« AnteriorContinuar »
3. sin (wc + y — z)= sin x cos y cosz + cos x siny cos z
cos x cos y sin z + sin æ siny sinz.
4. sin x + siny - sinz - sin (x + y - x)
= 4 sin }(x - 2) sin(y – z) sin }(x+y).
5. sin (y – z) + sin (z — «) + sin (x — y)
+ 4 sin }(y-2) sin }(2 — 2) sin } (x - y)=0.
49. Functions of Double Angles. — To express the trigonometric functions of the angle 2x in terms of those of the angle x.
Put y=x in (1) of Art. 42, and it becomes
sin 2x = sin x cos x + cos x sinx,
2 sin a COS X
Put y=x in (2) of Art. 42, and it becomes
Put y=æ in (1) and (3) of Art. 47, and they become
tan 2 x
2 tan x
cota -1 cot 2x =
Transposing 1 in (4), and dividing it into (1), we have
NOTE. – These seven formulæ are very important. The student must notice that x is any angle, and therefore these formulæ will be true whatever we put for x. Thus, if we write for x, we get
(8) 2 2
1 Given sin 45° ; find tan 221. Ans. V2 -1.
24 24 Given tan x=%; find tan 2 x, and sin 2 x.
50. To Express the Functions of 3x in Terms of the Functions of x.
Put y=2x in (1) of Art. 42, and it becomes
sin 3x =sin (2x + x)
= sin 2x cos x + cos2x sin x
= 2 sin x cos” x + (1 – 2 sin’x) sin x (Art. 49)
51. Functions of Half an Angle. - To express the functions of in terms of the functions of x.
By formulæ (3), (4), and (5) the functions of half an angle may be found when the cosine of the whole angle is given.
52. If the Cosine of an Angle be given, the Sine and the Cosine of its Half are each Two-Valued.
By Art. 51, each value of cos x (nothing else being known about the angle x) gives two values each for sin, and cos
2' one positive and one negative. But if the value of x be
= COS X.
given, we know the quadrant in which lies, and hence
2 we know which sign is to be taken. Thus, if « lies between 0° and 360°, lies between Coand °
2 180°, and therefore sin is positive ; but if æ lies between °,
2 360° and 720', lies between 180° and 360°, and hence °
2 sin is negative. Also, if « lie between 0° and 180°, cos
2 is positive; but if a lie between 180° and 360°, cos" is negative.
The case may be investigated geometrically thus :
Through M draw PQ perpendicular to OA; and draw OP, OQ. Then all angles whose cosines are equal to cos a are terminated either by OP or OQ, and the halves of these Á angles are terminated by the dotted lines Op, Oq, Or, or Os. The sines of angles ending at Op and Oq are the same, and equal numerically to those of angles ending at Or and Os; but in the former case they are positive, and in the latter, negative; hence we obtain two, and only two, values of sin from a given value
2 of cos X.
Also, the cosines of angles ending at Op and Os are the same, and have the positive sign. They are equal numerically to the cosines of the angles ending at Oq and Or, but the latter are negative; hence we obtain two, and only two, values of cos from a given value of cos x.
Also, the tangent of half the angle whose cosine is given is two-valued. This follows immediately from (5) of Art.