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Math Help - A few trig problems- help please (double angle and simplifying trig expressions)

  1. #1
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    A few trig problems- help please (double angle and simplifying trig expressions)

    Hey, I have been trying to figure out how to do these problems for quite some time, but I'm stuck. Help is very much appreciated.

    Verify:

    2- ((sin^2(x)/tan^2(x))= sin^2x-1

    I really cannot figure this out.

    Also, how do you solve this problem for theta? (this is one where I'm supposed to find the reference angle):

    2sin2u+sin=0

    and last thing, how do you find the cot(M+N)? Is it the same as the sum formula for cos but under 1?

    Thank you so much!
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    Quote Originally Posted by OnlyEternity View Post
    Also, how do you solve this problem for theta? (this is one where I'm supposed to find the reference angle):

    2sin2u+sin=0
    here's a kick start...

     2\sin 2u+\sin u=0

     2\times 2\sin u\cos u+\sin u=0

     4\sin u\cos u+\sin u=0

     4\sin u (\cos u+1)=0
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  3. #3
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    Quote Originally Posted by OnlyEternity View Post
    Hey, I have been trying to figure out how to do these problems for quite some time, but I'm stuck. Help is very much appreciated.

    Verify:

    2- ((sin^2(x)/tan^2(x))= sin^2x-1

    I really cannot figure this out.

    Also, how do you solve this problem for theta? (this is one where I'm supposed to find the reference angle):

    2sin2u+sin=0

    and last thing, how do you find the cot(M+N)? Is it the same as the sum formula for cos but under 1?

    Thank you so much!
    \frac{sin^{2}n}{tan^{2}n} = \frac{sin^{2}n}{\frac{sin^{2}n}{cos^{2}n}}= \frac{sin^{2}n}{1} \times\frac{cos^{2}n}{sin^{2}n}
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by OnlyEternity View Post
    Hey, I have been trying to figure out how to do these problems for quite some time, but I'm stuck. Help is very much appreciated.

    Verify:

    2- ((sin^2(x)/tan^2(x))= sin^2x-1

    I really cannot figure this out.

    Also, how do you solve this problem for theta? (this is one where I'm supposed to find the reference angle):

    2sin2u+sin=0

    and last thing, how do you find the cot(M+N)? Is it the same as the sum formula for cos but under 1?

    Thank you so much!
    cot(M+N)= \frac{1}{tan(M+N)}

    or,

    cot(M+N) = \frac{cos(M+N)}{sin(M+N)}
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    Quote Originally Posted by pickslides View Post
    here's a kick start...

     2\sin 2u+\sin u=0

     2\times 2\sin u\cos u+\sin u=0

     4\sin u\cos u+\sin u=0

     4\sin u (\cos u+1)=0
    Thanks! So after you get to this point, you solve for the reference angle and get:
    Sin's ref <= 0 and cos's=180
    so that means that u can = 0 degrees, 180 degrees or 360 degrees for sin, or 180 degrees for cos?

    Hopefully that makes sense

    Also:
    Quote Originally Posted by harish21 View Post
    \frac{sin^{2}n}{tan^{2}n} = \frac{sin^{2}n}{\frac{sin^{2}n}{cos^{2}n}}= \frac{sin^{2}n}{1} \times\frac{cos^{2}n}{sin^{2}n}
    I got that far, but it's at this point where I don't know what to do to make it = to sin^2(x)-1

    at that point I get 2-cos^2(x), but I need it to be sin^2(x)-1
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  6. #6
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    Quote Originally Posted by OnlyEternity View Post
    Sin's ref <= 0 and cos's=180
    so that means that u can = 0 degrees, 180 degrees or 360 degrees for sin, or 180 degrees for cos?
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    Quote Originally Posted by pickslides View Post
    Great, thanks a lot! It makes more sense now.
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  8. #8
    MHF Contributor harish21's Avatar
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    2-cos^{2}x= 2-(1-sin^{2}x)=2-1+sin^{2}x = 1+sin^{2}x

    But your answer says the RHS= sin^{2}x-1, for which the equation would be false!
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