What is the tangent of-sin(x/2) ?

The tangent of-sin(x/2) is y=-cos((a_{0})/2) 1/2 x-sin((a_{0})/2)+1/2 a_{0}cos((a_{0})/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of-sin(x/2)

tangent of

- sin (\frac{x}{2})

y = - cos (\frac{a _{0}}{2}) \frac{1}{2} x - sin (\frac{a _{0}}{2}) + \frac{1}{2} a_{0} cos (\frac{a _{0}}{2})

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - sin (\frac{a _{0}}{2}))

Find the slope of

y = - sin (\frac{x}{2}) : \frac{d y}{d x} = - cos (\frac{x}{2}) \frac{1}{2}

Compute the slope of

y = - sin (\frac{x}{2})

x = a_{0} : m = - cos (\frac{a _{0}}{2}) \frac{1}{2}

Find the line with slope m=

- cos (\frac{a _{0}}{2}) \frac{1}{2}

and passing through

(a_{0}, - sin (\frac{a _{0}}{2})) : y = - cos (\frac{a _{0}}{2}) \frac{1}{2} x - sin (\frac{a _{0}}{2}) + \frac{1}{2} a_{0} cos (\frac{a _{0}}{2})

y = - cos (\frac{a _{0}}{2}) \frac{1}{2} x - sin (\frac{a _{0}}{2}) + \frac{1}{2} a_{0} cos (\frac{a _{0}}{2})

x \to - 1 lim (10 x (14 x - 3))

\int \frac{( arctan ( x ) ) ^{5}}{1 + x ^{2}} d x

\int \frac{x ^{4} + 1}{x ^{3} + x ^{2}} d x

\frac{\partial}{\partial y} (tan (1 + 2 x^{2} y^{4} z^{2}))

f (x) = 3 sin (x) + ln (5 x)

What is the tangent of-sin(x/2) ?
The tangent of-sin(x/2) is y=-cos((a_{0})/2) 1/2 x-sin((a_{0})/2)+1/2 a_{0}cos((a_{0})/2)