2DTrigAddnFormula

toc = Compound Angle Formulas =

NOTE: Maths Methods students are not expected to have the compound angle formulas memorised (and therefore they will not be used in Methods Exam 1). They are expected to be able to use the formulas in Exam 2.

NOTE: Specialist Maths students have the compound angle formulas on their formula sheet.

Adding Angles in Trigonometry
Given any two angles, A and B, it should be noted that

math . \qquad \sin(A +B) \textbf { does NOT equal } \sin(A) + \sin(B) math

The following addition rules apply (they are called the ** Compound Angle Formulae **):

math \\ . \qquad \sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B) \\. \\ . \qquad \sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B) math

and

math \\ . \qquad \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B) \\. \\ . \qquad \cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B) math

and

math \\ . \qquad \tan(A+B) = \dfrac {\tan(A) + \tan (B)}{1 - \tan(A)\tan(B)} \\. \\ . \\ . \qquad \tan(A-B) = \dfrac {\tan(A) - \tan (B)}{1 + \tan(A)\tan(B)} math


 * Note: **
 * The derivation of these rules is explained in the Specialist Maths text.
 * You do not need to be able to derive these rules. (Just use them)

** Example **
Find sin(x + 30º)


 * Solution: **

math \\ . \qquad \text{Using:} \sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B) \\. \\ . \qquad \sin(x+30^\circ) = \sin(x)\cos(30^\circ) + \cos(x)\sin(30^\circ) \\. \\ . \qquad \qquad \qquad \quad = \dfrac{\sqrt{3}}{2} \sin(x) + \dfrac{1}{2} \cos(x) math

Exact Values
We can use the addition formula to get exact values of angles other than 30°, 45° and 60°.

Look for combinations of adding or subtracting these angles.

** Example **
Find the exact value of sin(15°)


 * Solution: **

... ... 15° = 45° – 30° so

math \\ . \qquad \sin(15^\circ) = \sin(45^\circ - 30^\circ) \\. \\ . \qquad \qquad \qquad = \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ) math . math \\ . \qquad \qquad \qquad =\dfrac {1}{\sqrt{2}} \times \dfrac {\sqrt{3}}{2} - \dfrac {1}{\sqrt{2}} \times \dfrac {1}{2} \\ .\\ . \qquad \qquad \qquad =\dfrac {\sqrt{3}}{2\sqrt{2}} - \dfrac {1}{2\sqrt{2}} \\. \\ . \qquad \qquad \qquad =\dfrac {\sqrt{3}-1}{2\sqrt{2}} \qquad \qquad \left\{ \times \dfrac{\sqrt{2}}{\sqrt{2}} \right\} \\. \\ . \qquad \qquad \qquad = \dfrac{ \sqrt{6} - \sqrt{2} }{4} math

Simplifying Trig Expressions
Look for either the left side OR the right side of one of the compound angle formulas

** Example **
Expand and Simplify math . \qquad \sqrt{2} \cos \left( x - \dfrac {\pi}{4} \right) math


 * Solution: **

math \\ . \qquad \sqrt{2} \cos \left( x - \dfrac {\pi}{4} \right) \\. \\ . \qquad =\sqrt{2} \left( \cos(x)\cos \left( \dfrac {\pi}{4} \right) + \sin(x)\sin \left( \dfrac {\pi}{4} \right) \right) \\. \\ . \qquad =\sqrt{2} \left( \cos(x) \left( \dfrac {1}{\sqrt{2}} \right) + \sin(x) \left( \dfrac {1}{\sqrt{2}} \right) \right) \\. \\ . \qquad = \cos (x) + \sin (x) math

Proving Trig Identities
(Specialist Maths only)

Start with the left side and simplify or change it using the trig identites until you end up with the right side.

** Example **
math \text {Prove that } \dfrac {\cos(A-B)}{\cos(A)\sin(B)} = \tan(A) + \cot{B} math

math \\ \text{LHS} = \dfrac {\cos(A-B)}{\cos(A)\sin(B)} \\. \\ . \qquad = \dfrac {\cos(A)\cos(B) + \sin(A)\sin(B)}{\cos(A)\sin(B)} \\. \\ . \qquad = \dfrac {\cos(A)\cos(B)}{\cos(A)\sin(B)} + \dfrac{\sin(A)\sin(B)}{\cos(A)\sin(B)} \\. \\ . \qquad = \dfrac {\cos(B)}{\sin(B)} + \dfrac {\sin(A)}{\cos(A)} \\. \\ . \qquad = \cot(B) + \tan(A) \\. \\ . \qquad = RHS math
 * Solution: **

math \text {So } \dfrac {\cos(A-B)}{\cos(A)\sin(B)} = \tan(A) + \cot{B} \text { as required.} math

It is important to use this formatting when proving an identity.

**See also** www.mathsonline.com.au Y12Extension --> Trigonometry --> Further Trigonometry --> Lesson 2

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