REDUCTION FORMULA. Learning Outcomes and Assessment Standards

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Lesson 5 REDUCTION FORMULA Learning Outcomes and Assessment Standards Learning Outcome 3: Space, shape and measurement Assessment Standard Derive the reduction formulae for: sin(90 o ± α), cos(90 o ±

1 Lesson 5 REDUCTION FORMULA Learning Outcomes and Assessment Standards Learning Outcome 3: Space, shape and measurement Assessment Standard Derive the reduction formulae for: sin(90 o ± α), cos(90 o ± α) sin(80 o ± α) cos(80 o ± α) tan(80 o ± α) sin(360 o ± α) cos(360 o ± α) tan(360 o ± α) sin(-α) cos(-α) tan(-α) Overview In this lesson you will: Learn to reduce all angles to co-terminal angles in the first quadrant. Simplify trigonometric expressions by writing ratios in terms of sin α and cos α. Prove more trigonometric identities by examining the left-hand side and the right-hand side. DVD Lesson The horizontal reduction formulae: Sin 90 (80 α) (α < 90 ) 80 (80 + α) (360 α) Tan 270 Let s try some: All Cos Here we look at angles in terms of the horizontal line 80 /360. Remember that the CAST rule still applies in the quadrants. So every angle will be reduced by this horizontal reduction formulae to an angle that lies in the first quadrant. We do this by looking at the CAST rule, and the size of the angle. sin25 (25 lies in the second quadrant) = sin (80 55 ) (the horizontal reduction formula in the 2nd Q) = sin 55 (since the CAST rule says that sin is positive in 2nd Q) cos (80 + q) (80 + q lies in the 3rd Q and cos is negative here) = cos q(80 q) (80 q in 2nd Q; tan negative here) = tan q sin (80 q) (80 q in 2nd Q; sin is positive here) = sin q 60 Thinking of negative angles: How do we measure the angle (a 80 )? Instead of learning them by rote, let us unpack them visually.

2 We know positive angles are measured anti-clockwise, and negative angles are measured clockwise. So a 80 will be: a = anti-clockwise, then 80 becomes 80 clockwise. This tells us we are in the 3rd quadrant. Here tan is positive! So sin (a 80 ) = sin a and cos (a 80 ) = cos a tan (a 80 ) = tan a Let s try more: tan 24 (24 is in 3rd Q: by horizontal reduction) = tan ( ) (In 3rd Q tan is positive) = tan 34 sin(a 360 ) which is in the first Q = sin a tan 2 ( a) in the 4th Q: tan is negative = ( tan a) 2 anything that is squared is positive = tan 2 a cos ( a + 80 ) 2nd Q; cos is negative = cos a sin ( a 80 ) 2nd Q; sin is positive = sin a tan ( 50 ) 3rd Q ; reduction (80 + a); ; tan is negative ; 30 away from 80 = tan 30 Example Simplify the following cos(80 + θ)sin(θ 80 ) cos( θ)tan(80 + θ) We look at one factor at a time to make sense of each one () cos (80 + θ) (80 + θ in 3rd Q here cos is negative) = cos θ 6

3 (2) sin (θ 80 ) 3rd Q where sin is negative = sin θ 62

4 (3) cos ( q) 4th Q where cos is positive = cos q (4) tan (80 + θ) 3rd Q where tan is positive So: cos(80 + θ) sin (q 80 ) = ( cos q)( sin q) cos( q) tan (80 + θ) (cos q)(tan q) = sin q (tan q = sin q cos q ) sin q cos q = cos q Now do Activity no A. REMEMBER Ask yourself: Which quadrant am I in? Example 2 Prove cos x sin x = 2sin 2 (80 + x) 2tan(80 + x) + 2sin( x)cos x RHS 2sin 2 x 2tan x + 2( sin x)(cos x) = 2sin2 x cos x 2sin xcos x ) ( 2sin x = 2sin2 x 2sin x 2sin xcos2 x cos x = 2sin2 x cos x 2sin x( cos 2 x) = 2sin2 x cos x 2sin x sin 2 x = cos x sin x = LHS Now do Activity no B. DVD Vertical Reduction formulae As the name suggests, we reduce angles in terms of the vertical line. The CAST rule still applies here, but we now have to work with the complementary ratios. Here is how they work: In ABC, C= ^ 90 has been given. So A ^ + B ^ = 90 since all angles in a triangle add to 80. We call A ^ and Bcomplements ^ of one another, or we say they are complimentary angles. Also notice that sin α = b = cos (90 a) c sin α = cos (90 a) and cos a = a = sin(90 α) cos a = sin(90 α) c Outside of the triangle we will see that for angles expressed as a vertical reduction: sine becomes cosine and cosine becomes sine So sin cos 63

5 Let s try some: sin(90 o + α) ((90 + a) in 2nd Q: sin is positive here; because of the 90 sin becomes cos.) = cos α cos(90 o + α) ((90 + a) in 2nd Q: cos is negative here; cos becomes sin because of 90.) = sin a cos(α 90 o ) 4th Quadrant; cos is positive here and cos becomes sin = sin α cos( 90 α) in the 3rd Q: cos is negative and cos becomes sin = sin α sin( a 90 o ) 3rd Q: sun is negative and sin cos = cos α Example 3 Simplify tan(80 + x)cos(90 x) cos(80 x) sin(90 x) sin(90 + x) Again: tan (80 + x) [3rd Q: tan positive] = tan x cos (90 x) [st Q: cos positive; cos sin] = sin x sin (90 x) [st Q: sin positive; sin cos] = cos x cos (80 x) [2nd Q: cos negative] = cos x sin (90 + x) [2nd Q: sin positive; sin cos] = cos x (tan x)(sin x) = ( cos x) cos x cos x = cos sin x x sin x cos x + ( tan x = sin x cos x ) = sin2 x + cos 2 x cos 2 x = cos 2 x (sin2 x + cos 2 x = ) 64

6 Example 4 Prove that cos2 (90 x) + sin 2 (90 + x) tan(80 + x) sin(x 90 ) = sin x LHS sin2 x + cos 2 x (tan x)( cos x) = ( sin x cos x ( cos x) ) = sin x = RHS Now do Activity 2 nos A and B. More complementary angles If α + β = 90 o, we say α and β are complementary angles we know sin α = cos (90 o α) so sin 20 o = cos 70 o cos 40 o = sin 50 o cos 0 and = and sin 20 sin 80 cos 70 = DVD Example 5 If α and β are complementary angles and sin α = 3, find tan β. 5 Step : sin α = 3 5 But α and β are complementary angles so cos b = 3 5. Step 2: Draw α 5 β 3 y = 3 r = 5 By Pythagoras x = 4 So tan β = 4 3 Example 6 If sin 50 = p, find in terms of p a) cos 40 b) cos 50 65

7 Using a diagram sin 50 = p Using identities (a) cos 40 = cos (90 50 ) = sin 50 According to Pythagoras x = p 2 So: (a) cos 40 = p = p (b) cos 50 = p 2 (b) cos 50 : cos sin 2 50 = cos 2 50 = sin 2 50 cos 50 = p 2 Now do Activity 3. Activity Individual summative assessment A. Simplify: ) sin2 (80 + A) cos (80 + A) 2) sin(360 q)sin( 80 q) cos(80 q) cos( q)tan(80 q)sinq 3) tan( q)cos(80 q) cos 2 ( q) sin(80 q) 4) sin2 (80 + q) + cos 2 ( q) tan(360 q)cos( q) 5) tana sin(80 A) + sin 2 ( A cos(80 + A) 6) tan(80 + q) + sin( q) ( + cos(80 + q) 7) sin(80 q)sin q cos2 (360 q) tan(80 + q) + tan( q) B. Prove the following identities: ) sin(80 A) + sina = 2 cos 2 A 2) 2 cos(80 x) sin( x) = 2 tan(80 + x) + tan 2 ( x) 3) ( cos q tan q ) 2 = + sin(360 q) sin(q 80 ) 4) cos(360 q) tan(80 q) = cos(360 q) + sin( q) Activity 2 66 A. Simplify ) sin(90 q)(360 q)cos(90 + q) tan( q)cos(360 q)tan(80 + q) 2) cos(90 + a)cos( a)sin( a) sin(a 90 )tan(360 a)cos a 3) cos(90 θ) tanθ cos( θ) cos( θ)sin(θ 90 )

An angle in the Cartesian plane is in standard position if its vertex lies at the origin and its initial arm lies on the positive x-axis.

An angle in the Cartesian plane is in standard position if its vertex lies at the origin and its initial arm lies on the positive x-axis. Learning Goals 1. To understand what standard position represents. 2. To understand what a principal and related acute angle are. 3. To understand that positive angles are measured by a counter-clockwise