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Thus the formula has been extended to the case where one of the angles is obtuse and less than 180°. similar way the formula for cos(x+y) is extended to this

case.

By continuing this method both formulas are proved to be true for all positive values of x and y.

Any negative angle y is equal to a positive angle y', minus some multiple of 360°. The functions of y are equal to those of y', and the functions of (x+y) are equal to those of (x+y'). 89

Therefore, the formulas being true for (x+y'), are true for (x+y).

A repetition of this reasoning shows that the formulas are true when both angles, x and y, are negative.

33. Substituting the angle -y for y in formula (11), it becomes

sin(x-y)=sin x cos (− y) + cos x sin (—y). But cos (−1) = cos y, and sin (—y)= — sin y. Therefore, sin (x − y) = sin x cos y − cosx sin y. Substituting (y) for y in formula (13), it becomes cos(x-y)= cos x cos (―y)—sin x sin (—y),

= cos x cosy + sin x siny.

Therefore, cos (x − y) = cos x cos y + sin x sin y.*

823

(12)

(14)

EXERCISES

34. (1.) Prove geometrically where x and y are acute and positive: sin(x−y)=sin x cosy— cosx siny,

cos(xy)cos x cosy + sin r siny.

* Formulas (12) and (14) are proved geometrically in § 34. The geometric proof is complicated by the fact that OD and DP are functions of -y, while the functions of y are what we use.

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Hint.—Angle AOQ=x, angle POQ=y, and angle AOP=(x−y).
Draw PD perpendicular to OQ.

Then DP=sin(—y)=—siny; but DP is negative, therefore PD taken as positive is equal to sin y:

OD=cos(-y)=cosy.

Angle HPD-angle AOQ=x. their sides being perpendicular.
Draw DH perpendicular to SP, DE perpendicular to OA.

sin(x-1)=SP=ED-PH.

From right triangle OED, ED=(sin x) × OD=sin x cos y.
From right triangle DHP, PH=(cos x) × PD=cosx sin y.
Therefore, sin (xy)=sin x cos y-cos x sin y.

Cos(x-1)=OS=OE+DH.

From right triangle OÊD,

OE=(cos x) × OD=cos x cosy.

From right triangle DHP, DH=(sin x) × PD=sin x sin y.
Therefore,

cos (x-1)=cos x cos y + sin x sin y.

(2.) Find the sine and cosine of (45°+ x), (30°— x), (60°+ x), in terms of sin r and cos x.

(3.) Given sin x=3, sin y=f3, x and y acute; find sin (x+y) and sin (x —y).

(4.) Find the sine and cosine of 75° from the functions of 30° and 45°. Hint.

75°=(45°+30°).

(5.) Find the sine and cosine of 15° from the functions of 30° and 45°. (6.) Given x and y, each in the second quadrant, sin x = = }, siny = 1; find sin(x+y) and cos (x —y).

(7.) By means of the above formulas express the sine and cosine of (180° - x), (180°+x), (270°— x), (270°+x), in terms of sin x and cos r. (8.) Prove sin (60°+45°)+cos (60°+45°) = cos 45°.

(9.) Given sin 45° = √√2, cos 45° = √2; find sin 90° and cos 90o. (10.) Prove that_sin (60° + x)— sin (60° — x) = sin x.

TANGENT OF THE SUM AND DIFFERENCE OF TWO ANGLES

35. Tan(x+y)=

=

sin (r+y) _ sint cosy+cost sin y
cos(x+y) cosx cos y − sin x sing

Dividing each term of both numerator and denominator of the right-hand side of this equation by cosa cosy, and

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In a similar way, dividing formula (12) by formula (14), we

obtain

tana - tany

tan (x − y):

1+tanæ tany

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FUNCTIONS OF TWICE AN ANGLE

36. An important special case of formulas (11), (13), and (15) is when y=x; we then obtain the functions of 2x in terms of the functions of x.

From (II), sin(r+r)=sinx cosx+cosr sin r.

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FUNCTIONS OF HALF AN ANGLE

37. Equations (19) and (20) are true for any angle; therefore for the angle r.

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FORMULAS FOR SUMS AND DIFFERENCES OF FUNCTIONS

38. From formulas (11)–(14), we obtain

Let

then

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sin (x+y)—sin (x − y)=2 cosx siny;
cos(x+y)+cos(x-1)=2 cosx cosy;
cos(x+y)-co's (x-y)=-2 sinx siny.
u=(x+y) and v=(x−y);

x=(u+v), y={(u —v).

Substituting in the above equations, we obtain sinu + sinv = 2 sin√(u+v) cos} (u — v);

sinu – sinv = 2 cos) (u +v) sin1⁄2 (u — v);

(25)

(26)

cosu+cosv=2 cos1⁄2 (u+v) cos1⁄2 (u – v) ;
cosu—cosv=−2 sin}(u+v) sin (u — v).

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(28)

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39. Express in terms of functions of x, by means of the formulas

of this chapter,

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(5.) The sine and cosine of (45° —x); of (45° +x).

(6.) Given tan 45° = 1, tan 30o =- § √3; find tan 75°; tan 15o.

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(9.) Prove cos (30°+y) — cos (30° — y) = — siny.

(10.) Prove sin 3x= 3 sin x - 4 sin3x.

Hint.-Sin 3x=sin(x+2x).

(11.) Prove cos 3x=4 cos3 x

3 cos x.

(12.) If x and y are acute and tan x=1, tan y=, prove that (x+y)=45°.

(13.) Prove that tan (x+45°):

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=

I + tan r
Itan x

(14.) Given siny and y acute; find siny, cosy, and tany. (15.) Given cos x = - and x in quadrant II; find sin 2x and

COS 2.x.

(16.) Given cos 45° = √2; find the functions of 2210. (17.) Given tan x = 2 and x acute; find tanr.

(18.) Given cos 30° = √3; find the functions of 15o. (19.) Given cos 90° = 0; find the functions of 45°.

(20.) Find sin 5r in terms of sin x.

(21.) Find cos 5r in terms of cos x.

(22.) Prove sin(x+y+z)=sin x cos y cos z+cos x sin y cos z+cos x cos y sin - sin x siny sin z.

Hint.-Sin (x+y+z)=sin(x+y) cos z+cos(x+y) sin z.

(23.) Given tan 2x=3 tan x; find x.

(24.) Prove sin 32° + sin 28° = cos 2o.

(25.) Prove tan x + cot x = 2 CSC 2x.

(26.) Prove (sin x + cos (27.) Prove (sin x cos

x)2 = 1 + sin x.
x)2 = 1

sin x.

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