# What is the equation of the tangent line of f(x)=sin x + sin^2 x at x=0?

May 17, 2016

$y = x$

#### Explanation:

This problem can be tackled in 3 steps:

1. Figure out the gradient of the line using the derivative of the function and the given $x$ value.
2. Use the given $x$ value and $f \left(x\right)$ to find $C$.
3. Put values into the standard equation of a straight line.

The equation of a line $\to y = m x + C$

1. We'll begin by working out $m$, the gradient of the tangent. To do this we simply have to find $f ' \left(x\right)$, the derivative.

$f \left(x\right) = \sin \left(x\right) + {\sin}^{2} \left(x\right)$
$f ' \left(x\right) = \cos \left(x\right) + 2 \cos \left(x\right) \sin \left(x\right)$

Where the chain rule has been used on the ${\sin}^{2} \left(x\right)$ term.

Since we want to find the tangent at $x = 0$ then the gradient can be given by:
$f ' \left(0\right)$
$= \cos \left(0\right) + 2 \cos \left(0\right) \sin \left(0\right) = 1$

Thus $m = 1$.
Step 1 is complete.

2. We must now find $C$.

Use $f \left(0\right)$
$= \sin \left(0\right) + {\sin}^{2} \left(0\right) = 0$

Thus $\left\{0 , 0\right\}$ Are the co -ordinates which the line intercepts $f \left(x\right)$ so we know the line passes through $\left\{0 , 0\right\}$.

So, using $y = 0 , x = 0 \mathmr{and} m = 1$ we can substitute these into our equation of a straight line to obtain:

$y = m x + C$
$\left(0\right) = \left(1\right) \left(0\right) + C$
$C = 0$

Thus we have our value for $C$. Step 2 is complete.

3. We can now write:

$y = m x + C$
$y = 1 \left(x\right) + \left(0\right)$ So the final equation is:
$y = x$

Indeed we see that if we plot both on the same graph we get a line intersection at a tangent exactly where we would expect: