Trigonometric functions and circular measure - the Australian ...

A guide for teachers – Years 11 and 12 • {23}Exercise 8Find sin15 ◦ in surd form.From the Pythagorean identity and the double angle formula for cosine, we havecos 2 θ + sin 2 θ = 1cos 2 θ − sin 2 θ = cos2θ.Adding these equations and dividing by 2 we obtain cos 2 θ = 1 2(cos2θ + 1), while subtractingthem and dividing by 2 we obtain sin 2 θ = 1 2(1 − cos2θ). These formulas are veryimportant in integral calculus, as discussed in the module The calculus of trigonometricfunctions.Exercise 9Use the double angle formulas for sine and cosine to show thattan2θ =2tanθ1 − tan 2 , for tanθ ≠ ±1.θSummary of double angle formulassin2θ = 2sinθ cosθcos2θ = cos 2 θ − sin 2 θ= 2cos 2 θ − 1= 1 − 2sin 2 θtan2θ =2tanθ1 − tan 2 , for tanθ ≠ ±1θTrigonometric functions of compound anglesIn the previous subsection, we derived formulas for the trigonometric functions of doubleangles. That derivation, which used triangles, was only valid for a limited range ofangles, although the formulas remain true for all angles.In this subsection we find the expansion formulas for sin(A + B), sin(A − B), cos(A + B)and cos(A − B), which are valid for all A and B. The double angle formulas can be recoveredby putting A = B = θ.These formulas are quite central in trigonometry. In the module The calculus of trigonometricfunctions, they are used to find, among other things, the derivative of the sinefunction.