Trigonometric identities and their uses · Grade 11 · 5 units
In Trigonometric identities and their uses (Grade 11 · 5 units) you practise 6 skills: Simplifying with the basic identities · The sine and cosine of a sum and a difference · The double angle identities · Equations solved with identities · Angles in plane figures by the identities · Angles and lengths in solids. A trigonometric identity holds for every angle. Besides the identity that the squares of the sine and the cosine add up to one, there are identities for the sine and the cosine of the sum and of the difference of two angles, and the double angle identities follow from them: the sine of a double angle is twice the sine times the cosine, and the cosine of a double angle is the square of the cosine minus the square of the sine. There are also identities that turn a sum of two sines or of two cosines into a product. The identities simplify expressions, solve equations and give angles and lengths in plane figures. In solids, such as a pyramid and a cone, this is only a first look, with no solid geometry theorems: find a right triangle inside the solid and solve it.
A model proof: the cosine of the difference of two angles. On the unit circle mark two points: the point of the angle alpha, whose coordinates are the cosine and the sine of alpha, and the point of the angle beta, whose coordinates are the cosine and the sine of beta. The angle between them at the centre is alpha minus beta. Work out the square of the distance between the two points in two ways. From the coordinates, it is the square of the difference of the cosines plus the square of the difference of the sines. Open the brackets: the square of the sine of an angle plus the square of its cosine is one, so this is two minus twice the sum of the cosine of alpha times the cosine of beta and the sine of alpha times the sine of beta. By the cosine rule in the triangle made by the centre and the two points, whose two sides from the centre are radii of length one, the square of the distance is two minus twice the cosine of alpha minus beta. The two ways agree, so the cosine of alpha minus beta equals the cosine of alpha times the cosine of beta, plus the sine of alpha times the sine of beta. From the identity we proved: the cosine of a sum. Put minus beta in place of beta. The cosine of minus beta is the cosine of beta, and the sine of minus beta is minus the sine of beta. So the cosine of alpha plus beta equals the cosine of alpha times the cosine of beta, minus the sine of alpha times the sine of beta. From the identity we proved: the sine of a sum. The sine of an angle is the cosine of ninety degrees minus the angle, so the sine of alpha plus beta is the cosine of ninety degrees minus alpha, minus beta. By the identity we proved, that is the cosine of ninety degrees minus alpha times the cosine of beta, plus the sine of ninety degrees minus alpha times the sine of beta, that is the sine of alpha times the cosine of beta, plus the cosine of alpha times the sine of beta. From the identity we proved: the sine of a difference. Put minus beta in place of beta in the identity for the sine of a sum: the sine of alpha minus beta equals the sine of alpha times the cosine of beta, minus the cosine of alpha times the sine of beta. The double angle identities follow from the identities for a sum: put beta equal to alpha.
Tip: Before dividing an equation by a sine or a cosine, check whether it can be zero. Take out the common factor instead, and no solution is lost.
What you learn
Simplifying with the basic identities
For example: Simplify the expression (1 − cos²x) ÷ sin x .
- cos x
- tan x
- 1
- sin x
Answer and solution
The answer: sin x
By sin²x + cos²x = 1 , the numerator 1 − cos²x equals sin²x . Cancel sin x and sin x is left.
The sine and cosine of a sum and a difference
For example: What is the exact value of sin 75° ?
- (√6 − √2) ÷ 4
- (√3 + 1) ÷ 2
- (√2 − √6) ÷ 4
- (√6 + √2) ÷ 4
Answer and solution
The answer: (√6 + √2) ÷ 4
Write 75° = 45° + 30° . By the identity, sin 75° equals sin 45° cos 30° plus cos 45° sin 30° , that is √6/4 + √2/4 . So the value is (√6 + √2) ÷ 4 .
The double angle identities
For example: The angle α is acute, with sin α = 3/5 and cos α = 4/5 . What is sin 2α ?
Answer and solution
The answer: 24/25
Put in the values: 2 × 3/5 × 4/5 = 24/25 , that is sin 2α = 24/25 .
Equations solved with identities
For example: How many solutions does the equation sin 2x = sin x have for 0° ≤ x < 360° ?
Answer and solution
The answer: 4
By sin 2x = 2 sin x cos x , bring everything to one side and take out the common factor: (sin x)(2 cos x − 1) = 0 . The factor sin x is 0 when x = 0°, 180° . The factor 2 cos x − 1 is 0 when x = 60°, 300° . Together there are 4 solutions: 0°, 60°, 180°, 300° .
Angles in plane figures by the identities
For example: The square ABCD has side 2. The point E is on the side BC and the point F on the side CD , with BE = DF = 1 . What is sin ∢EAF ?
Answer and solution
The answer: 3/5
In the triangle ABE the hypotenuse is √5 , so sin θ = 1 ÷ √5 and cos θ = 2 ÷ √5 . The angle ∢EAF is 90° − 2θ , so sin ∢EAF equals cos 2θ . By the identity cos 2θ = cos²θ − sin²θ , the value is 3/5 , so sin ∢EAF = 3/5 .
Angles and lengths in solids
For example: A right pyramid has a square base with side 4, and its height is 9. What is the tangent of the angle between the height of a side face (from the top to the midpoint of a base side) and the base?
Answer and solution
The answer: 9/2
The height comes down to the centre of the square, and from there to the midpoint of a side is half a side: 2. So the tangent is 9/2 .
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 5 units (grades 10 to 12).