543 5.2 Trigonometric Functions x y 0 –2 2 2 x = –√5 y = 2 r = 3 (–√5, 2) –√5 u 3 2 Figure 25 These have rationalized denominators. 1cos u22 and cos2 u are equivalent forms. Remember both roots. Because y r = 2 3 and r = 3, it follows that y = 2. We must find the value of x. x2 + y2 = r2 Pythagorean theorem x2 + 22 = 32 Substitute. x2 + 4 = 9 Apply exponents. x2 = 5 Subtract 4. x = 25 or x = -25 Square root property: If x2 = k, then x = 2k or x = -2k. Because u is in quadrant II, x must be negative. Choose x = -25 so that the point A -25, 2B is on the terminal side of u. See Figure 25. cos u = x r = -25 3 = - 25 3 sec u = r x = 3 -25 = - 32 5 # 25 25 = - 325 5 tan u = y x = 2 -25 = - 22 5 # 25 25 = - 225 5 cot u = x y = -25 2 = - 25 2 csc u = r y = 3 2 S Now Try Exercise 141. Pythagorean Identities We now derive three new identities. x2 + y2 = r2 Pythagorean theorem x2 r2 + y2 r2 = r2 r2 Divide by r2. ax rb 2 + a y rb 2 = 1 Power rule for exponents; am bm = A a bB m 1cos u22 + 1sin u22 = 1 cos u = x r , sin u = y r sin2 U +cos2 U =1 Apply exponents; commutative property Starting again with x2 + y2 = r2 and dividing through by x2 gives the following. x2 x2 + y2 x2 = r2 x2 Divide by x2. 1 + a y xb 2 = a r xb 2 Power rule for exponents 1 + 1tan u22 = 1sec u22 tan u = y x , sec u = r x tan2 U +1 =sec2 U Apply exponents; commutative property Similarly, dividing through by y2 leads to another identity. 1 +cot2 U =csc2 U These three identities are the Pythagorean identities because the original equation that led to them, x2 + y2 = r2, comes from the Pythagorean theorem.
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