Understand · explore · practise
Understand why sin x differentiates to cos x and cos x to −sin x. Explore chord and tangent gradients, first-principles proofs, radian units and original worked practice.
Before you startPure 1 differentiation; radian measure; angle-addition formulas
01 / Read the slope, not the height
On y = sin x, the graph reaches height 1 at x = π/2, but it is momentarily flat there. The derivative records the slope at each input. It is a new function, not another name for the original height.
d/dx(sin x) = cos x
d/dx(cos x) = −sin x
These formulas use radians.
Choose a function and a point in the explorer. Compare the finite-step chord with the tangent, then shrink the step from both sides.
At x = π/3, sin x = √3/2 and the tangent gradient is cos(π/3) = 1/2.
With h = 0.2, the chord gradient is approximately 0.410.
Derivative of sin x is cos x, with x in radians.
Try small positive and negative steps. The chord approaches the green tangent from either side. The limit option displays the derivative; it does not calculate 0/0.
02 / Match the sign to the graph
Sine rises through the origin, so its slope there is +1. At its peak the slope is 0; at x = π it crosses downwards with slope −1. These values match cos x. Cosine starts flat at x = 0, then falls: its gradient is −sin x.
At x = 0: (sin x)′ = 1 and (cos x)′ = 0
The height and the slope need not agree.
At x = π/2: (sin x)′ = 0 and (cos x)′ = −1
Cosine is falling as it crosses the horizontal axis.
At x = π: (sin x)′ = −1 and (cos x)′ = 0
Keep the minus sign for the cosine derivative.
A student says the gradient of y = sin x at π/2 is 1 because sin(π/2) = 1. Correct the claim.
Evaluate the derivative, not the original function.
The height is 1. The gradient is cos(π/2) = 0, so the tangent is horizontal.
03 / Prove the sine rule from first principles
For f(x) = sin x, fix x and let the non-zero increment h tend to 0. The required radian limits are sin h/h → 1 and (cos h − 1)/h → 0. A numerical picture supports the idea; these limits complete the proof.
f′(x) = lim as h → 0 of [sin(x + h) − sin x]/h
This is the definition of the derivative.
sin(x + h) = sin x cos h + cos x sin h
Expand before trying to take the limit.
[sin(x + h) − sin x]/h
= sin x · (cos h − 1)/h + cos x · sin h/h
Group the two known limits.
f′(x) = sin x · 0 + cos x · 1 = cos x
x is fixed throughout the limit.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
What are the limits of sin h/h and (cos h − 1)/h as h → 0 in radians?
Use the small-angle relationships.
The limits are 1 and 0, respectively. Neither result is obtained by substituting h = 0 into the fractions.
04 / Prove the cosine rule
cos(x + h) = cos x cos h − sin x sin h
The sine product is subtracted.
[cos(x + h) − cos x]/h
= cos x · (cos h − 1)/h − sin x · sin h/h
Again isolate the standard radian limits.
d/dx(cos x) = cos x · 0 − sin x · 1
Take the limit with x fixed.
d/dx(cos x) = −sin x
The result matches the direction of the graph.
A proof ends with cos x · 0 + sin x · 1. Which earlier sign needs checking?
Write out cos(x + h).
cos(x + h) = cos x cos h − sin x sin h. The second term of the difference quotient must be negative, giving −sin x.
05 / Why radians matter
If u counts degrees, the same sine wave is written g(u) = sin(πu/180), where the inner angle is in radians. Its derivative with respect to u is (π/180)cos(πu/180). A change of one degree is a smaller input-angle change than one radian.
Radian input: derivative of sin x = cos x.
Degree input u: derivative of sin(πu/180) = (π/180)cos(πu/180).
Use radian mode when evaluating trigonometric functions in these calculus questions. A graph label or model may use another unit; convert it explicitly.
For g(u) = sin(πu/180), find g′(0). Here u is the angle in degrees.
Include the conversion factor.
g′(0) = π/180. This is the rate per degree; the corresponding rate per radian at the origin is 1.
06 / Differentiate a linear inner angle
For constant k and b, an increment h in x changes kx + b by kh. Applying the same first-principles limits produces the factor k. A constant phase b shifts where the slopes occur.
d/dx[sin(kx + b)] = k cos(kx + b)
d/dx[cos(kx + b)] = −k sin(kx + b)
5 × 2 cos(2x − π/3)
The first inner angle has derivative 2.
−3 × [−4 sin(4x + 1)]
The second term has two minus signs.
10 cos(2x − π/3) + 12 sin(4x + 1)
Keep each inner angle unchanged.
Differentiate sin(−2x), and check by using an identity first.
Sine is odd; cosine is even.
The direct derivative is −2cos(−2x) = −2cos 2x. Alternatively sin(−2x) = −sin 2x, which gives the same derivative.
What happens to the derivative when k = 0 in sin(kx + b)?
The function no longer depends on x.
It is the constant sin b, so the derivative is 0. The general formula also gives 0cos b = 0.
07 / Combine with familiar derivatives
d/dx(4 sin 3x) = 12 cos 3x
Coefficient 4 times the inner factor 3.
d/dx(−2 cos 5x) = 10 sin 5x
The cosine rule contributes a minus sign.
f′(x) = 12 cos 3x + 10 sin 5x
The constant 7 differentiates to 0.
Find f′(0) for f(x) = x² + 6 sin 2x − cos 3x.
Differentiate before substituting x = 0.
f′(x) = 2x + 12cos 2x + 3sin 3x, so f′(0) = 12.
For f(x) = a sin 2x + 3 cos x, the gradient at x = 0 is 8. Find a.
f′(0) = 2a.
f′(x) = 2a cos 2x − 3sin x. Thus 2a = 8 and a = 4.
08 / Find a tangent or normal
y = 3cos(π/3) + 1 = 5/2
The contact point is (π/6, 5/2).
dy/dx = −6sin 2x
Differentiate before evaluating.
m = −6sin(π/3) = −3√3
This is the tangent gradient.
y − 5/2 = −3√3(x − π/6)
Point–slope form keeps the answer exact.
The normal at this point has gradient 1/(3√3). At a horizontal tangent, the normal is vertical: write x = constant instead of dividing by zero.
Find the normal to y = sin 2x at x = 0.
The point is (0, 0), and the tangent gradient is 2.
The normal has gradient −1/2, so y = −x/2.
Find the normal to y = cos x at x = π.
The tangent gradient is zero.
The point is (π, −1). Its tangent is horizontal, so its normal is x = π.
09 / Find every stationary point in the interval
dy/dx = 2cos x − 1
Set the gradient to zero.
cos x = 1/2 gives x = π/3 or 5π/3
Both solutions lie in the interval.
Points: (π/3, √3 − π/3) and (5π/3, −√3 − 5π/3)
Substitute into y, not into dy/dx.
The derivative signs are +, −, +
The first point is a local maximum; the second is a local minimum.
Stationary points are not automatically the global extrema on a closed interval. Compare endpoint values as well if the question asks for the greatest or least value.
Find the stationary points of y = sin 2x for 0 ≤ x < 2π.
cos 2x = 0; the inner angle covers [0, 4π).
x = π/4, 3π/4, 5π/4, 7π/4. The points have y-values 1, −1, 1, −1 respectively. The first and third are maxima; the others are minima.
For y = 2 sin x − x on [0, 2π], is the local minimum at 5π/3 the global minimum?
Compare its value with y(2π).
Yes. At 5π/3 the value is −√3 − 5π/3 ≈ −6.968, below y(2π) = −2π ≈ −6.283 and y(0) = 0. The derivative sign pattern confirms there are no other interior candidates.
10 / Interpret a trigonometric rate
Suppose a model gives displacement s = 0.4 sin(3t) metres, with t in seconds and the phase 3t in radians. The velocity is ds/dt = 1.2 cos(3t) metres per second. A negative velocity means decreasing displacement, not a negative speed.
At t = π/3: s = 0 and ds/dt = −1.2 m/s.
This is a model with stated units. Its formula and interval must fit the situation before its predicted rates can be trusted.
For that model, find displacement and velocity at t = π/6.
The phase is π/2.
s = 0.4m and ds/dt = 0m/s. The greatest displacement is reached while the instantaneous velocity is zero.
Differentiate 1.2 cos(3t) once more and evaluate at t = π/6.
Keep the factor 3 and the cosine minus sign.
d²s/dt² = −3.6sin(3t), giving −3.6m/s². This is the acceleration predicted by the model.
11 / Check signs, units and the requested quantity
Section 1 of 11 · Read the slope, not the height