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There will be two solutions from the two values of $ +B given in (2)

Find from the tables the value of cos ß. Next find from the tables the magnitude of the angle a whose sine = cos B, and we get

sin(+B) = sina,
..$+B=n7 +(-1)" a (Art. 38)

.: p=-B+ Na + (-1)", where n is zero or any positive or negative integer.

In order that the solution may be possible, it is necessary to have cos B=, or < 1.

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(4)

NOTE. This example might have been solved by squaring both sides of the equation; but in solving trigonometric equations, it is important, if possible, to avoid squaring both sides of the equation. Thus, solve COB = k sine

(3) If we square both sides we get

cos2 0 = k· sin2 0 = k-(1 - cos2 ).
K2

k
... cog2 = ; or cos = +
1+ k2

11 + K2

k Now if a be the least angle whose cosine =

we get from (4)
V1 + KP
в = пп +а

(5) But (3) may be written

cotA=k.
..A= n + a

(6) (6) is the complete solution of the given equation (3), while (5) is the solution of both cos 0 = k sin 0, and also of cos 0 = - k sin 8. Therefore by squaring both members of an equation we obtain solutions which do not belong to the given equation.

EXAMPLES.

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1. Solve 0.7466898 sin 1.0498 cos 0

0.431689, when < 180°.

log b= 0.02112-*

log a = 1.87314

log tan ß= 0.14798.: ß= 125° 25' 20".

* The minus sign is written thus to denote that it belongs to the natural number and does not affect the logarithm. Sometimes the letter n is written instead of the minus sign, to denote the same thing.

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..$+B=19° 34' 40" or 160° 25' 20".
... φ

– 105° 50' 40" or 35° 0' 0".
2. Solve - 23.8 sin 6 + 19.3 cos = 17.5(<180°).
- 0 $
Ans. 4° 12'.7 or

- 106° 7'.9.

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4

6

3. Solve 2 sin + 2 cos 0 =V2. Ans. – +na+(-1)" 0

+17.. sin A + V3 cos 0 = 1.

+na+(-1)"

1

4.

4.

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85. Solve the equation

sin (a + x) = m sinx . in which « and m are given.

(1)

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which determines x + 10, and therefore x.

If we introduce an auxiliary angle, the calculation of equation (3) is facilitated.

Thus, let m=tan ; then we have by [(14) of Art. 61] m +1 tan $ +1

= cot ($ – 45°),

tan 6 -1 which in (3) gives

m-1

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tan $ = m,

This, with gives the logarithmic solution. The logarithmic solution of the equation

sin (a — x) = m sin x is found in the same manner to be

tan = m,

OC

tan/ x

and

=cot ($ + 45°) tan

=

2 which the student may show.

(

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Example. -Solve

sin (106° + x)=-1.263 sin x(x<180°). log tan = log m= log (-1.263)= 0.10140 -.

.::$=128° 22.3. $- 45°= 83° 22.3; log cot (-45°)= 9.06523 10= 53° 0'.0,

log tan fa= 10.12289

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From (1) we have tan(x + x)

+ tan x m +1 tan(a + a)-tana m 1

sin(x+ 2x)

[(21) of Art. 61] sino

m +1 .. sin(a + 2x) =

(2) -1 = cot($ - 459) sina . (Art. 85)

° where

tan =m.

sino

m

S

Example. - Solve tan (23° 16' + x)=.296 tan x. log tan = log m= log (296)= 1.47129.

::$= 16° 29'.3. 0 - 45°= -28° 30'.7; log cot(- 45°) = 10.26502a=23° 16'.1,

log sina= 9.59661 log sin(a + 2x)= 9.86163

) a + 2x = 226° 38'.9 or 313° 21'.1.

=

::: x=

- 101° 41'.5 or 145° 2'.6.

(1)

87. Solve the equation

tan(a + x). tan x = m in which a and m are given.

From (1) we have 1+tan (a + 2) tan x

1+m 1- tan(a + æ) tan x 1

m

COS C

(Ex. 4 of Art. 47)

cos (a + 2x)

1 m
.. cos(a + 2) = COS OC

1+m

tan(45° — 6) cosa [(16) of Art. 61] Example. - Solve tan(65° + x) tan x = 1.5196(a < 180°). log tan $ = log m = 0.18173.

where

tan •

= m.

.: p= 50° 39' 9''; 45° – $=-11° 39' 9"; log tan(45° -6)=9.31434 a= 65° 0' 0";

log cos Q = 9.62595

log cos(a + 2x)= 8.94029a + 2x = 95° or 265o. .:. 2x= 30° or 200°.

X = 15° or 100°.

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.

88. Solve the equations m sin (0 + x)= a

(1) msin (+2=b.

(2) for m and x, the other four quantities, 0, 0, a, b, being known. Expanding (1) and (2) by (Art. 44), we get m sin 0 cos x + m cos o sin æ= a

(3) m sin $ cos x + m cos sin x = = b.

(4) Multiplying (3) by sin 6 and (4) by sin 0, and subtracting the latter from the former, we have

m sin x (sin $ cos 0 cos o sin 6)= a sin $ b sin 0. ... m sin x=

a sin -b sin 0

sin(-6) To find the value of mcos x, multiply (3) and (4) by cos and cos 0, respectively, and subtract the former from the latter. Thus m cos x(sin $ cos 0 — cos o sin 6)=b cos 0 a cos 0. b cos 0 a cos •

(6) sin(0 - 0) Having obtained the values of m sin x and m cos x from (5) and (6), m and x can be calculated by Art. 83.

.. m cOS X =

.

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