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anonymous

  • one year ago

What is the exact value of cos(arcsin(1/4))? Kindly explain to me how. :)

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  1. Haseeb96
    • one year ago
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    first convert 1/4 into decimals ??@mathway

  2. Haseeb96
    • one year ago
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    1/4= 0.25

  3. Haseeb96
    • one year ago
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    now as we know sin 30 =0.5 so sin 15 = 0.25 now arcsin (0.25) = 15 degress okay ?? @mathway

  4. Haseeb96
    • one year ago
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    Did you get it till now

  5. anonymous
    • one year ago
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    Nothing comes up in my notifications. Sorry.

  6. Haseeb96
    • one year ago
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    oh where you are having difficulty ?

  7. anonymous
    • one year ago
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    Why did sin 30 become sin 15? Is it because we ant to get the half of it (to get 0.5)?

  8. anonymous
    • one year ago
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    And I didn't know sin 30 =0.5. Yikes!

  9. Haseeb96
    • one year ago
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    it is in table of trigonometry , you should remember the whole table yourself sin 30 degree = 0.5 so sin 15 degree = 0.25 okay ???

  10. anonymous
    • one year ago
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    Since arcsin1/4 is equals to 15, we now need to get the cos of 15?

  11. anonymous
    • one year ago
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    I was looking up for the tigro table, but I don't know which one is that. Could you please tell what it looks like? :)

  12. Haseeb96
    • one year ago
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    yes you are correct arcsin (1/4) = 15 we need to know what is the value for cos 15 we do same for cos 15 as we do for sin 15 I mean first we take our the value of cos 30 which is cos 30 =0.866 so what will be the value for cos 15 Tell me your self

  13. anonymous
    • one year ago
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    Is it the half of 0.866?

  14. Haseeb96
    • one year ago
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    yes

  15. anonymous
    • one year ago
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    So it is 0.433.

  16. Haseeb96
    • one year ago
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    yes you are correct Good job smart girl :)

  17. anonymous
    • one year ago
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    Lol I'm really sorry if I seem dumb to you haha. I'm taking Pre-Calc and never took Trigonometry, so this topic is kinda new to me. But thank you so much for the help @Haseeb96 ! :)

  18. Haseeb96
    • one year ago
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    oh well, I am here to sort out the problems of my friends Whenever you need help in math or physics you can tag me If i will be able to solve that , I would have tried my best. @mathway Welcome again :)

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