Activity for Wolgwangâ€
Type | On... | Excerpt | Status | Date |
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Answer | — |
A: How can I intuit $\dfrac{a - b}{c - d} \equiv \dfrac{{\color{red}{-}}(b - a)}{{\color{red}{-}}(d - c)} \equiv \dfrac{b - a}{d - c}$? (1) $$\dfrac{a - b}{c - d}$$ Image 1 As the image depicts, $(a - b)$ means $a$ steps forward and $b$ steps backward which results in net $(a - b)$ steps, and dividing it by $(c - d)$ gives the number of steps (of length $(c - d)$) in $(a - b)$. Intuitive Meaning of Division can be found her... (more) |
— | over 2 years ago |
Answer | — |
A: Prove $(\cos^3\theta+\sin^3\theta)^2= \cos^6\theta(1+\tan^3\theta)^2$ Simplifying $LHS$: \begin{align} (\cos^3\theta+\sin^3\theta)^2&=\cos^6\theta+\sin^6\theta+2\cos^3\theta\sin^3\theta\\&=\cos^6\theta\bigg(1+\dfrac{\sin^6\theta}{\cos^6\theta}+2\dfrac{\cos^3\theta\sin^3\theta}{\cos^6\theta}\bigg)\\&=\cos^6\theta(1+\tan^6\theta+2\tan^3\theta)\\&=\cos^6\theta[1^2+(... (more) |
— | almost 3 years ago |