31. If A be less than 45°, or half a right angle, cos A is greater than sin A. (Fig. Art. 27.)

Let NAC be less than 45o.

Then, since LNAC+ LNCA = 90°, 4 NCA is greater than 45°. And in every triangle the greater side is opposite to the greater angle, (Eucl. 1. 19.);

Similarly, for angles between 45° and 90o, it may be shewn that the cosine is less than the sine.

32. To find the sines, cosines, and tangents, of 45o, 30°, and 60o.

(2) Let ABC be an equilateral and equiangular triangle; therefore each of its angles is of two right angles, or 60o.

Let AD be perpendicular to BC;

B

.. BD = DC = 1BC = AB; and ▲ BAD = ▲ DAC = 30°;

33. The relations established in Art. 27, between different functions of the same angle, will frequently enable us to solve equations of the following kind.

Ex. 1. From the equation, (sin A) + 5 (cos A) = 3, required the value of sin A.

Since (cos A)2 = 1 − (sin A)3, the equation becomes

(sin 4)2 + 5 . {1 − (sin A)°} = 3 ;

.. 4 (sin 4) = 2, and sin A =

Ex. 2.

If x

be

(sin 4)2 = a cos A+ b, to determine cos A.

put

for cos A, (sin A)2 = 1 − x2, and 1 − x2 = av + b will be the equation to be solved.

from which, x and y may be determined.

m = cosec A − sin A

Ex. 4.

If

n = sec A cos A

4}, required

required the relation

which exists between m and n; that is, required to find an equation between m and n, in which no function of the angle A shall enter.

CHAPTER III.

GONIOMETRICAL FORMULÆ INVOLVING MORE THAN ONE ANGLE.

34. GIVEN the sines and cosines of two angles, to find the sines and cosines of the angles equal to their sum and to their difference.

BAC, CAD are two angles containing Ao and Bo respectively.

From D, any point in AD, draw DB, DC perpendiculars on AB and AC; and from C draw CE, CF perpendiculars on AB and DB.

Then FE is a rectangle; FB = CE, and FC BE.

=

D

B

▲ CDF = 90o — ▲ DCF = 4 FCA = A, since CF is parallel to AE.