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x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} …

prove\:\cot(2x)=\frac{1-\tan^2(x)}{2\tan(x)} prove\:\csc(2x)=\frac{\sec(x)}{2\sin(x)} prove\:\frac{\sin(3x)+\sin(7x)}{\cos(3x)-\cos(7x)}=\cot(2x)

Use the form acsc(bx−c)+ d a csc (b x - c) + d to find the variables used to find the amplitude, period, phase shift, and vertical shift. Since the graph of the function csc c s c does not have a maximum or minimum value, there can be no value for the amplitude. Find the period of csc(2x) csc (2 x). Tap for more steps...

\left(-\csc(2x^{1})\right)\cot(2x^{1})\times 2x^{1-1} The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0.

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x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int \int_{\msquare}^{\msquare} \lim \sum \infty \theta (f\:\circ\:g) f(x)

Simplify csc^2(x) - cot^2(x)

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Apr 19, 2016 · cscx = ± √2. x = csc−1(± √2) x = sin−1(± 1 √2) x = ± π 4 +2πn, ± 3π 4 +2πn. S = {± π 4 + 2πn, ± 3π 4 + 2πn}