# Calculus ( Find the equation of a tangent line help) ?

Find the equation of the tangent line to the graph of f(x) = csc 2x at the point where

x =π/8

. Exact trigonometric values must be used.

Update:

Relevance

f(x) = csc(2x) ← this is a function, i.e. a curve

f(x) = 1/sin(2x) → when: x = π/8

f(π/8) = 1/sin(π/4)

f(π/8) = 1/[(√2)/2] = 2/(√2) = (2 * √2)/(√2 * √2) = √2 → the curve passes throught A (π/8; √2)

...so the tangent line to the curve at x = π/8 too.

The function f looks like (u/v), so the derivative looks like: [(u'.v) - (v'.u)]/v² → where:

u = 1 → u' = 0

v = sin(2x) → v' = 2.cos(2x)

f'(x) = [(u'.v) - (v'.u)]/v² → where: u' = 0

f'(x) = - (v'.u)/v² → where: u = 1

f'(x) = - v'/v²

f'(x) = - 2.cos(2x)/sin²(2x) ← this is the derivative

…but the derivative is too the slope of the tangent line to the curve at x

f'(π/8) = - 2.cos(π/4)/sin²(π/4)

f'(π/8) = - 2.[(√2)/2]/[(√2)/2]²

f'(π/8) = - 2/[(√2)/2]

f'(π/8) = - 4/√2

f'(π/8) = - (4 * √2)/(√2 * √2)

f'(π/8) = - (4 * √2)/2

f'(π/8) = - 2√2 ← this is the slope of the tangent line to the curve at x = π/8

The typical equation of a line is: y = mx + y₀ → where m: slope and where y₀: y-intercept

The slope of the tangent line is (- 2√2), so the equation of this tangent line becomes: y = - (2√2).x + y₀

The tangent line passes through A, so the coordinates of this point must verify the equation of the tangent line.

y = - (2√2).x + y₀

y₀ = y + (2√2).x → you substitute x and y by the coordinates of the point A (π/8; √2)

y₀ = √2 + [(2√2) * (π/8)]

y₀ = √2 + [(π√2)/4]

→ The equation of the tangent line at x = π/8 is: y = - (2√2).x + √2 + [(π√2)/4)

• Given curve :

f(x) = csc 2x

f'(x) = -  2 (csc 2x * cot 2x )

Hence the tangent line has slope (m)  =  - 2 (csc 2x * cot 2x )

=>  m  =  - 2 [ csc π/4 * cot π/4 ]  =  - 2 [ √2 * 1 ]  =  - 2√2 = - 2.8284

Hence the equation is :

y = (- 2√2) x + b ............................... (1)

when x = π/8 ,   y  =  √2

From (1)    ---

√2  =   (- 2√2) *  π/8  +  b

=>  b  =  √2  +  (2√2) * π/8  =  1.41421 +  1.110720  =  2.52493

Equation of the tangent :

y  =   - 2.8284 x + 2.52493  ......................  Answer

• f(x) = csc(2x) = 1 / sin(2x)

f'(x) = -2cos(2x) / (sin(2x))^2

f'(π/8) = -2cos(π/4) / (sin(π/4))^2 = -√2 / (1/2) = -2√2

f(π/8) = √2

In point-slope form, the line's equation is:

y - √2 = -2√2(x - π/8)