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Trigonometric Ratios for Sum of Two Angles

If x=a sin 2 ...

Question

If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t) then find $\frac{d y}{d x} a t t = \frac{π}{4}$ .

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Solution

$x = a s i n 2 t (1 + c o s 2 t), y = b c o s 2 t (1 - c o s 2 t) \frac{d y}{d x} = 2 a c o s 2 t (1 c o s 2 t) + a s i n 2 t (- 2 s i n 2 t) = 2 a c o s 2 t + 2 a c o s^{2} 2 t - 2 a s i n^{2} 2 t = 2 a c o s 2 t + 2 a c o s 4 t \frac{d y}{d x} = - 2 b s i n 2 t (1 - c o s 2 t) + b c o s 2 t (2 s i n 2 t) = - 2 b s i n 2 t + 2 b s i n 2 t c o s 2 t + 2 b c o s 2 t s i n 2 t (2 s i n 2 t) = - 2 b s i n 2 t + 4 b s i n 2 t c o s 2 t = - 2 b s i n 2 t + 2 b s i n 4 t \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- 2 b s i n 2 t + 2 b s i n 4 t}{2 a c o s 2 t + 2 a c o s 4 t} \Rightarrow \frac{d y}{d t} = \frac{- 2 b s i n 2 t + 2 b s i n 4 t}{2 a c o s 2 t + 2 a c o s 4 t} \Rightarrow {∣ ∣ \frac{d y}{d t} ∣ ∣}_{t = \frac{π}{4}} = \frac{- 2 b s i n \frac{2 π}{4} + 2 b s i n \frac{4 π}{4}}{2 a c o s \frac{2 π}{4} + 2 a c o s \frac{4 π}{4}} \Rightarrow {∣ ∣ \frac{d y}{d t} ∣ ∣}_{t = \frac{π}{4}} = \frac{- 2 b s i n \frac{π}{4} + 2 b s i n π}{2 a c o s \frac{π}{2} + 2 a c o s π} \Rightarrow {∣ ∣ \frac{d y}{d t} ∣ ∣}_{t = \frac{π}{4}} = \frac{- 2 b}{- 2 a} = \frac{b}{a} ∴\Rightarrow {∣ ∣ \frac{d y}{d t} ∣ ∣}_{t = \frac{π}{4}} = \frac{b}{a}$

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Similar questions

x = a sin 2 t (1 + cos 2 t)

y = b cos 2 t (1 - cos 2 t)

\frac{d y}{d x}

t = \frac{π}{4}

x = a sin 2 t (1 + cos 2 t)

y = b cos 2 t (1 + cos 2 t)

\frac{d y}{d x}

t = \frac{π}{4}

.

x = a sin 2 t (1 + cos 2 t)

y = b cos 2 t (1 - cos 2 t)

\frac{d y}{d x}

t = \frac{π}{4}

.

x = a s i n 2 t (1 + c o s 2 t)

y = b c o s 2 t (1 - c o s 2 t),

then find the values of

\frac{d y}{d x}

t = \frac{π}{4}

t = \frac{π}{3} .

x = a sin 2 t (1 + cos 2 t)

y = b cos 2 t (1 - cos 2 t)

t = \frac{π}{4}, (\frac{d y}{d x}) = \frac{b}{a} .

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Trigonometric Ratios for Sum of Two Angles

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