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Year 12 Maths Advanced (2027) Differential calculus

Tangents & Normals to Trigonometric Curves

20 practice questions 0 video lessons Theory + worked examples

Learn tangents and normals to trigonometric curves for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-04). Using the derivatives of sine, cosine and tangent in radians, you can find the slope of a trig curve at any point and write lines through that point.

You will learn to evaluate the gradient at an exact angle, write the tangent and perpendicular normal, and find where these lines cross the axes — careful work with exact values that brings together trigonometric derivatives and straight-line equations.

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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.

Tangent to y = sin x at the originThe tangent to y = sin x at (0,0) is the line y = x. x (0,0) tangent y=x y=sin x
The tangent to \(y=\sin x\) at \((0,0)\) is \(y=x\).
\[y-y_0=m(x-x_0),\quad m=f'(x_0)\]
tangent y minus y nought equals m times (x minus x nought), with m equal to the derivative at x nought

Method

  1. Find the point using radian values.
  2. Find the gradient \(m=f'(x_0)\).
  3. Write the tangent; a gradient of \(0\) means a horizontal line.
Example 1 — Tangent at origin
Find the tangent to \(y=\sin x\) at \(x=0\).
Solution
\(m\)\(=\)\(\cos0=1\)
\(y\)\(=\)\(x\)
Example 2 — Horizontal
Find the tangent to \(y=\cos x\) at \(x=0\).
Solution
\(m\)\(=\)\(-\sin0=0\)
\(y\)\(=\)\(1\)
Example 3 — Tangent at x=π
Find the tangent to \(y=\sin x\) at \(x=\pi\).
Solution
\(\text{point}\)\((\pi,0)\)
\(m\)\(=\)\(\cos\pi=-1\)
\(y\)\(=\)\(-x+\pi\)
Example 4 — Peak
Find the tangent to \(y=\sin x\) at \(x=\dfrac{\pi}{2}\).
Solution
\(m\)\(=\)\(\cos\dfrac{\pi}{2}=0\)
\(y\)\(=\)\(1\)

Common pitfalls

Radians throughout. Use radian values of \(\sin,\cos\) for both point and gradient.
Zero gradient means horizontal. At a peak or trough the tangent is \(y=\) constant.
Evaluate exactly using exact trig values where possible.

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.