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How Sin(90° + θ) =sin θ of triangle P'OM'

  1. Jan 24, 2016 #1
    Hello everyone. I'm learning Trigonometry right now with myself and at current about how to find the trigonometric ratio of the angle (90° + θ) in terms of θ. I'm quite confused in the Figure. How sin(90° + θ) become equal to sin (90° - θ) of triangle P'OM'. I know that triangle P'OM' and POM are congruent.

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  3. Jan 24, 2016 #2


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    ##\sin(90^o+\theta) = \cos \theta = \cos(-\theta) = \sin(90^o-\theta)##.
  4. Jan 24, 2016 #3
    Have you heard of a unit circle? Look it up.
  5. Jan 24, 2016 #4
    yes I know about it. But I didn't able to understand how to find the trigonometric ratio of the angle (90° + θ) in terms of θ from it.
  6. Jan 24, 2016 #5
    Thankyou for the answer. Can you please show me this with figure please. It will help me more.
  7. Jan 24, 2016 #6
    Sorry. Just a minute. I do believe I am wrong. I mistook 90 +θ for 180-θ and said it's equal to θ it's not. Just realised. Really, Really sorry. I have deleted that post. Read what follows.

    But, basically you can still do it using a unit circle. For your θ, sin 90+θ is if move over θ from the y axis. So, sin 90+θ becomes (-y,x) [ Remember that (x,y) is your original point.] So if you reflect that across the y axis you get (y,x) which is sin 90-θ. The following image shows the relationship between θ and 90+θ


    And this shows the relationship between θ and 90-θ. So you can see from both these images what I'm talking about.
  8. Jan 24, 2016 #7
    How does it proof that angle Sin(90+θ) = sin(90-θ) of triangle P'OM'?
  9. Jan 24, 2016 #8
    your stuff is quite similar to the unit circle. Look at the first diagram on the left in the image you posted .Angle AOP is θ. So angle POP' is 90+θ. So if you reflect triangle P'OM across the y axis, then angle AOP' will be 90-θ . It's easier to see that the two are equal if OP, OP' etc are the radii of a unit circle. Hence the unit circle.
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