# If x = cos t(3 - 2cos^2t) and y = sin t (3 - 2sin^2t), then find the value dy/dx of  at t = π/4.  ​

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If x = cos t(3 - 2cos2t) and y = sin t (3 - 2sin2t), then find the value of $\frac{dy}{dx}$ at t = $\frac{\pi}{4}$.

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Given,

x = cos t(3 - 2cos2t)

Differentiating both sides with respect to t, we get

Differentiating both sides with respect to t, we get

∴ $\frac{dy}{dx}]_{t=\frac{\pi}{4}}$

$= cot\,\frac{\pi}{4}=1$