Tangents & Normals to Trigonometric Curves
Tangents and normals to trigonometric curves use the trig derivatives to find the gradient at a point, then form the tangent \(y-y_0=m(x-x_0)\) and its perpendicular normal (radians).
Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Differential calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.
Theory
Tangents to a trig curve use the derivatives \(\sin'=\cos\), \(\cos'=-\sin\) (radians). This Year 12 Mathematics Advanced topic (MAV-12-04) finds tangents and normals to trigonometric curves.
At \((x_0,y_0)\) on \(y=f(x)\): the tangent has gradient \(m=f'(x_0)\), equation \(y-y_0=m(x-x_0)\); the normal has gradient \(-\dfrac1m\).
A gradient of \(0\) gives a horizontal tangent, which happens at a maximum or minimum.
Method
- Find the point using radian values.
- Find the gradient \(m=f'(x_0)\).
- Write the tangent; a gradient of \(0\) means a horizontal line.
| \(m\) | \(=\) | \(\cos0=1\) |
| \(y\) | \(=\) | \(x\) |
| \(m\) | \(=\) | \(-\sin0=0\) |
| \(y\) | \(=\) | \(1\) |
| \(\text{point}\) | \((\pi,0)\) | |
| \(m\) | \(=\) | \(\cos\pi=-1\) |
| \(y\) | \(=\) | \(-x+\pi\) |
| \(m\) | \(=\) | \(\cos\dfrac{\pi}{2}=0\) |
| \(y\) | \(=\) | \(1\) |
Common pitfalls
Frequently asked questions
How do you find the tangent to a trig curve?
Find the point, work out the gradient by substituting into the derivative (in radians), then write the straight line through the point with that gradient.
What is the tangent to y = sin x at the origin?
The gradient is cos 0 which is 1, so the tangent is y equals x.
What does a horizontal tangent tell you?
A gradient of zero means the tangent is horizontal, which happens at a maximum or minimum of the curve.
Why must the angle be in radians?
The trig derivatives such as the derivative of sin x being cos x only hold when the angle is measured in radians.