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Compound Angle Formulae 1. Addition Formulae cos( A B) cos A cos B sin 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) sin A cos B cos A sin B Example: sin( x 30)o sin x cos30 cos x sin 30 3 sin x cos x 2 2. Formulae Involving Double Angle (2A) sin 2 A 2sin A cos A cos 2 A cos 2 A sin 2 A 2cos 2 A 1 1 2sin 2 A Two further formulae derived from the cos 2 A formulae. cos 2 A 12 (1 cos 2 A) sin 2 A 12 (1 cos 2 A) Mixed Examples: 4 Given that A is an acute angle and tan A , calculate sin 2 A and cos 2 A. 3 sin A 4 cos A 3 sin 2 A cos 2 A 1 sin 2 A ( 43 sin A)2 1 sin A Similarly: sin 2 A 2sin A cos A cos A 16 25 3 5 4 5 Substitute form the tan (sin/cos) equation +ve because A is acute 3-4-5 triangle !!! 24 25 cos 2 A cos 2 A sin 2 A 9 16 25 7 25 A is greater than 45 degrees – hence 2A is greater than 90 degrees. Find the exact value of sin 75o. 2 sin(75o ) sin(45 30) sin(75o ) sin 45cos30 cos 45sin 30 45 1 o 1 Prove that sin( ) tan tan cos cos sin( ) sin cos cos sin cos cos cos cos sin sin cos cos tan tan 30o 1 1 3 1 1 1 3 2 2 22 2 2 2 Q.E.D. 3 For the diagram opposite show that cos LMN 5 . 5 M cos LMN cos( ) 3 2 Length of LM 18 3 2 Length of MN 10 1 3 1 1 2 10 2 10 2 2 20 4 5 5 5 10 3 3 cos( ) cos cos sin sin 1 5 L 1 N (A Higher Question) Show that, for the triangle ABC in the diagram, a a b c sin a sin b sin c b sin . cos( ) The sine rule b From the diagram: C a b sin sin( [ 2 ]) a b sin( 2 [ ]) The sum of the angles of a triangle=180 sin( 2 ) cos b cos( ) b sin cos( ) As required A 2 c a B Prove that, cos 4 sin 4 cos 2 . x 2 y 2 ( x y )( x y ) cos 4 sin 4 (cos 2 ) 2 (sin 2 ) 2 (cos 2 sin 2 )(cos 2 sin 2 ) cos 2 sin 2 1 cos 2 sin 2 cos( ) cos cos sin sin cos 2 TRIGONOMETRIC EQUATIONS Double angle formulae (like cos2A or sin2A) often occur in trig equations. We can solve these equations by substituting the expressions derived in the previous sections. Use sin2A = 2sinAcosA when replacing sin2A cos2A = 2cos2A – 1 cos2A = 1 – 2sin2A if cosA is also in the equation if sinA is also in the equation when replacing cos2A Solve: cos 2 x o 4sin x o 5 0 for 0 x 360o. cos2x and sin x, so substitute 1-2sin2 (1 2sin 2 x) 4sin x 5 0 6 4sin x 2sin 2 x 0 cp. w. 6 4 z 2 z 2 0 (6 2sin x)(1 sin x) 0 sin x 1 or sin x 3 x 90o 0 sin x 1 for all real angles Solve: 5cos 2 x o cos x o 2 for 0 x 360o. cos 2x and cos x, so substitute 2cos2 -1 5(2cos 2 x 1) cos x 2 10cos 2 x cos x 3 0 (5cos x 3)(2cos x 1) 0 3 1 cos x or cos x 5 2 x 51.3 or x 360 51.3 o 308.7o s a t c x 90 60 150o or x 270 60 210 o All S_ Talk C*&p ?? 2 y 0.6 y0 y 0.5 3 2 The diagram shows the graphs of f ( x) a sin bx o and g ( x) c sin x o for 0 x 360o. y y 4 y f ( x) 2 360o 0 -2 xx y g ( x) -4 Three problems concerning this graph follow. i) State the values of a, b and c. y y 4 f ( x) a sin bx o The max & min values of asinbx are 3 and -3 resp. The max & min values of sinbx are 1 and -1 resp. a3 2 g ( x) c sin x o The max & min values of csinx are 2 and -2 resp. c2 360o 0 -2 -4 f(x) goes through 2 complete cycles from 0 – 360o b2 y f ( x) y g ( x) xx ii) Solve the equation f ( x) g ( x) algebraically. From the previous problem we now have: f ( x) 3sin 2 x and g ( x) 2sin x Hence, the equation to solve is: 3sin 2 x 2sin x Expand sin 2x 3(2sin x cos x) 2sin x 6sin x cos x 2sin x 0 Divide both sides by 2 3sin x cos x sin x 0 Spot the common factor in the terms? sin x(3cos x 1) 0 Is satisfied by all values of x for which: sin x 0 or cos x 1 3 iii) find the coordinates of the points of intersection of the graphs for 0 x 360o. From the previous problem we have: sin x 0 or sin x 0 cos x cos x 1 3 Hence: x 0o or x 70.5o x x 180o 360o or x x (360 70.5)o 289.5o or 1 3