FURTHER TRIGONOMETRIC IDENTITIES AND EQUATIONS


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1 Mathematics Revision Guides Further Trigonometric Identities and Equations Page of 7 M.K. HOME TUITION Mathematics Revision Guides Level: AS / A Level AQA : C4 Edexcel: C3 OCR: C3 OCR MEI: C4 FURTHER TRIGONOMETRIC IDENTITIES AND EQUATIONS Version :. Date: Examples 4, 83 and 78 are copyrighted to their respective owners and used with their permission.
2 Mathematics Revision Guides Further Trigonometric Identities and Equations Page of 7 Further Trigonometric identities. The Pythagorean Identities (revision. For all angles cos + sin = (This is the Pythagorean identity tan sin cos From the definitions of the reciprocal functions of sin x = cosec x (the cosecant of x, sometimes abbreviated to csc x cosx tan x = sec x (the secant of x = cot x (the cotangent of x it is possible to obtain several new identities. Because the cotangent is the reciprocal of the tangent, we have cot cos sin By dividing the Pythagorean identity cos + sin = throughout by cos we have + tan sec A similar division of the identity cos + sin = throughout by sin gives cot cosec
3 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 3 of 7 Compound Angle Identities. The following identities hold true for all angles A and B. sin (A + B = sin A cos B + cos A sin B. cos (A + B = cos A cos B  sin A sin B. sin (A  B = sin A cos B  cos A sin B. cos (A  B = cos A cos B + sin A sin B. The formula for tan (A +B can be worked out as sin Acos B cos Asin B sin( A B sin Acos B cos Asin B tan( A B cos Acos B cos Acos B cos( A B cos Acos B sin Asin B cos Acos B sin Asin B cos Acos B cos Acos B This formidable expression simplifies to tan (A + B = tan A + tan B tan A tan B The expression for tan (A B can be obtained similarly. tan (A  B = tan A  tan B + tan A tan B In each case the results for (AB can be derived from those for (A+B by reversing the + and  signs a useful revision hint!
4 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 4 of 7 Double and Half Angles. By taking the compound angle formulae and replacing B with A, we obtain the double angle identities. sin (A + A = sin A cos A + cos A sin A = sin A cos A. cos (A + A = cos A cos A  sin A sin A = cos A  sin A. tan (A + A = tan A + tan A = tan A tan A tan A tan A The formula for cos A can be written in two other useful variant forms by using cos A+ sin A=. cos A  sin A = cos A ( cos A = cos A. cos A  sin A = (sin A  sin A =  sin A. The double angle formulae are thus: sin (A = sin A cos A cos (A = cos A  sin A = cos A =  sin A tan (A = tan A tan A By replacing A with ½A in the last two formulae for cos A, we have the half angle identities: cos A = cos ½A cos ½A = + cos A cos ½A = ½( + cos A cos A =  sin ½A sin ½A = cos A  sin ½A =  cos A sin ½A = ½(  cos A The above identities are useful in further calculus, especially in integration.
5 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 5 of 7 Using and Proving Trigonometric Identities. The above identities can be manipulated to give rise to new ones. Using and proving identities is a skill which improves with practice. Example (: Find the values of sin 75 and cos 75, leaving the results as surds. From the results, also find the values of tan 75 and tan 5. (Note that tan 5 = cot 75 We treat 75 as and apply the identity sin ( = sin 45 cos 30 + cos 45 sin 30. Since sin 45 = cos 45 = or ( 3 or 3., the expression for sin 75 becomes (cos 30 + sin 30 To find cos 75, we substitute into the identity cos ( = cos 45 cos 30  sin 45 sin 30. This expression simplifies into or ( 3 or 3. (cos 30  sin 30 To find tan 75, we could either divide the result for cos 75 into the one for sin 75, or we could substitute into the compound angle tangent formula. The question asks for the former method. Dividing sin 75 by cos 75 we have tan 75 = 3 3 = = = =. (rationalising the denominator To find tan 5, we can divide the result for sin 75 into cos 75, or find the reciprocal of tan 75. = = = =. (rationalising the denominator
6 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 6 of 7 cos A cos A sin A Example (: Prove that cos A sin A. (Copyright Letts Educational, Revise AS and A Mathematics (004 ISBN , Example Proving Identities p.33 The first step is to start with the expression on one side of the identity, and then to replace parts of it with an equivalent form. The LHS is more complex, and so we can rewrite it using the double angle result for cos A. The LHS thus becomes two squares, and so it can be rewritten cos A sin A cos A sin A (cos A sin A(cos A sin A (cos A sin A, but then the top line can be recognised as the difference of After taking out cos A sin A as a factor, the expression becomes cos A + sin A, thus LHS = RHS.. Example (3: Prove that csc sec csc A sin A A A (Copyright Letts Educational, Revise AS and A Mathematics (004 ISBN , Example Proving Identities, p.33 (The software on the computer uses csc for cosec. Since cosec A = sin A, we can multiply both top and bottom of the LHS by sin A to give sin, which is equivalent to A cos A (using cos A +sin A =, and finally to sec A as on the RHS (using sec A = cosa. Example (4: Prove that sin tan tan. (Copyright Letts Educational, Revise AS and A Mathematics (004 ISBN , Trigonometric Equations, Progress Check Ex., p.34 Rewrite the RHS as tan sec (using + tan sec tan cos (using sec = cos sin cos cos (using tan = sin cos sin cos, which is the same as sin LHS = RHS
7 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 7 of 7 Example (5: Use the result of the previous example to express cos A in terms of tan A. Since sin A tan A, it follows that cos A sin A cos A tan A cos A sin A tan A. tan A tan A cos A tan A tan A tan cos A tan A A (previous example plus double angle formula These examples are also often written as halfangle identities involving tan ½ If t = tan ½then the other ratios for can be obtained in terms of t, as per the diagram. The standard doubleangle formula is t tan t Using Pythagoras, the length of the hypotenuse is t ( t From this result, it follows that 4 4t t t 4 t t t. sin t t and cos t t.
8 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 8 of 7 Example (6: Show that sin P + sin Q = sin ½(P+Q cos ½(PQ. Hint: use P = A + B and Q = A  B Take the two compound angle sine identities and add them together: sin (A + B + sin (A  B = (sin A cos B + cos A sin B + (sin A cos B  cos A sin B = sin A cos B. Substituting P = A + B and Q = A B, we have A = ½(P+Q and B = ½(PQ. The summed expression above therefore becomes sin P + sin Q = sin ½(P + Q cos ½(P  Q. Had we subtracted the second sine identity from the first, we would have had sin (A + B  sin (A  B = (sin A cos B + cos A sin B  (sin A cos B  cos A sin B = cos A sin B, or sin P  sin Q = cos ½(P + Q sin ½(P  Q. The cosine identities are treated similarly. cos (A + B + cos (A  B = (cos A cos B  sin A sin B + (cos A cos B + sin A sin B = cos A cos B. cos P + cos Q = cos ½(P + Q cos ½(P  Q. cos (A + B  cos (A  B = (cos A cos B  sin A sin B  (cos A cos B + sin A sin B =  sin A sin B. (watch the minus sign here! cos P  cos Q = sin ½(P + Q sin ½(P  Q. Example (7: Express cos 3x in terms of cos x. cos 3x = cos (x + x = cos x cos x sin x sin x = ( cos x  (cos x ( sin x cos x(sin x (Use double angle formulae = ( cos 3 x cos x ( sin x cos x = ( cos 3 x cos x ( (  cos x(cos x (Use cos x +sin x = = ( cos 3 x cos x ( cos x  cos 3 x = 4 cos 3 x 3 cos x. Note: The corresponding working for sin 3x is similar. sin 3x = sin (x + x = sin x cos x + cos x sin x = ( sin x cos x(cos x + ( sin x(sin x (Use double angle formulae = ( sin x cos x + (sin x  sin 3 x = ( (sin x(  sin x + (sin x  sin 3 x (Use cos x +sin x = = ( sin x sin 3 x + (sin x  sin 3 x = 3 sin x  4 sin 3 x.
9 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 9 of 7 Example (8: Prove that sin tan cos sec. (Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN , Exercise 4D, Q.3a sin tan cos sec sin sin cos cos sec sin cos sec cos Multiplying both sides by cos we have the standard identity sin cos, because sec and cos are reciprocals of each other. Example (9: Prove that sin cos csc. cos sin (The software on the computer uses csc for cosec. (Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN , Exercise 4D, Q.8a sin cos csc cos sin sin ( cos ( cos (sin csc sin cos cos ( cos (sin csc cos ( cos (sin csc ( cos ( cos (sin csc csc sin LHS = RHS because cosec and sin are reciprocals of each other. csc cot tan Example (0: Prove that cos. (Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN , Exercise 4D, Q.8b cos sin sin cos cos sin sin cos sin cos We begin with cot tan. Therefore sin cos. cot tan csc cot tan cot tan Hence csc (csc (sin cos and finally csc cos cot tan (because cosec and sin are reciprocals of each other.
10 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 0 of 7 Example (: Prove that cos 4 sin 4 cos. (Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN , Exercise 4D, Q.8d cos 4 sin 4 cos cos 4 sin 4 cos (cos sin cos sin cos ((cos sin cos cos cos (Both LHS and RHS are expressions for cos. Example (: Prove that sec 4 sec tan 4 tan. (Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN , Exercise 4D, Q.8e Since sec tan we havesec 4 sec tan tan tan 4 sec 4 sec tan tan 4 tan = tan 4 tan LHS = RHS. Example (3: Prove the identity: tan( tan 3 60 tan( 60. 3tan (Copyright OCR, GCE Mathematics Paper 473, Jun. 007., Q.9, part (i. We begin with tan( tan tan60 60 tan tan60 tan( tan tan60 60 tan tan60. Multiplying, we have tan tan60 tan tan60 tan( 60 tan( 60 tan tan60 tan tan60 tan tan 60 tan( 60 tan( 60 tan tan 60 (using difference of squares Because tan 60 = we can substitute 3 for tan 60 ; thus tan( tan 3 60 tan( 60 3tan.
11 Mathematics Revision Guides Further Trigonometric Identities and Equations Page of 7 Solving more trigonometric equations. Identities may be used to rewrite an equation in a form that is easier to solve. Example (4 : Solve the equation sin x + cos x  = 0 for  x. We must first use sin x =  cos x to change the equation into a quadratic in cos x. Thus sin x + cos x  = 0 (cos x + cos x  = 0  cos x + cos x  = 0 cos x cos x = 0 This factorises at once into (cos x( cos x = 0. cos x = 0 when x = or x = . cos x =, or cos x = 0.5, when x = 3 or x = 3. The solutions to the equation are therefore x = and 3 (illustrated below. It is important not to simply cancel out cos x from the equation, because there are solutions where cos x = 0, i.e. where x =. (Graph for illustration only students are not expected to sketch it!
12 Mathematics Revision Guides Further Trigonometric Identities and Equations Page of 7 Example (5: Solve the equation cos x + sin x + = 0 for  x. This equation can be turned into a quadratic in sin x by using the identity cos x =  sin x. This gives  sin x + sin x + = 0, or  sin x + sin x + = 0. Using the general formula, we substitute x = sin x, a = , b = and c =. x b b 4ac a sin 7 4 x. The two solutions are sin x =.8 or sin x = to 3 d.p. The positive value can be rejected at once, since no angle has a sine greater than. The principal value of the other result is c and it falls within the range  x Using the fact that sin (x = sin x, another possible solution is c, but it falls outside the range. Since the sine function has a period of the other solution in the range is ( or.46 c.
13 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 3 of 7 Example (6: Solve the equation sec = 4 cosec for 0 < <, giving results in radians to 3 decimal places. 4 sin sec = 4 cosec 4 cos sin cos tan 4. The principal solution is =.36 c, but because the tangent function has a period of, another solution is (.36 + c or c. Example (7: Given that sin 3 = 3 sin  4 sin 3 (see Example 6, use this result to solve, for 0 < < 90, the equation 3 sin 6 cosec = 4. (Copyright OCR, GCE Mathematics Paper 473, Jan. 006., Q.9 part iii. Firstly we substitute =to obtain3 sin 3 cosec = 4, for 0 < < 80 3sin 3 3 sin 3 cosec = 4 4 sin (cosec is reciprocal of sin 3 3(3sin 4sin 4 3(34 sin = 4 sin = 4 sin = 0 sin 5 sin 5 sin, or 40., (There is no need to consider the negative value since sin is positive for 0 < < 80 Finally, we must remember that =so we must divide by to give = 0., Example (8: Use the identity tan( tan 3 60 tan( 60 from Example (3, to solve, 3tan for 0 < < 80, the equation tan( + 60 tan(60 = 4sec 3. (Copyright OCR, GCE Mathematics Paper 473, Jun. 007., Q.9, part (ii. tan 3 4sec 3 3tan tan 3 4(tan 3 3 tan (using sec = + tan tan 3 4 tan 3tan tan 3 (4 tan ( 3tan 4 tan 3 4 tan tan 3tan 4 tan 3tan 4 tan tan 3 0 (bringing RHS over to left tan tan 4 0 tan 4 tan 4 = tan  ( 3 3 = 37., The negative value is outside the range, but must be included since the tangent function repeats every 80, and ( or 4.8 is also within the range. Solutions are = 37., 4.8.
14 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 4 of 7 Solution of trigonometric equations of the form a cos b sin = c. (Harmonic Form Equations of the form above can be rewritten in the form R cos (or R sin (for suitable choices of angles and The four compound angle identities are : R cos( + = R cos cos  R sin sin R cos(  = R cos cos + R sin sin R sin(+ = R sin cos + R cos sin R sin(  = R sin cos  R cos sin Any expression of the form a cos b sin can be rewritten to match the above forms. Example (9: Express 3 cos 4 sinin the form R cos (Give in degrees, and from there solve 3 cos 4 sin.5 for R cos(  = 3 cos 4 sin is the closest match : R cos cos R sin sin 3 cos 4 sin Equating sine and cosine terms we have: R cos cos 3 cos Rcos 3. R sin sin sin R sin 4. Squaring the righthand sides gives R cos = 9 and R sin = 6 R (cos + sin = 5. The bracketed expression is simply, therefore R = When any expression of the form a cos b sin is rewritten in the form R cos (or R sin (then R = a b. To find we see that R sin 4, or that tan = R cos 3 3 cos 4 sin5 cos (53. To solve 5 cos (53..5 for 0 360, first divide by 5 to give cos ( for Two values of (53. satisfying the condition are 60 and 60 ; 3. or 6.9. The second value is out of the required range for but by adding 360 to it we obtain 353., which is within the range. 4, and thus
15 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 5 of 7 Example (0: Express sin  cosin the form R sin (Give in radians. R sin(  = sin  cos is the closest match : R sin cos  R cos sin sin  cos Equating sine and cosine terms we have: R sin cos sin R cos =. R cos sin cos R sin. Hence R = 5 and tan =, and thus.07 c. sin  cos5 sin (.07 c Example (: Express 5 sin cosin the form R sin (Give in radians. R sin( + = 5 sin cos is the closest match. R sin cos R cos sin 5 sin cos R cos 5; R sin R = 5 3 and tan 5 c 5 sin cos3 sin (.76 c Example (: The water temperature ( C of a swimming pool is modelled on the equation W = 7  sin (5t . cos(5t, where t is the time in hours after midnight. (The multiple of 5 enters into the questions because the period of 360 degrees needs modifying into a period of 4 hours: 4 5 = 360. i Express the model equation in the form W = 7  R sin (5t +, where is positive. ii Show that the maximum temperature of the water cannot exceed 30 C according to the model. iii Find the time of day when the water reaches this maximum temperature. iv The management decide that the water temperature should not fall below 6 C during the opening hours of between 0730 and 00. Does the given model satisfy the requirements? i We must rearrange the brackets in the formula first: W = 7  sin (5t . cos(5t W = 7  ( sin (5t +. cos(5t Then continuing in the usual way we have R sin(5t + = sin(5t. cos(5t. R sin(5t cos R cos(5t sin sin(5t. cos(5t. Equating sine and cosine terms we have: R sin(5t cos sin(5t R cos =. R cos(5t sin. cos(5t R sin.. Hence R = ( ( or.973, and tan = the model equation is given by W = sin (5t =., hence ii Since the sine function cannot values outside the range  to, the temperature of the water cannot exceed ( C, or 9.97 C according to this model.
16 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 6 of 7 iii This maximum temperature is reached when sin (5t =  (remember, we re subtracting!, i.e. when 5t = 70 5t =.3 t = 4.8 hours after midnight, or at about 449. iv (A sketch graph is useful here. The temperature of the water is at a maximum at about 500, and so it must take a minimum at about 0300, because this is a sine function with a 4hr period. By sketching, it can be seen that the water temperature exceeds 6 from about 0800 to 00. We must solve the equation sin (5t = sin (5t = sin (5t = sin (5t = (5t = 9.654, t = 8.07,.6 The answers in degrees must finally be converted into hours after midnight by first adding 360 to the negative value, giving 33.93, and then dividing both results by 5. Hence the water temperature = 6 C when t =.3 hours after midnight, or 08, or when t = 7.5 hours after midnight, or 073. By looking at the graph, the water temperature is above 6 between 073 and 08, which practically coincide with the opening hours!
17 Mathematics Revision Guides Further Trigonometric Identities and Equations Page 7 of 7 The angle between two straight lines. Example (3: Two straight lines have the equations x + y 5 = 0 and 3x y  = 0 respectively. What is the acute angle between them? Each of the lines makes a particular angle with the xaxis, measured anticlockwise. Thus is the angle between the line x + y 5 = 0 and the x axis, while is the angle between the line 3x  y  = 0 and the x axis. The gradient of the line x + y  5 = 0 is also the tangent of, and its value is ½. Similarly the gradient of the line 3x y  = 0 is the tangent of, namely 3. The acute angle between the lines is, and, by the triangle laws, = . Taking tangents we have tan = tan (  or tan tan tan tan tan Substituting tan = ½ and tan = 3, we can calculate the acute angle between the two lines using ( 3 3 tan 7 = 8.9 or.49 c. ( (3 Also note that the constant terms in the equations of the lines do not affect the result. Thus the acute angle between the lines x + y  6 = 0 and 3x y + 5 = 0 would also be 8.9 or.49 c. The general formula for the angle between two straight lines (not in scope of syllabus, given their gradients m and m, is tan m m m m. (It is not important which gradient is chosen as m ; if tan comes out negative, then disregarding the sign will still give the acute angle result.
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