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Math Help - Trig Functions & Unit Circle

  1. #1
    Member realintegerz's Avatar
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    Trig Functions & Unit Circle

    I have nearly no clue on how to do these kind of problems, I somewhat understand how to move the values by pi and pi/2 (kinda off sometimes) and I do know that cosine is x and sine is y

    One question I want to ask first is, when adding or subtracting by pi/2, do the sine and cosine values flip and become negative or does that happen when adding pi or ?

    1) Suppose cos(5) = 0.28 and sin (5) = -0.96, find an angle alpha such that cos(alpha) = 0.28 and sin(alpha) = 0.96 without using a calculator

    2) If cos(t) = -3/5 and pi< t <3pi/2, find tan(t) without a calculator

    3) Find a value of alpha where -3pi< alpha < -2pi

    Any help is appreciated as always, I'll be sure to give you a thanks for your input
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  2. #2
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    Quote Originally Posted by realintegerz View Post
    One question I want to ask first is, when adding or subtracting by pi/2, do the sine and cosine values flip and become negative or does that happen when adding pi or ?
    It depends on what value you are looking, but generally adding \pi will result in a negative result(assuming you had a positve answer to begin with, and vice versa). so for instance if you have \cos\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2}, now if you add \pi you will basically have \frac{11\pi}{4} =\frac{3\pi}{4}, so taking \cos \left( \frac{3\pi}{4} \right) = -\frac{\sqrt{2}}{2}. now if instead of adding \pi we decide to add \frac{\pi}{2} we get \frac{7\pi}{4}+\frac{\pi}{2} = \frac{9\pi}{4}=\frac{\pi}{4} so taking \cos\left(\frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}=\cos\left(\frac{7\pi}{4}\right) you'll find more information here Unit circle - Wikipedia, the free encyclopedia

    Quote Originally Posted by realintegerz View Post
    1) Suppose cos(5) = 0.28 and sin (5) = -0.96, find an angle alpha such that cos(alpha) = 0.28 and sin(alpha) = 0.96 without using a calculator
    Since we are given \cos(5) = 0.28 then \cos(\alpha) = 0.28 \Rightarrow \alpha =\arccos(0.28) \Rightarrow \alpha =5.

    now for the sine function, we are give \sin(5) = -0.96 then \sin(\alpha) = 0.96 \Rightarrow \alpha =\arcsin(0.96) \Rightarrow \alpha =-5 since sine is an odd function.

    Quote Originally Posted by realintegerz View Post
    2) If cos(t) = -3/5 and pi< t <3pi/2, find tan(t) without a calculator
    if we take \tan(\arccos(x)) we get \frac{\sqrt{1-x^2}}{x} now replacing x by -\frac{3}{5} yields the desired result.
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