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# Find an equation of the tangent line to the curve at the given point. y = sin(sin(x)), (4π, 0)?

**Solution:**

y = sin(sin(x)) [Given]

From the chain rule

y’ = cos (sin x) cos x

At x = 4π it takes the value

y’ (4π) = cos (sin 4π) cos 4π

y’ (4π) = cos (0) (-1)

y’ (4π) = - 1

We know that the equation of the tangent is

y - y\(_1\) = m (x - x\(_1\))

Substituting the values

y - 0 = - 1 (x - 4π)

y = - x + 4π

Therefore, the equation of the tangent line is y = - x + 4π.

## Find an equation of the tangent line to the curve at the given point. y = sin(sin(x)), (4π, 0)?

**Summary: **

An equation of the tangent line to the curve at the given point y = sin(sin(x)), (4π, 0) is y = - x + 4π.

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