Sketch Trigonometric Functions  sine and cosine
The sketching of sine and cosine functions of the form y = a sin( k ( x  d)) + c and y = a cos( k ( x  d)) + c are discussed with detailed examples.
Graphing Parameters
Amplitude = a
Period = 2π/k
Horizontal Shift (translation) = d , to the left if ( d) is positive and to the right if ( d) is negative.
Vertical Shift (translation) = c , up if c is positive and down if c is negative.
Unit Circle
In order to sketch transformed sine and cosine functions, we need to know how to sketch basic sine and cosine functions. The unit circle (radius = 1) gives the values of sin(x) and cos(x) at 5 key points which can be used to graph more complex sine and cosine functions. The coordinates of any point on the unit circle gives the cosine and sine of the angle in standard position corresponding to that point.
Examples
to the rotation of 0 corresponds the point (1,0) = (cos(0),sin(0))
to the rotation of π/2 (or 90°) corresponds the point (0,1) = (cos(π/2),sin(π/2))
to the rotation of π (or 180°) corresponds the point (1,0) = (cos(π),sin(π))
to the rotation of 3π/2 (or 270°) corresponds the point (0,1) = (cos(3π/2), sin(3π/2))
to the rotation of 2 π (or 360°) corresponds the point (1,0) = (cos(2π),sin(2π)) as shown below .
Sketching Sine and Cosine Functions: Examples with Detailed Solutions

Sketch the graph of y = 2 cos(x) + 1 over one period.
Solution
Graphing Parameters
Amplitude = 2 = 2
Period = 2π
Vertical Shift (translation) = 1 , up 1 unit.
Horizontal Shift (translation) = 0
We start by skeching y = cos(x) using the values of x and y from the unit circle (blue graph below).
x = 0 π/2 π 3π/2 2π
y = 1 0  1 0 1
We then sketch y = 2 cos(x) streching y = cos(x) by 2 (green graph below) and finally y = 2 cos(x) + 1 by shifting up 1 unit (red graph below).

Sketch the graph of y =  2 sin(x)  1 over one period.
Solution
Graphing Parameters
Amplitude = 2 = 2
Period = 2π
Vertical Shift (translation) =  1 , down 1 unit.
Horizontal Shift (translation) = 0
We start by skeching y = sin(x) using the values of x and y from the unit circle (blue graph below).
x = 0 π/2 π 3π/2 2π
y = 0 1 0  1 0
We then sketch y =  2 sin(x) streching y = sin(x) by 2 and reflecting it on the xaxis(green graph below) and finally y =  2 sin(x)  1 by shifting down 1 unit (red graph below).

Sketch the graph of y = 3 cos(2 x + π/3)  1 over one period.
Solution
Graphing Parameters
Amplitude = 3 = 3
Period = 2π/2 = π
Vertical Shift (translation) =  1 , down 1 unit.
Horizontal Shift: Because of the term π/3, the graph is shifted horizontally. We first rewrite the given function as: y = 3 cos [ 2( x + π/6)]  1 and we can now write the shift as being equal to π/6 to the left.
We start by skeching 3 cos(2 x) with minimum and maximum values  3 and + 3 over one period = π (blue graph below).
We then sketch y = 3 cos(2 x)  1 translating the previous graph 1 unit down (green graph below). We now shift the previous graph π/6 to the left (red graph below) so that the sketched period starts at  π/6 and ends at  π/6 + π = 5π/6 which is one period = π.

Sketch the graph of y =  0.2 sin(0.5 x  π/6) + 0.1 over one period.
Solution
Graphing Parameters
Amplitude =  0.2 = 0.2
Period = 2π/0.5 = 4π
Vertical Shift (translation) = 0.1 , up 0.1 unit.
Horizontal Shift: Because of the term  π/6, the graph is shifted horizontally. We first rewrite the given function as: y =  0.2 cos [ 0.5( x  π/3)] + 0.1 and we can now write the shift as being equal to π/3 to the right.
We start by skeching  0.2 sin(0.5 x) with minimum and maximum values  0.2 and +  0.2 over one period = 4 π (blue graph below).
We then sketch y =  0.2 sin(0.5 x) + 0.1 translating the previous graph 0.1 unit up (green graph below). We then shift the previous graph π/3 to the right (red graph below) so that the sketched period starts at π/3 and ends at π/3 + 4π which is one period = 4π.

Sketch the graph of y = 2 cos(2 x  60°)  2 over one period.
Solution
Solution
Graphing Parameters
Amplitude = 2 = 2
Vertical Shift (translation) =  2 , down 2 units.
Period = 360/2 = 180°
Horizontal Shift: Because of the term  60°, the graph is shifted horizontally. We first rewrite the given function as:y = 2 cos[2( x  30°)]  2 and we can now write the shift as being equal to 30° to the right.
We start by skeching y = 2 cos(2 x) with minimum and maximum values  2 and + 2 over one period = 180° (blue graph below).
We then sketch y = 2 cos(2 x)  2 translating the previous graph 2 units down (green graph below). We then shift the previous graph 30° to the right (red graph below) so that the sketched period starts at 30° and ends at 30° + 180° = 210° which is one period = 180°.

Sketch the graph of y =  2 sin(x/3 + π/3)  1 over one period.
Solution
Graphing Parameters
Amplitude =  2 = 2
Period = 2π/(1/3) = 6π
Vertical Shift (translation) =  1 , down 1 unit.
Horizontal Shift: Because of the term π/3, the graph is shifted horizontally. We first rewrite the given function as: y =  2 sin[(1/3)(x + π)]  1 and we can now write the shift as being equal to π to the left.
We start by skeching  2 sin(x/3) with minimum and maximum values  2 and + 2 over one period = 6 π (blue graph below).
We then sketch y =  2 sin(x/3)  1 translating the previous graph 1 unit down (green graph below). We then shift the previous graph π to the left (red graph below) so that the sketched period starts at π and ends at 5π which is one period = 6π.


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