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Section 5.3 Notes Page 1 5.3 Double Angle and Half Angle Formulas If we have either a double angle 2 or a half angle 2 then these have special formulas: sin(2 ) 2 sin cos cos(2 ) cos 2 sin 2 cos(2 ) 2 cos 2 1 cos(2 ) 1 2 sin 2 tan(2 ) sin cos tan 2 2 2 There are three formulas for cos(2 ) 2 tan 1 tan 2 1 cos 2 You will choose plus or minus depending on what quadrant 1 cos 2 You will choose plus or minus depending on what quadrant 1 cos 1 cos You will choose plus or minus depending on what quadrant There are better formulas for tan tan 2 sin , 1 cos tan 2 2 2 2 2 is. is. is. that don’t involve a plus or minus: 1 cos sin EXAMPLE: Compute sin , tan , csc , sec , cot , sin(2 ) , cos(2 ) , tan(2 ) , sin 2 , cos 2 , and tan 2 if you are given cos 0.8 and 90 180 . Round decimals to two decimal places. We need to draw a triangle for this one. We are given that the triangle should be drawn in the second quadrant. 0.8 We can rewrite our problem as cos . We know the adjacent side is –0.8. The hypotenuse is 1. Once 1 we label our triangle we find the third side by using the Pythagorean theorem. 1 0.6 -0.8 Section 5.3 Notes Page 2 Now we can get our first 5 trigonometric functions by reading off our triangle: csc sin 0.6 tan 0.6 0.75 0.8 1 1.67 0.6 cot sec 1 1.25 0.8 0.8 1.33 0.6 Next I will find sin(2 ) by using its formula: sin(2 ) 2 sin cos . We already know sine and cosine, so we will substitute in those decimals: sin(2 ) 2(0.6)(0.8) 0.96 . Next I will find cos(2 ) by using its formula: cos(2 ) 2 cos 2 1 . Notice I had a choice of three formulas to use. Any of them would give you the correct answer. We already know cos 0.8 , so we will substitute in this decimal: cos(2 ) 2(0.8) 2 1 0.28 . 2 tan . We already know tan 0.75 , so we 1 tan 2 2(0.75) 1.5 will substitute in this decimal: tan(2 ) 3.43 . 2 0.4375 1 (0.75) Next I will find tan(2 ) by using its formula: tan(2 ) For the half angle formulas I need to determine which quadrant 2 is in. To do this let’s first start with our given statement 90 180 . If I divide everything by two we get: 45 in the first quadrant, so sine, cosine, and tangent of 2 2 90 . This tells us that 2 is should all be positive. 1 cos . We chose a positive because is in the 2 2 2 2 second quadrant. We already know cos 0.8 , so we will substitute in this decimal: 1 (0.8) 1.8 sin .9 0.95 . 2 2 2 Now I will find sin 1 cos . We chose a positive because is in the 2 2 2 2 second quadrant. We already know cos 0.8 , so we will substitute in this decimal: 1 (0.8) 0.2 cos .1 0.32 . 2 2 2 Now I will find cos by using its formula: sin by using its formula: cos sin . We already know 2 2 1 cos 0.6 cos 0.8 , and sin 0.6 so we will substitute in these decimals: tan 3. 2 1 (0.8) Finally I will find tan . I have three formulas to choose. I will choose: tan Section 5.3 Notes Page 3 EXAMPLE: Compute sin , tan , csc , sec , cot , sin(2 ) , cos(2 ) , tan(2 ) , sin if you are given cot 2 , cos 2 , and tan 2 1 and 180 270 . 2 We need to draw a triangle first. We are given that the triangle should be drawn in the third quadrant. We know the adjacent side is 1 and the opposite side is 2. However because we are in the third quadrant we need to make the 1 and 2 negative. Once we label our triangle we find the third side by using the Pythagorean theorem. This will give us 5 . -1 -2 5 Now we can get our first 5 trigonometric functions by reading off our triangle: sin 2 csc 5 2 5 2 5 5 cos 1 5 5 5 sec 5 tan 2 Next I will find sin(2 ) by using its formula: sin(2 ) 2 sin cos . We already know sine and cosine, so we 2 5 5 20 4 . will substitute in those fractions: sin(2 ) 2 5 5 25 5 Next I will find cos(2 ) by using its formula: cos(2 ) 2 cos 2 1 . Notice I had a choice of three formulas to use. Any of them would give you the correct answer. We already know cos 5 , so we will substitute 5 2 2 3 5 5 1 . This simplifies to: 2 1 1 . in this fraction: cos(2 ) 2 5 5 25 5 Next I will find tan(2 ) by using its formula: tan(2 ) substitute in this number: tan(2 ) 2(2) 4 . 2 3 1 (2) 2 tan . We already know tan 2 , so we will 1 tan 2 For the half angle formulas I need to determine which quadrant 2 Section 5.3 Notes Page 4 is in. To do this let’s first start with our given statement 180 270 . If I divide everything by two we get: 90 is in the second quadrant, so sine of 2 2 135 . This tells us that should be positive, and the cosine and tangent of 2 2 should be negative. 1 cos . We chose a positive because is in the 2 2 2 2 second quadrant. We already know cos 0.8 , so we will substitute in this decimal: Now I will find sin by using its formula: sin 5 1 5 sin 2 2 5 5 5 5 5 . 2 10 1 cos . We chose a negative because is in the 2 2 2 2 second quadrant. We already know cos 0.8 , so we will substitute in this decimal: Now I will find cos by using its formula: cos 5 5 5 1 5 5 5 5 . cos 2 2 2 10 Finally I will find tan . I have three formulas to choose. I will choose: tan sin . We already know 1 cos 2 2 cos 0.8 , and sin 0.6 so we will substitute in these decimals: 2 5 2 5 5 2 5 5 2 5 5 5 10 5 10 10 5 10 5 1 . 5 tan 5 5 5 25 5 20 2 2 5 5 5 5 5 5 5 1 5 5 EXAMPLE: Compute sin(22.5 ) and tan(22.5 ) using a half-angle formula. 45 1 cos 45 . Then we know that is 45 degrees, so now we use: sin . It We can write this as sin 2 2 2 is positive since 22.5 is in the first quadrant. You get: sin 45 sin 45 For tan(22.5 ) we will use: tan 2 1 cos 45 2 1 2 2 2 2 2 2 2 2 2 4 2 2 . 2 2 2 2 2 2 2 2 2 2 1. 2 2 2 2 2 2 2 2 1 2 2 2 2 Section 5.3 Notes Page 5 EXAMPLE: Rewrite 1 2 sin 2 12 using a double angle formula. Then find its exact value. This one is written in the form 1 2 sin 2 , which is a form of cos 2 . In this problem, we will let 12 . Therefore we will substitute this into the formula cos 2 : cos 2 . Simplifying gives us cos . Looking 6 12 at our table we see this has an exact value of 3 2 . EXAMPLE: Establish the identity: cot tan cos 2 cot tan First we want to change these into sines and cosines. cos sin sin cos cos 2 cos sin sin cos Now get common denominators on the top and bottom. cos cos sin sin cos sin cos sin cos 2 cos cos sin sin cos sin cos sin Multiply and write over a single denominator cos 2 sin 2 cos sin cos 2 cos 2 sin 2 cos sin We will get rid of the double fractions and use cos 2 sin 2 1 cos 2 sin 2 cos 2 The left side is the identity for cos 2 . cos 2 cos 2 EXAMPLE: 4 sin cos 1 2 sin 2 sin 4 First we will use the identity 1 2 sin 2 cos 2 . Now our problem becomes: 4 sin cos cos 2 sin 4 We know that sin 2 2 sin cos . We want to rewrite our problem as the following: 2 2 sin cos cos 2 sin 4 Now we can use the identity sin 2 2 sin cos . 2 sin 2 cos 2 sin 4 We know that 2 sin 2 cos 2 is the same as sin 2 2 sin 4 . sin 4 sin 4