How do you find the equation of the tangent line $y = sin x$ $y=sinx$ at $(\frac{π}{6}, \frac{1}{2})$ $(pi/6, 1/2)$ ?

Question

How do you find the equation of the tangent line $y = sin x$ $y=sinx$ at $(\frac{π}{6}, \frac{1}{2})$ $(pi/6, 1/2)$ ?

2 Answers

Impact of this question

47515 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-31T03:34:26

$y - \frac{1}{2} = \frac{\sqrt{3}}{2} (x - \frac{π}{6})$ $y-1/2=sqrt3/2(x-pi/6)$

Explanation:

The slope of the tangent line to a function $y$ $y$ at $x = a$ $x=a$ is found by calculating the value of $\frac{d y}{d x}$ $dy/dx$ , the derivative of $y$ $y$ , at $x = a$ $x=a$ .

Where

$y = sin x$ $y=sinx$

the derivative is given by

$\frac{d y}{d x} = cos x$ $dy/dx=cosx$

The slope of the tangent line to $y$ $y$ at $x = \frac{π}{6}$ $x=pi/6$ is found by evaluating the derivative of $y$ $y$ at $x = \frac{π}{6}$ $x=pi/6$ :

$m = \frac{d y}{d x} ∣_{x = \frac{π}{6}} = cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$ $m=dy/dx|_(x=pi/6)=cos(pi/6)=sqrt3/2$

The slope of the tangent line is $\frac{\sqrt{3}}{2}$ $sqrt3/2$ . Writing the equation of the line that passes through $(\frac{π}{6}, \frac{1}{2})$ $(pi/6,1/2)$ with slope $\frac{\sqrt{3}}{2}$ $sqrt3/2$ in point-slope form, we get:

$y - y_{1} = m (x - x_{1})$ $y-y_1=m(x-x_1)$

$y - \frac{1}{2} = \frac{\sqrt{3}}{2} (x - \frac{π}{6})$ $y-1/2=sqrt3/2(x-pi/6)$

Answer 2 · 2017-08-31T03:59:02

Differentiate y and evaluate $\frac{d y}{d x}$ $dy/dx$ at $x = \frac{π}{6}$ $x=pi/6$

The equation of the tangent line would then be $y = a x + b$ $y=ax+b$ , where $a$ $a$ is the derivative at $x = \frac{π}{6}$ $x=pi/6$ and $b$ $b$ can be solved by setting $y = \frac{1}{2}, x = \frac{π}{6}$ $y=1/2, x=pi/6$

The equation would be $y = \frac{\sqrt{3}}{2} x + \frac{1}{2} - \frac{\sqrt{3} π}{12}$ $y=sqrt(3)/2x+1/2-(sqrt(3)pi)/12$

Explanation:

Let the equation of the tangent line be $y = a x + b$ $y=ax+b$

$\frac{d y}{d x} = cos x$ $dy/dx=cos x$

$a = \frac{cos π}{6} = \frac{\sqrt{3}}{2}$ $a=cos pi/6=sqrt(3)/2$

$:. y=sqrt(3)/2x+b$

$1/2=sqrt(3)/2*pi/6+b$

$b=1/2-(sqrt(3)pi)/12$

Hence the equation of the tangent line is $y=sqrt(3)/2x+1/2-(sqrt(3)pi)/12$

You can verify this answer visually too

graph{(y-sqrt(3)/2x-1/2+(sqrt(3)pi)/12)(y-sin(x))=0 [-1.259, 1.781, -0.477, 1.04]}

The reason the equation of a tangent line is as shown above is because in a linear function, $y=ax+b$ , a represents the gradient / slope of the line and b represents the y-intercept.

By definition, the gradient of a tangent line is equal to the slope of a curve at the point where the tangent line meets the curve.

Hence, $a=dy/dx$ when $x=pi/6$ and the rest of the equation can be derived through algebra

Science

Math

Humanities

... and beyond

How do you find the equation of the tangent line $y = sin x$ $y=sinx$ at $(\frac{π}{6}, \frac{1}{2})$ $(pi/6, 1/2)$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the equation of the tangent line y=sinxy=sinxy=sinx at (pi/6, 1/2)(π6,12)(pi/6, 1/2)?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you find the equation of the tangent line $y = sin x$ $y=sinx$ at $(\frac{π}{6}, \frac{1}{2})$ $(pi/6, 1/2)$ ?