# Core 4 maths question, please help!find the coordinates of the point of intersectino of the tangents to y=sin x and y=tan x at the points where x=1/3 pi I know that dy/dx of sin x is cos x, and...

find the coordinates of the point of intersectino of the tangents to y=sin x and y=tan x at the points where x=1/3 pi

I know that dy/dx of sin x is cos x, and that dy/dx of tan x is sec^2 x. Any further help would be greatly appreciated. Thank you

hala718 | Certified Educator

We need to find the point of intersection of the tangent lines of the curves:

y= sinx

y= tanx

We will find the equation of the tangent line of the curve y= sinx at the point x= pi/3

First we will find the y-coordinate

==> y= sin(pi/3)= sqrt3/2

Then the pint (pi/3, sqrt3/2) is on the tangent line.

Now we will find the slope.

==> The slope is the derivative of y at x= pi/3

==> y' = cosx

==> y'(pi/3)= cos(pi/3)= 1/2

Then the slope is m= 1/2

==> y- y1= m (x-x1)

==> y- sqrt3/2 = (1/2) (x- pi/3)

==> y- sqrt3/2 = (1/2)x - pi/6

==> y= (1/2)x - pi/6 + sqrt3/2............(1)

Now we will find the equation of the tangent line for the curve y= tanx.

==> x= pi/3 ==> y= tan(pi/3)= sqrt3 ==> Then the point (pi/3, sqrt3) is on the line.

Now we will find the slope.

==> y'= sec^2 x

==> y'(pi/3)= sec^2 (pi/3)= 1/cos^2 (pi/3)= 1/(1/2)^2= 4

Then the slope is m= 4.

==> y-y1= m(x-x1)

==> y- sqrt3 = 4(x- pi/3)

==> y= 4x - 4pi/3 + sqrt3 .................(2)

Now we will find the intersection points between the tangent lines (1) and (2).

==> 4x - 4pi/3 + sqrt3 = (1/2)x - pi/6 + sqrt3/2

Now we will combine like terms.

==> (7/2)x = 4pi/3 - pi/6 + sqrt3/2 - sqrt3

==> (7/2)x = 7pi/6 - sqrt3/2

==> x= (2/7)*[ 7pi/6 - sqrt3/2)

==> x= pi/3 - sqrt3/7 = 0.8 ( approx.)

==> y= 4x + sqrt3 - 4pi/3

==> y= 4(0.8) + sqrt3 - 4pi/3

==> y= 3.2 + sqrt3 - 4pi/3= 0.74 ( approx.)

Then, the intersection point of the tangent lines is (0.8, 0.74)