Finding the Equation of a Tangent: y=2sinx at P(5pie/6,1)

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How do i find the equation of the tangent to the graph given by the equation at the point
P(5pie/6),1) y=2sinx
The answer in the back of the textbook says.
sqroot3X + y - 1 - 5sqroot3pie/6 = 0
 
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what do you need to define a line? gradient and intercept...
 
Gradient can be given by the derivative, so find the derivative. And the y-intercept is quite straight forward.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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