(d) sec ( 227 ) u 180 . 360 u. in .NET framework Making QR-Code in .NET framework (d) sec ( 227 ) u 180 . 360 u.

(d) sec ( 227 ) u 180 . 360 u.
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(a) sin 232 sin (232 180 ) sin 52 232 is in quadrant III, so the sine is negative and R (b) cos 312 cos (360 312 ) cos 48 312 is in quadrant IV, so the cosine is positive and R (c) tan 912 tan [192 2(360 )] tan 192 tan (192 180 ) tan 12 Since 912 360 , we find the coterminal angle first. 192 is in quadrant III, so the tangent is positive and R
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(d) sec ( 227 )
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sec (133 360 ) sec 133 sec (180 133 ) sec 47 Since 227 < 0 , we find a coterminal angle first. 133 is in quadrant II, so the secant is negative and R
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When finding the value of a trigonometric function by using a calculator, a reference angle is unnecessary. The function value is found as indicated in Sec. 4.5. However, when an angle having a given function value is to be found and that angle is to be in a specific quadrant, a reference angle is usually needed even when using a calculator.
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6.4 Angles with a Given Function Value
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Since coterminal angles have the same value for a function, there are an unlimited number of angles that have the same value for a trigonometric function. Even when we restrict the angles to the interval of 0 to 360 , there are usually two angles that have the same function value. All the angles that have the same function value also have the same reference angle. The quadrants for the angle are determined by the sign of the function value. The relationships from Sec. 6.3 are used to find the angle u, once the reference angle is found from a table (see Sec. 4.4) or a calculator (see Sec. 4.6).
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CHAPTER 6 Reduction to Functions of Acute Angles
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EXAMPLE 6.5 Find all angles
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between 0 and 360 when: 0.3256, (c) tan 1.2799
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(a) sin u
0.6293,
(b) cos
(a) Since sin 0.6293 is positive, solutions for u are in quadrants I and II because the sine is positive in these quadrants. sin R 0.6293; thus R 39 . In quadrant I, R , so 39 . In quadrant II, R 180 , so 180 R 180 39 141 . 39 and 141 . (b) Since cos 0.3256 is negative, solutions for are in quadrants II and III because the cosine is negative in these quadrants. cos R 0.3256; thus R 71 . In quadrant II, R 180 , so 180 R 180 71 109 . In quadrant III, R 180 , so 180 R 180 71 251 . 109 and 251 . 1.2799 is negative, solutions for u are in quadrants II and IV because the tangent is negative in (c) Since tan these quadrants. tan R 1.2799; thus R 52 . In quadrant II, R 180 , so 180 R 180 52 128 . In quadrant IV, R 360 , so 360 R 360 52 308 . 128 and 308 .
EXAMPLE 6.6 Find all angles u when (a) sin u
0.2079 and (b) tan u
(a) sin R 0.2079; thus R 12 . The sine is negative in quadrants III and IV. In quadrant III, 180 R 180 12 192 . In quadrant IV, 360 R 360 12 348 . All angles coterminal with these values of are needed, so 192 n360 and 348 integer. (b) tan R 0.5543; thus R 29 . The tangent is positive in quadrants I and III. In quadrant I, R 29 . In quadrant III, 180 R 180 29 209 . All angles coterminal with these values of are needed, so 29 n360 and 209 integer.
n360 where n is any
n360 where n is any
SOLVED PROBLEMS
6.1 Derive formulas for the functions of
in terms of the functions of .
In Fig. 6.1, and are constructed in standard position and are numerically equal. On their respective terminal sides the points P(x, y) and P1(x1, y1) are located so that OP OP1. In each of the figures, the two triangles y. Then are congruent and r1 r, x1 x, and y1 sin ( u) cos ( u) tan ( u) y1 r1 x1 r1 y1 x1 x r y x y r y r cosu y x tan u sin u cot ( u) sec ( u) csc ( u) x1 y1 r1 x1 r1 y1 r x r y x y x y sec u r y csc u cot u