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How to prove this?
It's not for homework, but I'd really like to get an explanation of this problem:
Prove sin(x-20)+cos(x+70) = 0
Can one of you guys go through this step by step?
2 Answers
- RogueLv 77 years ago
sin(x − 20°) + cos(x + 70°) = 0
given sin(α − β) = sin(α)cos(β) − cos(α)sin(β)
=> sin(x)cos(20°) − cos(x)cos(20°) + cos(x+70°) = 0
given cos(α + β) = cos(α)cos(β) − sin(α)sin(β)
=> sin(x)cos(20°) − cos(x)sin(20°) + cos(x)cos(70°) − sin(x)sin(70°) = 0
=> sin(x)cos(20°) − sin(x)sin(70°) + cos(x)cos(70°) − cos(x)sin(20°) = 0
=> sin(x)(cos(20°) − sin(70°)) + cos(x)(cos(70°) − sin(20°)) = 0
given sin(90° − θ) = cos(θ)
=> sin(x)(cos(20°) − cos(20°)) + cos(x)(cos(70°) − cos(70°)) = 0
=> 0sin(x) + 0cos(x) = 0
=> 0 = 0
QED
- Anonymous7 years ago
sin(x-20) + cos(x+70)
= sin(x-20) + sin(90-(x+70)) (cosx = sin(90-x)
= sin(x-20) +sin(-x+20)
= sin(x-20) - sin(x-20) = 0 = R.H.S