator (this usually leads to more simplification).multiply numerator and denominator of $\frac{\cos \theta}{1 - \sin\theta }$ by $\cos\theta$ . $\begin{aligned} & =\dfrac{\cos\theta \cdot \cos \theta}{\cos\theta(1 - \sin\theta) } - \dfrac{\sin \theta}{\cos\theta} \\ & = \dfrac{\cos^2 \theta}{\cos\theta(1 - \sin\theta) } - \dfrac{\sin \theta}{\cos\theta} \end{aligned}$ multiply numerator and denominator of $\frac{\sin \theta}{\cos\theta}$ by $(1 - \sin\theta)$ $\begin{aligned}& = \dfrac{\cos^2 \theta}{\cos\theta(1 - \sin\theta) } - \dfrac{\sin \theta(1 - \sin\theta)}{\cos\theta(1 - \sin\theta)} \end{aligned}$ same denominators mean we can combine the fractions: $\begin{aligned}& = \dfrac{\cos^2 \theta - \sin\theta(1 - \sin\theta)}{\cos\theta(1 - \sin\theta)} \end{aligned}$ distribute in the numerator: $\begin{aligned}& = \dfrac{\cos^2 \theta - \sin\theta(1 - \sin\theta)}{\cos\theta(1 - \sin\theta)} \\ & = \dfrac{\cos^2 \theta - \sin\theta + \sin^2\theta}{\cos\theta(1 - \sin\theta)} \end{aligned}$ addition order does not matter; rearrange so that cos² θ is added on to sin² θ $\begin{aligned}& = \dfrac{\cos^2 \theta + \sin^2\theta - \sin\theta }{\cos\theta(1 - \sin\theta)} \\ & = \dfrac{\boxed{\cos^2 \theta + \sin^2\theta} - \sin\theta }{\cos\theta(1 - \sin\theta)} \end{aligned}$ use pythagorean identity to replace that boxed part with 1. $\begin{aligned}& = \dfrac{\boxed{1} - \sin\theta }{\cos\theta(1 - \sin\theta)} \\ & = \dfrac{1 - \sin\theta }{\cos\theta(1 - \sin\theta)} \end{aligned}$ the $(1-\sin\theta)$ cancel out and we are left with $= \dfrac{1}{\cos\theta}$ which is, by definition, secant because secant is the reciprocal of cosine. $= \sec \theta$

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