Find an expression for tan(arccos[x+1]) that does not involve trigonometric functions or their inverses

If anyone can help, I would really appreciate it.

Find an expression for tan(arccos[x+1]) that does not involve trigonometric functions or their inverses

If anyone can help, I would really appreciate it.

If anyone can help, I would really appreciate it.

- stapel_eliz
**Posts:**1670**Joined:**Mon Dec 08, 2008 4:22 pm-
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Draw a right triangle. The particular shape doesn't matter.

By definition, the arccosine returns an angle value. So arccos(x + 1) = @ for some angle @. (I'm using "@" to stand for "theta".) You need to find the tangent of that angle. So let's label the triangle with the angle:

We also know, by definition, that "arccos(x + 1) = @" means that "cos(@) = x + 1". In this case, it will be more useful to express this as "(x + 1)/1", because that gives us our "adjacent" (namely, x + 1) and our "hypotenuse" (namely, 1). So now we have:

Use the Pythagorean Theorem to find the value for "opposite", and then read off the value of the tangent.

right triangle:

/|

/ |

/ |

/ |

/ |

/ |

/ |

----------------------

By definition, the arccosine returns an angle value. So arccos(x + 1) = @ for some angle @. (I'm using "@" to stand for "theta".) You need to find the tangent of that angle. So let's label the triangle with the angle:

right triangle:

/|

/ |

/ |

/ |

/ |

/ |

/@ |

----------------------

We also know, by definition, that "arccos(x + 1) = @" means that "cos(@) = x + 1". In this case, it will be more useful to express this as "(x + 1)/1", because that gives us our "adjacent" (namely, x + 1) and our "hypotenuse" (namely, 1). So now we have:

right triangle:

/|

/ |

1 / |

/ |

/ |

/ |

/@ |

----------------------

x + 1

Use the Pythagorean Theorem to find the value for "opposite", and then read off the value of the tangent.