Trigonometric Functions Help. (1 Viewer)

[Lemiixem]

Q4. Find the equation of the tangent to the curve y = sin (Pie - x) at the point (Pie/6 , 1/2), in exact form.

I have trouble with the exact form part.

 

[Carrotsticks]

y' = - cos (pi - x)

Substitute x = pi/6 for the gradient.

y' = -cos (pi - pi/6) = - cos (5pi/6) = - (-sqrt(3)/2) = sqrt(3)/2

Point gradient formula is: y - y_1 = m (x - x_1)

Substitute (x_1 , y_1) = (pi/6 , 1/2) and m = sqrt(3)/2

y - 1/2 = sqrt(3)/2 (x - pi/6)

Multiply both sides by 2:

2y - 1 = sqrt(3) (x - pi/6)

Arrange to make it either general form (ax+by+c=0) or gradient/intercept form (y=mx+b)

And exact form means leave exact numbers (root 3 in this case) instead of having decimal points.

 

[Lemiixem]

- cos (5pi/6) = - (-sqrt(3)/2) = sqrt(3)/2

I dont understand this step.

 

[Carrotsticks]

I dont understand this step.

I used the exact value of cos 5pi/6.

Remember that 5pi/6 = 150 degrees, so I'm using the exact value of cos 150 degrees = -sqrt(3)/2