HIGHER MATHEMATICS Unit 2 - Outcome 3 Trigonometry

cos( ) = cos(105) = cos(A + B) = cosAcosB - sinAsinB cos60 ⁰ cos45 ⁰ - sin60 ⁰ sin45 ⁰ )( 1 2 x 1 22 () - 33 2 x 1 22 1 2222 - 33 2222 cos( ) = 1 -  3 2222 x 22 22  2 -  6 4 Using the fact that 105 = Show that the exact value of cos 105 ⁰  2 -  6 4 is. Ex.3 Question 1

Acute angles X and Y are such that sinX = 8 17 and tanY = 3 4, show that cos(X + Y) = X Y Ex.3 Question 2

cos(X + Y) = cosY = 4 5 cosX = sinY = 3 5 sinX = 8 17 cos(X + Y) = cosXcosY - sinXsinY ( x ) ( 8 x ) 3 5 cos(X + Y) = cos(X + Y) = cos(X + Y) = 36 85

Acute angles X and Y are such that sinX = 1 55 and tanY = 3. Find the exact value of sin(X – Y). X Y 1 5  10 Ex.3 Question 3

cosY = 4  10 cosX = 2 55 sinY = 3  10 sinX = 1 55 sin(X - Y) = sinXcosY - cosXsinY sin(X - Y) = ( 1 55 x ) 1  10 - ( 2 55 x ) 3 sin(X - Y) = 1  sin(X - Y) = -5  50 sin (X – Y) ? 22 = = -5 5252

cosx ⁰ cos70 ⁰ – sinx ⁰ sin70 ⁰ = 0.45 Solve the equation cosx ⁰ cos70 ⁰ – sinx ⁰ sin70 ⁰ = 0.45 for 0 ⁰ ≤ x ≤ 360 ⁰. cos (0.45) = 63.3 ⁰ cos(x + 70) = 0.45 C AS T (x + 70) =63.3 ⁰, ⁰, ⁰, ⁰ x = -6.7 ⁰, ⁰, ⁰, ⁰ x = ⁰, ⁰ Ex.3 Question 4

= 2sinAcosA A 1  15 4 Given that 0 ⁰ < A < 90 ⁰, and that tanA = 1  15 find the exact values of sin2A, cos2A and sin4A. cosA = 1515 4 sinA = 1 4 sin2A = 2  ( 1 4 )  15 4 ) ( 2 sin2A =  15 8 = Ex.3 Question 5

A 1  15 4 Given that 0 ⁰ < A < 90 ⁰, and that tanA = 1  15 find the exact values of sin2A, cos2A and sin4A. cosA = 1515 4 sinA = 1 4 = cos²A – sin²A cos2A = ( 1515 4 ) 2 ) 1 4 ( 2 – – cos2A = 7 8 = Ex.3 Question 5 cont.

Given that 0 ⁰ < A < 90 ⁰, and that tanA = 1  15 find the exact values of sin2A, cos2A and sin4A. cos2A = 7 8 = sin2A =  15 8 = = 2sin2Acos2A sin4A = (  15 8 ) 7 8 ) ( 2 sin4A = 14   = Ex.3 Question 5 cont.

sin2x – cosx = 0 Solve the equation sin2x – cosx = 0 for 0 ≤ A ≤ 2.  2sinxcosx – cosx = 0 cosx(2sinx – 1) = 0 sin2x cosx = 0 or sinx = ½ Ex.3 Question 6

C AS T A = 90 ⁰ or 270 ⁰ A =  2 33 2 or C AS T A = 30 ⁰ or 150 ⁰ A =  6 55 6 or A =,,,  6  2 55 6 33 2 cosx = 0 sinx = ½

(2cosx + 1)(cosx + 3) = 0 cos2x + 7cosx = -4 Solve the equation cos2x + 7cosx = -4 for 0 ≤ x ≤ 2.  2cos²x cosx + 4 = 0 or cosx = -3 cos2x + 7cosx + 4 = 0 2cos²x + 7cosx + 3 = 0 cosx = ½ - cos2x 2X² + 7X + 3 = 0 (2X + 1)(X + 3) = 0 Ex.3 Question 7

C AS T x = 120 ⁰ or 240 ⁰ x = 22 3 44 3 or x =, 22 3 44 3 cosx = -3 cosx = ½ - NO SOLUTIONS cos (½) = 60 ⁰

3cos2x + sinx - 2 = 0 Solve the equation 3cos2x + sinx - 2 = 0 for 0 ⁰ ≤ x ≤ 360 ⁰. 3 – 6sin²x + sinx - 2 = 0 3(1 – 2sin²x) + sinx - 2 = 0 -6sin²x + sinx + 1 = 0 cos2x (3sinx + 1)(2sinx - 1) = 0 6sin²x - sinx - 1 = 0 6X² - X - 1 = 0 (3X + 1)(2X - 1) = 0 sinx = - ⅓ or sinx = ½ n Ex.3 Question 8

C AS T x = ⁰ or ⁰ C AS T x = 30 ⁰ or 150 ⁰ x = 30 ⁰, 150 ⁰, ⁰, ⁰ sinx = - ⅓ sinx = ½ sin ( ⅓ ) = 19.5 ⁰