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An Insight into Trigonometric Identities and Trigonometric Equations Elementary Trigonometric Results Recall definition of sine, cosine and tangent of an angle. Let P( x, y ) be a point on a circle with radius r = x 2 + y 2 in the Cartesian Plane uuuv uuuv As usual, ∠XOP is measured in the direction from OX to OP , taking anticlockwise as positive. Let θ = ∠XOP x sin θ y y Then sin θ = , cos θ = , tan θ = = cos θ x r r Relations between trigonometric functions of angles of opposite signs Observe the following diagram where points A and A’, B and B’ are reflection of each other in the x-axis (think of A' to A as well as A to A' ) B(c,d) A(a,b) A’(a,-b) B’(c,-d) We can summarize the change of coordinates corresponding to change of sign of angle (namely a reflection) as this : ⎛x⎞ ⎛ x ⎞ ⎜ ⎟ → ⎜ ⎟ as θ → −θ ，for all values of θ , positive or negative ⎝ y⎠ ⎝−y⎠ We thus have the result sin(−θ ) = − sin θ , cos(−θ ) = cos θ and tan(−θ ) = − tan θ ……….. (1) Relations between trigonometric functions of angles by a difference of one right angle The following diagram shows how coordinates are changed as a result of adding one right angle to a given one. The various cases are: 1 ∠XOP is changed to ∠XOQ , ∠XOQ is changed to ∠XOR ,……… Y Q(-b,a) P(a,b) X O R(-a,-b) S(b,-a) We can summarize the change of coordinates as a result of the addition of one right angle as this: ⎛ x ⎞ ⎛−y⎞ π ⎜ ⎟ → ⎜ ⎟ as θ → θ + , where θ is in any quadrant 2 ⎝ y⎠ ⎝ x ⎠ We thus have x y If cos θ = and sin θ = , r r π π −y x then sin(θ + ) = = cos θ , cos(θ + ) = = − sin θ 2 r 2 r So, π π π sin(θ + ) = cos θ , cos(θ + ) = − sin θ and tan(θ + ) = − tan θ 2 2 2 for all values of θ ………………………………………………………… (2) (The reader should not just take ∠XOP as an example of θ , instead, think about ∠XOQ, ∠XOR and ∠XOS as well ) With results (1) and (2), which are valid for all values of θ , we deduce all results related to complimentary and supplementary angles. For example, cos(1800 − θ ) = cos(900 + 900 − θ ) = - sin(900 − θ ) = - cos( −θ ) = - cosθ which is true for all values of θ Formulae for addition and subtraction of angles. 2 Y C B p q O A X The above figure shows two angles p and q of any size, given by ∠AOC and ∠AOB respectively, where A, B and C are points on a circle, centre O, radius r . (It need not be p > q as depicted) Let OA be lying on the OX axis and so the angles are measured using OX as initial direction. We find C(r cos p, r sin p ) , B(r cos q, r sin q ) By using the cosine rule, we have, for all possible positions of B and C, 2(OB)(OC) cos( p − q ) = (OB) 2 + (OC) 2 − (BC) 2 So, 2r 2 cos( p − q ) = r 2 + r 2 − [(r cos p − r cos q ) 2 + (r sin p − r sin q ) 2 ] = 2r 2 − r 2 [cos 2 p − 2 cos p cos q + cos 2 q] − r 2 [sin 2 p − 2sin p sin q + sin 2 q ] = 2r 2 − r 2 [1 + 1 − 2 cos p cos q − 2sin p sin q ] = 2r 2 [cos p cos q + sin p sin q] So, cos( p − q) = cos p cos q + sin p sin q , for all values of p and q ……………… (3) From results (1) , (2) and (3) we can establish all formulae related to additions and subtractions of angles, For example cos( p + q) = cos( p − −q) = cos p cos − q + sin p sin − q = cos p cos q − sin p sin q Trigonometric Equations Let us investigate the three simple equations sin θ = a, where -1 ≤ a ≤ 1 cos θ = b, where -1 ≤ b ≤ 1 tan θ = c, for all values of c 3 First note that all three equations have two solutions in any interval of length 2π , for example, we can find two solutions in each of (0.5π , 2.5π ), (-0.9π , 1.1π ) etc So, to begin with, it is usually convenient to obtain two in the range of [−π , π ] Next, we note that if α is a solution, say to sin θ = a, where -1 ≤ a ≤ 1 , then sin α = a, sin(α + 2π ) = a, sin(α − 2π ) = a, sin(α + 2nπ ) = a, where n is any integer So, if α , β are two solutions to sin θ = a, where -1 ≤ a ≤ 1 , in the interval [−π , π ] , then the complete solution set is θ = α + 2nπ , β + 2nπ , though of course, it works out to be θ = nπ + (−1) n (α ) , which is equivalent to θ = nπ + (−1) n ( β ) (More discussions are left as an exercise) Exercise (without answers attached) Recall these results: sin(−θ ) = − sin θ , cos(−θ ) = cos θ and tan(−θ ) = − tan θ , for all values of θ …………………………. π π (1) π sin(θ + ) = cos θ , cos(θ + ) = − sin θ and tan(θ + ) = − tan θ 2 2 2 ……………………… for all values of θ cos( p − q ) = cos p cos q + sin p sin q , for all values of p and q ……………… (2) (3) 1. Testify the rule : ⎛ x ⎞ ⎛−y⎞ 0 ⎜ ⎟ → ⎜ ⎟ as θ → θ + 90 , where P ( x, y ) is the point on a circle of radius y x ⎝ ⎠ ⎝ ⎠ r , θ = ∠XOP , for the following cases 2. By using Results (1) and (2) above, express the following in terms of sin θ , cosθ and tanθ only (i) (iv) 3. cos(1800 − θ ) (iii) cos(1800 + θ ) (vi) tan(1800 − θ ) tan(1800 + θ ) By using Results (1) and (2) above, express the following in terms of sin θ , cosθ , tanθ and cotθ only (i) (iv) 4. sin(1800 − θ ) (ii) sin(1800 + θ ) (v) sin(2700 − θ ) (ii) sin(2700 + θ ) (v) cos(2700 − θ ) (iii) cos(2700 + θ ) (vi) tan(2700 − θ ) tan(2700 + θ ) By using Results (1), (2) and (3) above, obtain the formulae sin( p + q) = sin p cos q + cos p sin q and sin( p − q ) = sin p cos q − cos p sin q 4 5. α , β are two different solutions of the equation sin θ = a, where a ≤ 1 State a relation between α and β if (i) α and β are in the same quadrant (ii) α and β are in different quadrant (iii) there is no information about the quadrants of α and β . 6. α , β are two different solutions of the equation cos θ = a, where a ≤ 1 State a relation between α and β 7. α , β are two different solutions of the equation tan θ = a What is the values of α − β ? 8 In the solution for the equation cos(2 x − 700 ) = 0.5 , values of x required for are in the interval (−1500 , 5700 ) How many possible values of x are there? 5 Exercise (with answers attached) Recall these results: sin(−θ ) = − sin θ , cos(−θ ) = cos θ and tan(−θ ) = − tan θ , …………………………. for all values of θ π π (1) π sin(θ + ) = cos θ , cos(θ + ) = − sin θ and tan(θ + ) = − tan θ 2 2 2 for all values of θ ……………………… cos( p − q) = cos p cos q + sin p sin q , for all values of p and q ……………… (2) (3) 1. Testify the rule : ⎛ x ⎞ ⎛−y⎞ 0 ⎜ ⎟ → ⎜ ⎟ as θ → θ + 90 , where P ( x, y ) is the point on a circle of radius y x ⎝ ⎠ ⎝ ⎠ r , θ = ∠XOP , for the following cases (i) (ii) (iii) θ = 2400 θ = −2400 θ = 5000 2. By using Results (1) and (2) above, express the following in terms of sin θ , cosθ and tanθ only (i) sin(1800 − θ ) (ii) cos(1800 − θ ) (iii) tan(1800 − θ ) (iv) sin(1800 + θ ) (v) cos(1800 + θ ) (vi) tan(1800 + θ ) Answers sin θ − tan θ (i) (ii) − cos(θ ) (iii) (iv) − sin θ (v) − cos θ (vi) tan θ 3. By using Results (1) and (2) above, express the following in terms of sin θ , cosθ , tanθ and cotθ only (i) (iv) Answers: (i) (iv) sin(2700 − θ ) (ii) sin(2700 + θ ) (v) cos(2700 − θ ) (iii) cos(2700 + θ ) (vi) tan(2700 − θ ) tan(2700 + θ ) − cos θ − cos θ − sin θ sin θ cot θ − cot θ (ii) (v) (iii) (vi) 4. By using Results (1), (2) and (3) above, obtain the formulae sin( p + q) = sin p cos q + cos p sin q and sin( p − q) = sin p cos q − cos p sin q Solution π π π π π sin( p + q ) = − cos( + p + q) = −[cos( + p ) cos q − sin( + p ) sin q ] 2 2 2 = sin p cos q + cos p sin q π sin( p − q ) = − cos( + p − q) = −[cos( + p ) cos − q − sin( + p ) sin − q 2 2 2 6 = sin p cos q − cos p sin q 5. α , β are two different solutions of the equation sin θ = a, where a ≤ 1 State a relation between α and β if α and β are in the same quadrant (i) α and β are in different quadrant (ii) (iv) Answers: (i) (ii) (iii) 6. β = 2nπ + α , where n ∈ β = 2nπ + π − α , where n ∈ π − α β = π − α , 2π + α , 3π − α ,.... -π − α , -2π + α , -3π − α ,...... So, β = nπ + (−1) n α , where n ∈ α , β are two different solutions of the equation cos θ = a, where a ≤ 1 State a relation between α and β Answer 7. there is no information about the quadrants of α and β . β = 2nπ ± α , where n ∈ α , β are two different solutions of the equation tan θ = a What is the values of α − β ? Answer α − β = nπ , for n ∈ In the solution for the equation cos(2 x − 700 ) = 0.5 , values of x required for are in the interval (−1500 , 5700 ) How many possible values of x are there? Answer The interval length for x is 5700 − −1500 = 7200 = 2 × 3600 So, the interval length for (2 x − 700 ) is 4 × 3600 So, there are 8 possible values for (2 x − 700 ) , and therefore also 8 possible values for x . 8 7