Trigonometric functions and circular measure - the Australian ...

{26} • Trigonometric functions and circular measureSolution1 cos75 ◦ = cos(45 ◦ + 30 ◦ )= cos45 ◦ cos30 ◦ − sin45 ◦ sin30 ◦= 1 32 2 − 1 12 2= 1 4(6 −2)2 sin75 ◦ = sin(45 ◦ + 30 ◦ )= sin45 ◦ cos30 ◦ + cos45 ◦ sin30 ◦= 1 32 2 + 1 12 2= 1 4(6 +2)3 cos105 ◦ = cos(45 ◦ + 60 ◦ )= cos45 ◦ cos60 ◦ − sin45 ◦ sin60 ◦= 1 12 2 − 1 32 2= 1 4(2 −6)(Note that we could obtain cos105 ◦ directly from cos75 ◦ , since the two angles aresupplementary.)We can also find expansions for tan(A + B) and tan(A − B). Recalling that tanθ = sinθcosθ ,we can writesin(A + B)tan(A + B) =cos(A + B)=sin A cosB + cos A sinBcos A cosB − sin A sinB .Dividing the numerator and denominator by cos A cosB, we obtaintan(A + B) =1 −sin Acos A + sinBcosBsin A sinBcos A cosBSince tan(−B) = −tanB, we havetan(A − B) ==tan A − tanB1 + tan A tanB .Note carefully the pattern with the signs.tan A + tanB1 − tan A tanB .