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soty2013

  • 2 years ago

find the domain and range of (sin x + cos x)

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  1. soty2013
    • 2 years ago
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    how

  2. math>philosophy
    • 2 years ago
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    What is the range of each function?

  3. soty2013
    • 2 years ago
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    -1 to +1

  4. soty2013
    • 2 years ago
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    now ?

  5. math>philosophy
    • 2 years ago
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    now satellite73 will help you

  6. satellite73
    • 2 years ago
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    it might help to know that \(a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\) for suitable \(\theta\) that should help a great deal with the range

  7. soty2013
    • 2 years ago
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    asin(x)+bcos(x)=a2+b2−−−−−−√sin(x+θ) but how do you get this ?

  8. satellite73
    • 2 years ago
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    square root is on the outside it is \(\sqrt{a^2+b^2}\sin(x+\theta)\) and it is a consequence of the "addition angle formula"

  9. soty2013
    • 2 years ago
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    @satellite73 i am having very problem in it... can you explain the function chapter from the starting, i will be very thankful to you..

  10. satellite73
    • 2 years ago
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    i do not know exactly what you mean by "explain the function chapter" you could graph the function \[\sin(x)+\cos(x)\] using technology to see what you get or you could surmise that the largest the sum could be is if they were the same value, making \(\sin(x)=\frac{\sqrt{2}}{2}\) and also \(\cos(x)=\frac{\sqrt{2}}{2}\) and therefore the max would be \(\sqrt{2}\)

  11. satellite73
    • 2 years ago
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    or more simply you could use the formula i wrote above, telling you that \[\sin(x)+\cos(x)=\sqrt{2}\sin(x+\frac{\pi}{4})\] and now the range is more or less obvious

  12. soty2013
    • 2 years ago
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    Guru JEE thanks

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