Find the equation of the tangent where x=1/2pi on the curve y=3sin^2(2x)
The working would be greatly appreciated

Question

irishboy123 · Answer

$$y = 3 \sin^22x$$
$$\frac {dy}{dx} = 2(3)\sin2x \cos2x(2) = 12\sin(2x)\cos(2x) = 6\sin(4x)$$
simplification at end relies on:
sin (A + A) = sinAcosA + cosA sinA = 2 sinAcosA => sin4x = 2sin2x cos2x

anonymous · Answer

I did that much but the equation is y=3 
Since y=6sin(4x) at x=1/2π the gradient is zero, the value of the y-coordinate is also 0. I can't seem to find out a way to solve this.

irishboy123 · Answer

i agree with you.

in fact you don't need to do much other than a sketch to answer the original question.  and get the same "wrong" answer .  soz  

personally i'd stick with the answer we and the sketch arrive at as the correct answer to the question *as it is worded*.


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