How much will a person pay for 12.5 pounds of

Ответ:

wbrandi118

06.08.2020

Refresh if you see weird display errors. $\sin 2x = -2 \sin\left(x - \frac{\pi}{2}\right) \cos\left(x - \frac{\pi}{2}\right)$ for the left hand side, sine double angle identity can be used: $\text{lhs} = \sin 2x = 2\sin x \cos x$ for the right hand side, we have $\text{rhs} = -2 \sin\left(x - \frac{\pi}{2}\right) \cos\left(x - \frac{\pi}{2}\right)$ i will deal with $\sin\left(x - \frac{\pi}{2}\right)$ separately. we can use this sine difference identity: $\begin{aligned} \sin(a-b) & = \sin a \cos b - \cos a \sin b\\ \sin\left(x - \tfrac{\pi}{2}\right) & = \sin x \cos\tfrac{\pi}{2}-\cos x \sin \tfrac{\pi}{2} \end{aligned}$ from unit circle, etc, we have $\cos\frac{\pi}{2} = 0,\ \sin\frac{\pi}{2} = 1$ , therefore $\begin{aligned}\sin\left(x - \tfrac{\pi}{2}\right) & = \sin x \cos\tfrac{\pi}{2}-\cos x \sin \tfrac{\pi}{2} \\ & = \sin x (0) - \cos x (1)\\ & = -\cos x \end{aligned}$ now i will deal with $\cos\left(x - \frac{\pi}{2}\right)$ separately. we can use this cosine difference identity: $\begin{aligned} \cos(a-b) & = \cos a\cos b + \sin a \sin b \\ \cos\left(x - \tfrac{\pi}{2}\right) & = \cos x \cos\tfrac{\pi}{2} + \sin x \sin \tfrac{\pi}{2} \\ & = \cos x (0) + \sin x (1) \\ & = \sin x \end{aligned}$ bringing these two simplifications together: $\begin{aligned} \text{rhs} & = -2 \sin\left(x - \tfrac{\pi}{2}\right) \cos\left(x - \tfrac{\pi}{2}\right) \\ & = -2 (-\cos x)(\sin x) \\ & = 2\cos x \sin x \\ & = 2\sin x \cos x \\ & = \text{lhs} \end{aligned}$

0,0(0 оценок)

How much will a person pay for 12.5 pounds of bananas at a price of $1.52 per pound?

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question