Trigonometric Graphs

Check the value at zero and the direction of the first part of the curve.

1. The curve starts at (0,0), rises to a maximum of 1 at 90^\circ, crosses the axis at 180^\circ and reaches a minimum of -1 at 270^\circ.
2. This is y=\sin x^\circ.
3. y=\cos x^\circ starts at 1, not 0, and y=-\sin x^\circ goes down first.

Look at the output when the input is zero.

1. The curve starts at 1 when x=0, falls to -1 at 180^\circ and climbs back to 1 at 360^\circ.
2. This is y=\cos x^\circ.
3. y=-\cos x^\circ would start at -1, and y=\sin x^\circ starts at 0.

Look for separate increasing branches and gaps in the curve.

1. The graph is in separate pieces that rise more and more steeply, with breaks at 90^\circ and 270^\circ where it has no value.
2. This is y=\tan x^\circ, which repeats every 180^\circ.
3. The sine and cosine graphs are unbroken waves between -1 and 1.

Read the height of the highest point.

1. The highest point has output y=1.
2. The maximum value is 1.

Read the x-coordinate of the lowest point.

1. The lowest output is -1.
2. It occurs at x=180^\circ.

Include the endpoints of the stated interval.

1. The sine curve meets the axis at 0^\circ, 180^\circ and 360^\circ.
2. Both endpoints are included.

Find how far along the x-axis one branch must move to match the next.

1. The same tangent shape repeats after a horizontal move of 180^\circ.
2. The period is 180^\circ.

Imagine the horizontal line y=0.8 and read the x-values where it meets the curve.

1. Draw the horizontal line y=0.8. It meets the sine curve twice, both times between 0 and 180.
2. Reading down from the first crossing gives x\approx53.
3. The curve is symmetrical about x=90, so the second crossing is as far before 180 as the first is after 0: x\approx180-53=127.
4. Estimates: x\approx53 and x\approx127. (A calculator gives \sin^{-1}0.8=53.1^\circ.)

The cosine graph from 0 to 360 is symmetrical about x=180.

1. The graph of y=\cos x^\circ for 0\le x\le360 is symmetrical about the vertical line x=180.
2. x=70 is 110 to the left of 180, so the matching point is 110 to the right: 180+110=290.
3. So \cos290^\circ=0.342 as well. (It is not 110: the cosine curve is below the axis between 90 and 270, so \cos110^\circ is negative.)

Multiplying by -1 changes the sign of every y-value.

1. y=-\cos x^\circ is the graph of y=\cos x^\circ reflected in the x-axis: every y-value changes sign.
2. \cos x^\circ starts at 1, falls to -1 at 180 and returns to 1, so -\cos x^\circ starts at -1, rises to 1 at 180 and returns to -1.
3. Graph B has this shape. Graph D is y=\cos x^\circ itself, Graph A is y=\sin x^\circ and Graph C is y=\tan x^\circ.

Where is the minimum of y=\sin x^\circ, and what does multiplying by 2 do to the y-values?

1. y=\sin x^\circ has its minimum value, -1, at x=270.
2. Multiplying by 2 doubles every y-value but does not change the x-values, so the minimum of 2\sin x^\circ is 2\times(-1)=-2, still at x=270.
3. The minimum point is (270,-2).

Divide by 2 first, then use the symmetry of the sine graph.

1. Divide both sides by 2: \sin x^\circ=\frac12.
2. \sin30^\circ=\frac12, so one solution is x=30.
3. The sine graph is symmetrical about x=90, so the other solution is 180-30=150.
4. Between 180 and 360 the sine graph is below the axis, so there are no more solutions: x=30 or x=150.

The cosine graph repeats every 360^\circ.

1. On the graph of y=\cos x^\circ from 0 to 360, the lowest point, y=-1, is at x=180.
2. The graph repeats every 360, so the next lowest point is at 180+360=540.
3. 540+360=900 is more than 720, so the solutions are x=180 and x=540.

Look where the cosine curve crosses the x-axis.

1. The cosine curve crosses the axis at 90^\circ and 270^\circ.
2. These are all the solutions in the interval.

Use the tangent value at a familiar angle and its repeating period.

1. \tan45^\circ=1.
2. Tangent repeats every 180^\circ, giving the second solution 225^\circ.
3. The solutions are 45^\circ and 225^\circ.

First solve \sin A^\circ=0.5, where A=x-30, then add 30.

1. Let A=x-30. As x goes from 0 to 360, A goes from -30 to 330.
2. Solve \sin A^\circ=0.5: \sin30^\circ=0.5, and by the symmetry of the sine graph about 90, \sin150^\circ=0.5 too.
3. No other values between -30 and 330 work: the sine graph is below the axis from -30 to 0 and from 180 to 330.
4. Add 30 to get x: x=30+30=60 or x=150+30=180.

Isolate the cosine value, then use the two matching positions on the cycle.

1. Rearrange: 4\cos x^\circ=-2, so \cos x^\circ=-\frac12.
2. You know \cos60^\circ=\frac12. On the cosine graph, y=-\frac12 is below the axis, which happens between 90 and 270.
3. The curve from 90 to 180 is the curve from 0 to 90 turned upside down and reversed, so \cos(180-60)^\circ=-\frac12: x=120.
4. The graph is symmetrical about x=180, so the other solution is 180+60=240.

Use a reference angle and the period of tangent.

1. \tan45^\circ=1. The tangent graph has rotational symmetry about the origin, so \tan(-45)^\circ=-1.
2. -45 is not in the range, but the tangent graph repeats every 180^\circ: -45+180=135 and 135+180=315.
3. The solutions are x=135 and x=315.

Your calculator gives one angle; use the symmetry of the sine graph for a second, then use the fact that it repeats every 360^\circ.

1. Use a calculator: \sin^{-1}0.4=23.578\ldots^\circ, so x=23.6 is one solution.
2. The sine graph is symmetrical about x=90, so another solution is 180-23.578\ldots=156.42\ldots
3. The graph repeats every 360^\circ, so add 360 to each: 383.578\ldots and 516.42\ldots
4. To 1 decimal place: x=23.6, 156.4, 383.6 and 516.4.

The required inputs are where the two curves cross.

1. The solutions are the x-values where the two curves cross.
2. \sin45^\circ=\cos45^\circ=\frac{\sqrt2}2, so x=45 is one solution: the first crossing on the graph.
3. Both curves repeat their pattern after 180^\circ with the signs changed, so they cross again at 45+180=225, where both are -\frac{\sqrt2}2.
4. The solutions are x=45 and x=225.