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Trigonometric functions and circular measure - the Australian ...

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A guide for teachers – Years 11 <strong>and</strong> 12 • {47}5 cm5 cm30 º150 º4 cm 4 cmExercise 612 ac sinB = 1 2Exercise 7LHS =Exercise 8bc sin A =⇒ a sinB = b sin A =⇒asin A =1secθ + tanθ = 11cosθ + sinθcosθ= cosθ1 + sinθ = RHSbsinBThe double angle formula for cosine gives cos30 ◦ = 1 − 2sin 2 15 ◦ . Sosin 2 15 ◦ = 1 2= 1 2Hence sin15 ◦ = 1 2(1 − cos30◦ )(1 −3)2= 1 4(2 −3).√2 − 3. (Take <strong>the</strong> positive square root, as sin15 ◦ is positive.)An alternative method is to start from sin30 ◦ = 2sin15 ◦ cos15 ◦ . Proceed by squaring bothsides of <strong>the</strong> equation <strong>and</strong> using <strong>the</strong> Pythagorean identity.Exercise 9Using <strong>the</strong> double angle formulas, we havetan2θ = sin2θcos2θ = 2sinθ cosθcos 2 θ − sin 2 θ .If we assume that tanθ is defined, <strong>the</strong>n we can divide <strong>the</strong> numerator <strong>and</strong> denominatorby cos 2 θ to obtaintan2θ =2 sinθcosθ1 − sin2 θcos 2 θ= 2tanθ1 − tan 2 θ .Note that tan2θ is defined if <strong>and</strong> only if cos2θ = cos 2 θ − sin 2 θ ≠ 0, in which case tanθcannot equal ±1.

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