Topic name Homework Sheet 123 Name



WorkSHEET 6.2 Trigonometric equations Name: ___________________________

| |State the Pythagorean identity. |sin2x + cos2x = 1 |3 |

| |Write the Pythagorean identity with sin2x as the subject. |sin2x = 1 ( cos2x | |

| |Write the Pythagorean identity with cos2x as the subject. |cos2x = 1 ( sin2x | |

| | | | |

| |If sin [pic] = 0.7, and 0o < [pic] < 90o, find, correct to three |[pic] |4 |

| |decimal places: | | |

| |[pic] |[pic] | |

| | | | |

| |[pic] | | |

| |Find all possible values of sin x if cos x = 0.25. |sin2x = 1 ( cos2x |3 |

| | |= 1 ( (0.25)2 | |

| | |= 1 (0.0625 | |

| | |= 0.9375 | |

| | |sin x = ([pic] | |

| | |x = 0.968 or (0.968 | |

| | | | |

| |Find the exact value of sin x if cos x = [pic] and x is in the fourth |[pic] |3 |

| |quadrant. | | |

| | |Third side of triangle = [pic] | |

| | |From triangle sin x = [pic] but x is in the fourth quadrant, so sin x | |

| | |= ([pic]. | |

| | | | |

| |Find ao if 0o < [pic] < 90o and |sin ao = cos 67o |2 |

| |sin ao = cos 67o |= sin (90 – 67)o | |

| | |= sin 23o | |

| |cos ao = sin 8o |ao = 23o | |

| | | | |

| | |cos ao = sin 8o | |

| | |= cos (90 – 8)o | |

| | |= cos 82o | |

| | |ao = 82o | |

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| |If 0o ( a ( 90o and sin ao = [pic], find the exact value of : |[pic] |4 |

| |cos ao | | |

| |tan ao |Third side of triangle = [pic] | |

| |sin (90 – a)o |cos ao = [pic] | |

| |cos (180 + a)o |tan ao = [pic] | |

| | |sin (90 – a)o = cos ao | |

| | |= [pic] | |

| | |cos (180 + a)o = (cos ao | |

| | |= ([pic] | |

| | | | |

| |Solve the equation 2 sin2x = sin x over the domain 0 ( x ( 2[pic] | 2 sin2x = sin x |4 |

| | |2 sin2x ( sin x = 0 | |

| | |sin x (2 sin x ( 1) = 0 | |

| | |sin x = 0 2 sin x ( 1 = 0 | |

| | |x = 0, [pic] 2[pic] 2 sin x = 1 | |

| | |sin x = [pic] | |

| | |x = [pic], [pic] | |

| | |Solution: x = 0, [pic], (, [pic], 2[pic] | |

| |Solve the equation 2cos2x + [pic]cos x = 0 for the domain 0 ( x ( | 2cos2x + [pic]cos x = 0 |4 |

| |2[pic] |cos x (2 cos x + [pic]) = 0 | |

| | |cos x = 0 2 cos x + [pic] = 0 | |

| | |x = [pic], [pic] 2 cos x = ( [pic] | |

| | |cos x = [pic] | |

| | |x = [pic], [pic] | |

| | |Solution: x = [pic], [pic], [pic] or [pic] | |

| | | | |

| |Solve 2 sin2x = 3 sin x ( 1 in the domain 0 ( x ( 2[pic] | 2 sin2x = 3 sin x ( 1 |4 |

| | |2 sin2x ( 3 sin x + 1 = 0 | |

| | |(2 sin x ( 1)(sin x (1) = 0 | |

| | |2 sin x ( 1 = 0 sin x ( 1 = 0 | |

| | |2 sin x = 1 sin x = 1 | |

| | |sin x = [pic] x = [pic], [pic] | |

| | |x = [pic], [pic] | |

| | |Solution: x = [pic], [pic], [pic] or [pic] | |

| | | | |

| |Solve 2sin2x = 2 ( [pic]cos x in the domain 0 ( x ( 2[pic] | 2sin2x = 2 ( [pic]cos x |4 |

| | |2(1 ( cos2x) = 2 ( [pic]cos x | |

| | |2cos2x ( [pic]cos x = 0 | |

| | |cos x (2cos x ( [pic]) = 0 | |

| | |cos x = 0 2 cos x ( [pic] = 0 | |

| | |x = [pic], [pic] 2 cos x = [pic] | |

| | |cos x = [pic] | |

| | |x = [pic], [pic] | |

| | |Solution: x = [pic], [pic], [pic] or [pic] | |

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