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0580 · 6.3Exact trigonometric values

Cambridge IGCSE Mathematics (0580) · Topic 6, Trigonometry · 4:31

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In this video

  1. 0:00Exact values and 45 degrees. Explains exact trigonometric values and derives sine, cosine and tangent for 45 degrees using an isosceles right triangle.
  2. 1:23Values for 30, 60, 0 and 90 degrees. Derives exact values for 30 and 60 degrees, explains endpoint values, and introduces patterns for recalling the table.
  3. 3:00Using exact values in calculations. Works through exact-value calculations and a triangle problem, then reviews common errors and the geometric origins of the values.

Transcript: Exact trigonometric values

Exact values and 45 degrees

0:00The sine of thirty degrees is exactly one half, not merely a decimal close to it. Some angles have trigonometric values we can find using simple geometry. Two special triangles explain most of them. By the end, you will understand where these values come from and how to use them without a calculator. An exact value has no rounding error. Fractions and square roots can represent exact values, even when their decimal forms continue forever.

0:23For example, √2 is exact, but one point four one four is only an approximation. When a question asks for an exact answer, keep the fraction or square root. Remember that sine means opposite divided by hypotenuse. Cosine compares the adjacent side with the hypotenuse, while tangent means opposite divided by adjacent. Opposite and adjacent depend on the angle you choose. Similar triangles have the same ratios, so enlarging a triangle does not change its trigonometric values.

0:52Start with a right-angled triangle whose two shorter sides both have length one. The remaining angles are equal, so each is forty-five degrees. By Pythagoras, the hypotenuse squared equals 1 + 1, which is two. Its length is therefore √2. For either forty-five degree angle, the opposite and adjacent sides are equal. Sine and cosine therefore both equal 1/√2. Multiplying the numerator and denominator by √2 gives √2/2. Tangent equals one because its numerator and denominator are equal.

Values for 30, 60, 0 and 90 degrees

1:23For thirty and sixty degrees, begin with an equilateral triangle of side length two. Its angles are all sixty degrees. The perpendicular from the top vertex bisects both the base and the top angle. Each resulting right-angled triangle has hypotenuse two, short side one, and remaining side √3 by Pythagoras. Relative to thirty degrees, the opposite side is one and the adjacent side is √3. This gives sine one half, cosine √3/2, and tangent 1/√3.

1:52Rationalising the denominator turns 1/√3 into √3/3. Choosing sixty degrees swaps the opposite and adjacent sides. Its sine is √3/2, cosine is one half, and tangent is √3. Zero and ninety degrees need a slightly different picture because an ordinary triangle would collapse. Imagine a line of length one turning from horizontal to vertical. Its horizontal reach represents cosine, and its vertical height represents sine. At zero degrees, cosine is one and sine is zero, while ninety degrees reverses them.

2:25Tangent also equals sine divided by cosine because the hypotenuse cancels from their ratios. At zero degrees, this gives 0 ÷ 1, so tangent is zero. At ninety degrees, it would require division by zero, which is undefined. Tangent at ninety degrees is not one of the required exact values. The table brings all the values together. Sine increases across these angles, while cosine lists the same values in reverse.

2:49For sine, a useful pattern is the square roots of zero, one, two, three and four, each divided by two. That pattern helps recall the values, but the triangles explain why they work.

Using exact values in calculations

3:00An exam-style question asks you to calculate the sine of thirty degrees plus the cosine of sixty degrees exactly. Replace each trigonometric value with one half. Adding one half to one half gives one. No decimal approximation is needed because both substitutions are exact. Next, calculate the square of the sine of forty-five degrees, plus the cosine of sixty degrees. Substitute √2/2 and one half. Squaring √2/2 gives 2/4, because the numerator and the denominator are both squared.

3:29Simplifying gives ½ + ½, so the exact answer is one. Finally, triangle PQR has its right angle at Q, angle QPR of sixty degrees, and hypotenuse PR of twelve centimetres. Find the exact length x of side QR. The sine of sixty degrees equals x/12, so √3/2 = x/12. Multiplying both sides by twelve gives x = 12 × √3/2. The exact length is 6√3 centimetres. These methods must work without a calculator on Paper two, and remain useful on calculator Paper four.

4:01An exact answer must not be rounded to three significant figures. Watch for swapping sine and cosine, or taking the square root of the whole fraction 3/2. The denominator sits outside the square root. The essential idea is that exact values come from exact side ratios. Equal legs explain forty-five degrees, and a bisected equilateral triangle explains thirty and sixty degrees. Horizontal and vertical positions explain the endpoint values. Reconstructing these pictures lets you check the table rather than trust memory alone.