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Calculate sin75 without calculator

  1. Sep 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate sin75 without calculator!

    2. Relevant equations

    3. The attempt at a solution
    I constructed a triangle with angles 15, 75 and 90 then I defined sin75 by the sides. In a previous exercise the task was the same but with tg15 and I used the result in this execise too.
    I wrote the equation: tg15 = b/a and if I express b and put it in the Pythagorean theorem I get the right result. However, if I express a and put it in the Pythagorean theorem I got a wrong result and I don't know why is this happening. I looked through it and I haven't find any calculation error... The second way is a lot longer but I should work. :(


    Last edited: Sep 2, 2015
  2. jcsd
  3. Sep 2, 2015 #2
    Are you familiar with double angle and half angle formulas? How is sin75 related to cos15? What is cos(2θ) in terms of cosθ?

  4. Sep 2, 2015 #3

    Ray Vickson

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    You really should type out your work; most helpers here (me included) will ignore any of the details in your post.

    I am willing to offer a hint, though: if you look on-line you can find documents that give closed-form formulas for sines or cosines of all angles from 1 degree to 89 degrees in increments of 1 degree.
    Last edited: Sep 2, 2015
  5. Sep 2, 2015 #4
    I haven't learnt them but I have the formulas in my book.
  6. Sep 2, 2015 #5
    • Member warned about posting complete solutions
    Break the angle 75 into component angles which are more familiar e.g 15,30 ,45 , then apply sin(alpha+beta) formula...
    sin 75
    = sin(45 + 30)
    = sin45cos30 + cos45sin30
    = (½√2)(½√3) + (½√2)(½)
    = ¼(√6 + √2)
  7. Sep 2, 2015 #6

    Ray Vickson

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    PF helpers are not allowed to give complete solutions!
  8. Sep 2, 2015 #7
    Oh !! Sorry I didn't knew that , I will try to avoid it in the future.
  9. Sep 2, 2015 #8
    Thanks for your answer! :) But can you help me with explaining why my solution doesn't work? They seem completely equivalent for me.
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