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Section 6.5 Notes Page 1 6.5 Double Angle and Half Angle Formulas If we have either a double angle 2θ or a half angle θ 2 then these have special formulas: sin( 2θ ) = 2 sin θ cos θ cos(2θ ) = cos 2 θ − sin 2 θ cos(2θ ) = 2 cos 2 θ − 1 cos(2θ ) = 1 − 2 sin 2 θ tan( 2θ ) = sin θ cos tan 2 θ 2 θ 2 There are three formulas for cos( 2θ ) 2 tan θ 1 − tan 2 θ =± 1 − cos θ 2 You will choose plus or minus depending on what quadrant =± 1 + cos θ 2 You will choose plus or minus depending on what quadrant =± 1 − cos θ 1 + cosθ You will choose plus or minus depending on what quadrant There are better formulas for tan tan θ 2 = sin θ , 1 + cos θ tan θ 2 = θ 2 θ 2 θ 2 θ 2 is. is. is. that don’t involve a plus or minus: 1 − cos θ sin θ EXAMPLE: Compute sin θ , tan θ , cscθ , secθ , cot θ , sin( 2θ ) , cos( 2θ ) , tan( 2θ ) , sin θ if you are given cosθ = −0.8 and 90 ≤ θ ≤ 180 . Round decimals to two decimal places. o 2 , cos θ 2 , and tan θ 2 o 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 θ = . 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 θ -0.8 Section 6.5 Notes Page 2 Now we can get our first 5 trigonometric functions by reading off our triangle: sin θ = 0.6 tan θ = csc θ = 0 .6 = −0.75 − 0 .8 1 = 1.67 0 .6 cot θ = sec θ = 1 = −1.25 − 0 .8 − 0 .8 = −1.33 0 .6 Next I will find sin( 2θ ) by using its formula: sin( 2θ ) = 2 sin θ cos θ . We already know sine and cosine, so we will substitute in those decimals: sin( 2θ ) = 2(0.6)( −0.8) = −0.96 . Next I will find cos( 2θ ) by using its formula: cos(2θ ) = 2 cos 2 θ − 1 . Notice I had a choice of three formulas to use. Any of them would give you the correct answer. We already know cosθ = −0.8 , so we will substitute in this decimal: cos(2θ ) = 2(−0.8) 2 − 1 = 0.28 . 2 tan θ . We already know tan θ = −0.75 , so we 1 − tan 2 θ − 1 .5 2(−0.75) will substitute in this decimal: tan( 2θ ) = = = −3.43 . 2 0.4375 1 − (−0.75) Next I will find tan( 2θ ) by using its formula: tan( 2θ ) = For the half angle formulas I need to determine which quadrant θ 2 is in. To do this let’s first start with our given statement 90 o ≤ θ ≤ 180 o . If I divide everything by two we get: 45 o ≤ in the first quadrant, so sine, cosine, and tangent of 2 2 ≤ 90 o . This tells us that θ 2 is should all be positive. 1 − cosθ θ . We chose a positive because is in the 2 2 2 2 second quadrant. We already know cosθ = −0.8 , so we will substitute in this decimal: θ 1 − (−0.8) 1.8 sin = = = .9 = 0.95 . 2 2 2 Now I will find sin θ by using its formula: sin θ θ θ 1 + cos θ θ . We chose a positive because is in the 2 2 2 2 second quadrant. We already know cosθ = −0.8 , so we will substitute in this decimal: θ 1 + (−0.8) 0.2 cos = = = .1 = 0.32 . 2 2 2 Now I will find cos θ by using its formula: cos = sin θ . We already know 2 2 1 + cos θ θ 0.6 cosθ = −0.8 , and sin θ = 0.6 so we will substitute in these decimals: tan = = 3. 2 1 + (−0.8) Finally I will find tan θ θ = . I have three formulas to choose. I will choose: tan θ = Section 6.5 Notes Page 3 EXAMPLE: Compute sin θ , tan θ , cscθ , secθ , cot θ , sin( 2θ ) , cos( 2θ ) , tan( 2θ ) , sin θ 2 , cos θ 2 , and tan θ 2 1 if you are given cot θ = and 180 o ≤ θ ≤ 270 o . 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 θ -2 5 Now we can get our first 5 trigonometric functions by reading off our triangle: sin θ = − 2 csc θ = − 5 2 5 =− 2 5 5 cos θ = − 1 5 =− 5 5 secθ = − 5 tan θ = 2 Next I will find sin( 2θ ) by using its formula: sin( 2θ ) = 2 sin θ cos θ . We already know sine and cosine, so we ⎛ 2 5 ⎞⎛ 5 ⎞ 20 4 ⎟= ⎟⎜ − will substitute in those fractions: sin( 2θ ) = 2⎜⎜ − ⎟ 25 = 5 . ⎜ ⎟ 5 5 ⎠ ⎠⎝ ⎝ Next I will find cos( 2θ ) by using its formula: cos(2θ ) = 2 cos 2 θ − 1 . Notice I had a choice of three formulas to use. Any of them would give you the correct answer. We already know cos θ = − 5 , so we will substitute 5 2 ⎛ 5⎞ ⎟ − 1 . This simplifies to: 2⎛⎜ 5 ⎞⎟ − 1 = 2 − 1 = − 3 . in this fraction: cos(2θ ) = 2⎜⎜ − ⎟ 5 5 ⎝ 25 ⎠ ⎝ 5 ⎠ Next I will find tan( 2θ ) by using its formula: tan( 2θ ) = substitute in this number: tan( 2θ ) = 2( 2) 4 =− . 2 3 1 − ( 2) 2 tan θ . We already know tan θ = 2 , so we will 1 − tan 2 θ For the half angle formulas I need to determine which quadrant Section 6.5 Notes Page 4 θ 2 is in. To do this let’s first start with our given statement 180 o ≤ θ ≤ 270 o . If I divide everything by two we get: 90 o ≤ is in the second quadrant, so sine of θ 2 θ 2 ≤ 135 o . This tells us that should be positive, and the cosine and tangent of 2 2 should be negative. 1 − cosθ θ . We chose a positive because is in the 2 2 2 2 second quadrant. We already know cosθ = −0.8 , so we will substitute in this decimal: Now I will find sin θ by using its formula: sin ⎛ 5⎞ ⎟ 1 − ⎜⎜ − 5 ⎟⎠ θ ⎝ sin = = 2 2 θ θ θ = 5+ 5 5+ 5 5 = . 2 10 1 + cosθ θ . We chose a negative because is in the 2 2 2 2 second quadrant. We already know cosθ = −0.8 , so we will substitute in this decimal: Now I will find cos θ by using its formula: cos θ =− ⎛ 5⎞ 5− 5 ⎟ 1 + ⎜⎜ − ⎟ 5 θ 5− 5 ⎝ ⎠ =− 5 . cos = − =− 2 2 2 10 Finally I will find tan θ . I have three formulas to choose. I will choose: tan θ = sin θ . We already know 1 + cos θ 2 2 cosθ = −0.8 , and sin θ = 0.6 so we will substitute in these decimals: 2 5 2 5 − − θ 5 5 = − 2 5 ⋅ 5 = − 2 5 ⋅ 5 + 5 = − 10 5 − 10 = − 10 5 − 10 = − 5 − 1 . tan = = 2 5 5− 5 25 − 5 20 2 ⎛ 5− 5 5+ 5 5 ⎞ 5− 5 ⎟ 1 + ⎜⎜ − ⎟ 5 ⎝ 5 ⎠ EXAMPLE: Compute sin(22.5 o ) and tan(22.5 o ) using a half-angle formula. ⎛ 45 o ⎞ 1 − cos 45 o θ ⎟⎟ . Then we know that θ is 45 degrees, so now we use: sin = We can write this as sin ⎜⎜ . It is 2 2 ⎝ 2 ⎠ positive since 22.5 is in the first quadrant. You get: sin 45 o sin 45 o = = For tan(22.5 o ) we will use: tan 2 1 + cos 45 o θ 2 2 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 EXAMPLE: Establish the identity: cot θ − tan θ = cos 2θ cot θ + tan θ Section 6.5 Notes Page 5 First we want to change these into sines and cosines. cos θ sin θ − sin θ cos θ = cos 2θ cos θ sin θ + sin θ cos θ Now get common denominators on the top and bottom. ⎛ cos θ ⎞ cos θ sin θ ⎛ sin θ ⎞ − ⋅⎜ ⎜ ⎟⋅ ⎟ ⎝ cos θ ⎠ sin θ cos θ ⎝ sin θ ⎠ = cos 2θ ⎛ cos θ ⎞ cos θ sin θ ⎛ sin θ ⎞ + ⋅⎜ ⎜ ⎟⋅ ⎟ ⎝ cos θ ⎠ sin θ cos θ ⎝ sin θ ⎠ Multiply and write over a single denominator cos 2 θ − sin 2 θ cos θ sin θ = cos 2θ cos 2 θ + sin 2 θ cos θ sin θ We will get rid of the double fractions and use cos 2 θ + sin 2 θ = 1 cos 2 θ − sin 2 θ = cos 2θ The left side is the identity for cos 2θ . cos 2θ = cos 2θ EXAMPLE: (4 sin θ cos θ )(1 − 2 sin 2 θ ) = sin 4θ First we will use the identity 1 − 2 cos 2 θ = cos 2θ . Now our problem becomes: (4 sin θ cosθ ) cos 2θ = sin 4θ We know that sin 2θ = 2 sin θ cosθ . We want to rewrite our problem as the following: 2 ⋅ 2 sin θ cosθ ⋅ cos 2θ = sin 4θ Now we can use the identity sin 2θ = 2 sin θ cosθ . 2 sin 2θ cos 2θ = sin 4θ We know that 2 sin 2θ cos 2θ is the same as sin (2 ⋅ 2θ ) = sin 4θ . sin 4θ = sin 4θ