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COMPOUND ANGLE FORMULAE

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**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

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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

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**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:

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**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 =

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**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 +

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**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.)

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**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

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**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

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