PRELIMINARY ALGEBRA
drawn throughR, the point (0,sin(A+B)) in theOxysystem. That all the angles
marked with the symbol•are equal toAfollows from the simple geometry of
right-angled triangles and crossing lines.
We now determine the coordinates ofP in terms of lengths in the figure,expressing those lengths in terms of both sets of coordinates:
(i) cosB=x′=TN+NP=MR+NP=ORsinA+RPcosA= sin(A+B)sinA+cos(A+B)cosA;(ii) sinB=y′=OM−TM=OM−NR
=ORcosA−RPsinA= sin(A+B)cosA−cos(A+B)sinA.Now, if equation (i) is multiplied by sinAand added to equation (ii) multiplied
by cosA, the result is
sinAcosB+cosAsinB= sin(A+B)(sin^2 A+cos^2 A)=sin(A+B).Similarly, if equation (ii) is multiplied by sinAand subtracted from equation (i)
multiplied by cosA, the result is
cosAcosB−sinAsinB=cos(A+B)(cos^2 A+sin^2 A)=cos(A+B).Corresponding graphically based results can be derived for the sines and cosines
of the difference of two angles; however, they are more easily obtained by setting
Bto−Bin the previous results and remembering that sinBbecomes−sinB
whilst cosBis unchanged. The four results may be summarised by
sin(A±B)=sinAcosB±cosAsinB (1.18)cos(A±B)=cosAcosB∓sinAsinB. (1.19)Standard results can be deduced from these by setting one of the two anglesequal toπor toπ/2:
sin(π−θ)=sinθ, cos(π−θ)=−cosθ, (1.20)
sin( 1
2 π−θ)
=cosθ, cos( 1
2 π−θ)
=sinθ, (1.21)From these basic results many more can be derived. An immediate deduction,obtained by taking the ratio of the two equations (1.18) and (1.19) and then
dividing both the numerator and denominator of this ratio by cosAcosB,is
tan(A±B)=tanA±tanB
1 ∓tanAtanB. (1.22)
One application of this result is a test for whether two lines on a graphare orthogonal (perpendicular); more generally, it determines the angle between
them. The standard notation for a straight-line graph isy=mx+c,inwhichm
is the slope of the graph andcis its intercept on they-axis. It should be noted
that the slopemis also the tangent of the angle the line makes with thex-axis.