Download presentation

1
COMPOUND ANGLE FORMULAE

2
**Consider the expression: sin (A + B).**

Firstly note that sin (A + B) ≠ sin A + sin B ( This can easily be shown, e.g. let A = 30° and B = 60°). It can be shown that the compound angle formula for sin (A + B) is: . . . (1) sin (A + B) = sinA cosB + cosA sinB By letting B = – B in (1) sin (A – B) = sinA cos(–B) + cosA sin(–B) Since: sin(–B) = – sinB cos(–B) = cosB . . . (2) sin (A – B) = sinA cosB – cosA sinB

3
From (1): sin (A + B) = sinA cosB + cosA sinB Let A = 90 – A: sin (90 – A + B) = sin(90 – A) cosB + cos(90 –A) sinB sin (90 – (A – B)) = sin(90 – A) cosB + cos(90 –A) sinB sin(90 – A) = cosA cos(90 – A) = sinA Since: . . . (3) cos (A – B) = cosA cosB + sinA sinB Let B = – B in (3): cos (A + B) = cosA cos(–B) + sinA sin(–B) . . . (4) cos (A + B) = cosA cosB – sinA sinB

4
**sin (A + B) = sinA cosB + cosA sinB . . . (1)**

cos (A + B) = cosA cosB – sinA sinB (4) Divide (1) by (4): tan (A + B) = sinA cosB + cosA sinB cosA cosB – sinA sinB Divide each term by cosA cosB tan (A + B) = sinA cosB cosA sinB cosA cosB sinA sinB cosA cosB – + . . . (5) tan (A + B) = tanA + tanB 1 – tanA tanB . . . (6) tan (A – B) = tanA – tanB 1 + tanA tanB Now let B = – B:

5
**Example 1: Simplify sin(x + 90°).**

sin (A + B) = sinA cosB + cosA sinB Using: sin (x + 90) = sin x cos 90 + cos x sin 90 = sin x ( 0 ) + cos x ( 1 ) = cos x Example 2: Find in surd form cos15°. cos (A – B) = cosA cosB + sinA sinB Using: Let A = 45° and B = 30° cos (45 – 30) = cos45 cos30 + sin45 sin30 = 2 2 + 2 1 2 4 = + 4 + 4 =

6
**Example 3: Show that, in surd form tan75° = 2 +**

tan (A + B) = tanA + tanB 1 – tanA tanB Using: = 1 1 + 1 – × tan ( ) = tan45 + tan30 1 – tan45 tan30 Multiply each term by: = + 1 – 1 × + 1 – 1 = (Rationalise here) = 3 + 2 + 1 3 – 1 = 4 + 2 2 = 2 +

7
**Example 4: Solve the equation sin (x + 30°) = 2 cos (x + 60°) ; **

for 0 < x < 360°. sin (x + 30°) = 2 cos (x + 60°) sin x cos 30 + cos x sin 30 = 2{ cos x cos 60 – sin x sin 60 } 2 1 2 1 2 2 } sin x + cos x cos x = 2 { sin x – Multiply by 2: sin x cos x + = 2 cos x sin x – 2 = 3 sin x cos x Now, TanA is positive in the 1st and 3rd quadrants: 3 1 tan x = α = tan-1 3 1 α = 10.89° x = 10.9°, ° (1d.p.)

8
**Example 5: Prove the identity cot (A + B) = cot A + cot B **

tan (A + B) 1 tan (A + B) = tanA + tanB 1 – tanA tanB cot (A + B) = LHS = tan A + tan B 1 – tan A tan B = Divide each term by tan A tan B tan A tan B 1 = – 1 tan B 1 tan A 1 + cot B + cot A cot A cot B – 1 = = RHS

9
**The compound angle formulae are:**

Summary of key points: The compound angle formulae are: sin (A ± B) = sinA cosB ± cosA sinB cos (A ± B) = cosA cosB sinA sinB tan (A ± B) = tanA ± tanB 1 tanA tanB The following results also need to be known: sin(–B) = – sinB cos(–B) = cosB sin(90 – A) = cosA cos(90 – A) = sinA This PowerPoint produced by R.Collins ; Updated Mar. 2010

Similar presentations

OK

Bilborough College Maths - core 4 double angle formulae (Adrian)

Bilborough College Maths - core 4 double angle formulae (Adrian)

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt online downloader youtube Ppt on programmable logic array programmer Ppt on indian constitution for class 8 Ppt on the art of war by sun Ppt on coffee shop business plan Ppt on genetically modified crops Ppt on supreme court of india Ppt on electronic media Ppt on different forms of agriculture Ppt on entertainment industry in india