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Trigonometric Identities. Examples Prove that (1 – cos A)(1 + sec A)  sin A tan A L.H.S.(1 – cos A)(1 + sec A) = 1 + sec A – cos A –Cos A sec A = 1.

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Presentation on theme: "Trigonometric Identities. Examples Prove that (1 – cos A)(1 + sec A)  sin A tan A L.H.S.(1 – cos A)(1 + sec A) = 1 + sec A – cos A –Cos A sec A = 1."— Presentation transcript:

1 Trigonometric Identities

2 Examples Prove that (1 – cos A)(1 + sec A)  sin A tan A L.H.S.(1 – cos A)(1 + sec A) = 1 + sec A – cos A –Cos A sec A = 1 + sec A – cos A - 1 = sec A – cos A = sin A tan A = R.H.S.

3 Examples Prove that cot A + tan A  sec A cosec A L.H.S. = R.H.S.

4 Examples R.H.S. = R.H.S.

5 Solving equations Solve 2 tan 2 x – 7 sec x + 8 = 0 for 0  x  360  2 (sec 2 x – 1) – 7 sec x + 8 = 0 2 sec 2 x – 2 – 7 sec x + 8 = 0 2 sec 2 x – 7 sec x + 6 = 0 ( 2 sec x – 3)(sec x – 2)= 0 sec x = 3/2 or sec x = 2 cos x = 2/3 or cos x = ½ x = 48.2 or x = 60 or: x = 360 – 48.2 or x = complete solution: x = 48.2  or 60  or 300  or 

6 Solving equations Solve 2 cos x = cot x for 0  x  360  2 cos x = cos x/ sin x 2 cos x sin x = cos x 2 cos x sin x – cos x= 0 cos x(2 sin x – 1)= 0 cos x = 0 or sin x = ½ cos x = 0  x = 90  or 270  sin x = ½  x = 30  or 330  complete solution: x = 30  or 90  or 270  or 30 

7 Solving equations Solve 3 cot 2 x – 10 cot x + 3 = 0 for 0  x  2  ( 3 cot x - 1)(cot x – 3) = 0 cot x = 1/3 or cot x = 3  tan x = 3 or tan x = 1/3 tan x = 3  x = 1.24 c or 4.39 c tan x = 1/3  x = 0.32 c or 3.46 c complete solution: x = 0.32 c or 1.24 c or 3.46 c or 4.39 c

8 Solving equations Solve 5 cot 2 x – 2 cosec x + 2 = 0 for 0  x  2  5(cosec 2 x – 1) – 2 cosec x + 2 = 0 5cosec 2 x – 5 – 2 cosec x + 2 = 0 5cosec 2 x – 2 cosec x - 3 = 0 sin x = -5/3 not possible or sin x = 1  x =  /2

9 Additional formulae sin (A + B) = sin A cos B + sin B cos A sin (A - B) = sin A cos B - sin B cos A cos (A + B) = cos A cos B - sin A sin B cos (A - B) = cos A cos B + sin A sin B

10 Examples Find the exact value of sin 75  sin (A + B) = sin A cos B + sin B cos A sin ( ) = sin 30 cos 45 + sin 45 cos 30

11 Examples Express cos ( x +  /3) in terms of cos x and sin x cos (A + B) = cos A cos B - sin A sin B cos (x +  /3) = cos x cos  /3 - sin  /3 sin x

12 Examples L.H.S. = R.H.S.

13 Double angle formulae sin (A + B) = sin A cos B + sin B cos A sin (A + A) = sin A cos A + sin A cos A sin 2A = 2 sin A cos A cos (A + B) = cos A cos B - sin A sin B cos (A + A) = cos A cos A- sin A sin A cos (A + A) = cos 2 A - sin 2 A cos 2A = cos 2 A - sin 2 A cos 2A = 2cos 2 A - 1 cos 2A = 1 – 2sin 2 A

14 Double angle formulae

15 Examples Given that cos A = 2/3, find the exact value of cos 2A. cos 2A = 2cos 2 A - 1 Given that sin A = ¼, find the exact value of sin 2A. sin 2A = 2 sin A cos A A 4 1  15

16 Solving equations Solve cos 2A cos A = 0 for 0  x  2  =2 cos 2 A cos A = 0 =2 cos 2 A + 4 cos A + 2= 0 = cos 2 A + 2 cos A + 1 = 0 = (cos A + 1) 2 = 0 = cos A = - 1  A = 

17 Solving equations Solve sin 2A = sin A for -   x   =2sin A cos A = sin A =2 sin A cos A – sin A = 0 = sin A(2 cos A – 1) = 0  sin A = 0 or cos A = ½ sin A = 0  A = -  or 0 or  cos A = ½  A = -  /3 or  /3 Complete solution: A = -  or -  /3 or 0 or  /3 or 

18 Solving equations Solve tan 2A + 5 tan A = 0 for 0  x  2  Complete solution: A= 0.97, 2.27, 4.01, 5.41 c 0,  or 2  tan A = 0  A = 0 or  or 2  7 – 5tan 2 A = 0  tan A =   7/5  A = 0.97, 2.27, 4.01 or 5.41 c

19 Harmonic form If a and b are positive a sin x + b cos x can be written in the form R sin( x +  ) a cos x + b sin x can be written in the form R cos( x -  ) a sin x - b cos x can be written in the form R sin( x -  ) a cos x - b sin x can be written in the form R cos( x +  )

20 Examples Express 3 cos x + 4 sin x in the form R cos( x -  ) R cos( x -  ) = R cos x cos  + R sin x sin  3 cos x + 4 sin x = R cos x cos  + R sin x sin  R cos  = 3 [1] R sin  = 4 [2] [1] 2 + [2] 2 : R 2 sin 2 x + R 2 cos 2 x = R 2 (sin 2 x + cos 2 x ) = R 2 = = 25  R = 5 [2]  [1]: tan  = 4/3   = cos x + 4 sin x = 5 cos( x  )

21 Examples Express 12 cos x + 5 sin x in the form R sin( x +  ) R sin( x +  ) = R sin x cos  + R cos x sin  12 cos x + 5 sin x = R sin x cos  + R cos x sin  R cos  = 12 [1] R sin  = 5 [2] [1] 2 + [2] 2 : R 2 cos 2 x + R 2 sin 2 x = R 2 (cos 2 x + sin 2 x ) = R 2 = = 169  R = 13 [2]  [1]: tan  = 5/12   = cos x + 5 sin x = 13 sin( x  )

22 Examples Express cos x -  3 sin x in the form R cos( x +  ) R cos( x +  ) = R cos x cos  - R sin x sin  cos x -  3 sin x = R cos x cos  - R sin x sin  R cos  = 1 [1] R sin  =  3 [2] [1] 2 + [2] 2 : R 2 cos 2 x + R 2 sin 2 x = (  3 ) 2 R 2 (cos 2 x + sin 2 x ) = R 2 = = 4  R = 2 [2]  [1]: tan  =  3   = 60 cos x +  3 sin x = 2 cos( x + 60  )

23 Solving equations Solve 7 sin x + 3 cos x = 6 for 0  x  2  R sin( x +  ) = R sin x cos  + R cos x sin  7 sin x + 3 cos x = R sin x cos  + R cos x sin  R cos  = 7 [1] R sin  = 3 [2] R 2 =  R = 7.62 [2]  [1]: tan  = 3/7   = c (Radians) 7 sin x + 3 cos x = 7.62 sin( x ) 7.62 sin( x ) = 6  x = sin -1 (6/7.62) x = or x = c or c


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