10H - Trigonometric equations on an interval

10H - Trigonometric equations on an interval

A calculator gives one angle when you enter an inverse trigonometric function. How can you know that you have found every angle an equation allows? You will use symmetry, periods and the stated interval to solve equations in degrees and radians, including multiplied and shifted angles.

All angles with a given ratio

The equation sin⁡x=12\sin x=\tfrac12 has two solutions between 0∘0^\circ and 360∘360^\circ: 30∘30^\circ and 150∘150^\circ. The inverse sine button returns only 30∘30^\circ, because a function must return one value for an input. The graph shows why that one answer is insufficient for the equation.

[DIAGRAM: asset_name: Repeated sine intersections; asset_slug: sine-interval-solutions; caption: Four intersections of sine x with one half over two complete turns; description: The sine graph from zero to 720 degrees crosses the horizontal line y equals one half at 30, 150, 390 and 510 degrees. The two turns are shaded in different pale colours and each intersection is labelled.]
Four intersections of sine x with one half over two complete turns

Reference angle

A reference angle is an acute angle used with symmetry to locate angles with the same magnitude of trigonometric ratio. The signs of sine, cosine and tangent determine which quadrants to use.

For a reference angle α\alpha, positive sine occurs at α\alpha and 180∘−α180^\circ-\alpha; positive cosine at α\alpha and 360∘−α360^\circ-\alpha; positive tangent at α\alpha and 180∘+α180^\circ+\alpha. Negative values occupy the other two quadrants. At axes, use the actual graph values rather than treating 0∘0^\circ as an acute reference angle.

Periods and solution families

Sine and cosine repeat every 360∘360^\circ, or 2π2\pi radians. Tangent repeats every 180∘180^\circ, or π\pi radians.

If sin⁡α=k\sin\alpha=k, all sine solutions are x=α+360∘nx=\alpha+360^\circ n or x=180∘−α+360∘nx=180^\circ-\alpha+360^\circ n.
If cos⁡α=k\cos\alpha=k, all cosine solutions are x=±α+360∘nx=\pm\alpha+360^\circ n.
If tan⁡α=k\tan\alpha=k, all tangent solutions are x=α+180∘nx=\alpha+180^\circ n, where nn is any integer.

Replace 360∘360^\circ by 2π2\pi and 180∘180^\circ by π\pi for radians. List each distinct angle only once.

Use a permitted scientific calculator for inverse values; symbolic algebra, stored formula retrieval and symbolic differentiation or integration facilities are prohibited. First set degree or radian mode to match the question. Keep extra digits until the final answer. Sine and cosine values must lie between −1-1 and 11; otherwise there are no real solutions. Tangent is undefined at 90∘+180∘n90^\circ+180^\circ n.

Worked example 1: negative sine

Question. Solve sin⁡x=−32\sin x=-\tfrac{\sqrt3}{2} for 0∘≤x≤360∘0^\circ\le x\le360^\circ.

Plan. Use reference angle 60∘60^\circ, then choose the quadrants where sine is negative.

x=180∘+60∘=240∘,x=360∘−60∘=300∘.x=180^\circ+60^\circ=240^\circ, \qquad x=360^\circ-60^\circ=300^\circ.

The interval contains one complete turn. Neither endpoint has the required sine value.

Answer. x=240∘,300∘x=240^\circ,300^\circ.

Check. Both unit-circle heights are −3/2-\sqrt3/2. In this turn sine is negative only in the third and fourth quadrants, with one crossing in each.

Now use tangent and an interval containing negative angles. Its shorter period changes the number of solutions.

Worked example 2: tangent across several half-turns

Question. Solve tan⁡x=1\tan x=1 for −180∘≤x<360∘-180^\circ\le x<360^\circ.

Plan. Start at 45∘45^\circ, then add or subtract whole tangent periods.

x=45∘+180∘n.x=45^\circ+180^\circ n.

The values n=−1,0,1n=-1,0,1 give −135∘,45∘,225∘-135^\circ,45^\circ,225^\circ. The next values, −315∘-315^\circ and 405∘405^\circ, are outside the interval.

Answer. x=−135∘,45∘,225∘x=-135^\circ,45^\circ,225^\circ.

Check. Each angle differs from 45∘45^\circ by a multiple of 180∘180^\circ, so each has tangent 11. Exactly three integers nn give angles inside the interval; the neighbouring family members are outside.

The same method works when the angle is not a special exact value. Retain the calculator's full value until you have found all the angles.

Worked example 3: a calculator angle

Question. Solve cos⁡x=0.4\cos x=0.4 for 0∘≤x<360∘0^\circ\le x<360^\circ, giving answers to one decimal place.

Plan. Use inverse cosine in degree mode for the reference angle, then reflect in the horizontal axis of the unit circle.

α=cos⁡−1(0.4)=66.421821…∘,360∘−α=293.578178…∘.\alpha=\cos^{-1}(0.4)=66.421821\ldots^\circ, \qquad 360^\circ-\alpha=293.578178\ldots^\circ.

Cosine is positive in the first and fourth quadrants. Only now round the two angles.

Answer. x=66.4∘,293.6∘x=66.4^\circ,293.6^\circ, to one decimal place.

Check. The unrounded angles both give cosine 0.40.4. The rounded answers give approximately 0.400350.40035, consistent with the requested precision. The graph has one crossing in each of those two quadrants.

Choose the correct symmetry and period in the next equation.

Continue this lesson.

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