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    I don't know how to do these trig questions:

    1) sec^4x - tan^4x = 2sec^2x - 1

    2) (cosx + secx)^2 = tan^2x + cos^2x + 3

    3) cos^4x - sec^4x = cos^2x - sin^2x

    4) cose^2x(tan^2x - sin^2x) = tan^2x

    Thanks in advance, the main problem I have with these is that I don't know what to do when there are powers bigger that 2 in the question, so if anyone had any advice I'd be greatful for it.
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    (Original post by Bebop)
    I don't know how to do these trig questions:

    1) sec^4x - tan^4x = 2sec^2x - 1

    2) (cosx + secx)^2 = tan^2x + cos^2x + 3

    3) cos^4x - sec^4x = cos^2x - sin^2x

    4) cose^2x(tan^2x - sin^2x) = tan^2x

    Thanks in advance, the main problem I have with these is that I don't know what to do when there are powers bigger that 2 in the question, so if anyone had any advice I'd be greatful for it.



    2) (cosx + secx)^2 = tan^2x + cos^2x + 3

    LHS: cos^2x + 2sinxcosx + sin^2x

    2sinxcosx = sin2x
    cos^2x - sin^2x = cos2x

    oh never mind
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    sorry, just try and use the double angle formulaes for them.
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    These question come before the double angle forumla is taught in the textbook so I think you have to use another method.
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    (Original post by Bebop)
    3) cos^4x - sec^4x = cos^2x - sin^2x
    Is difference of 2 squares so
    LHS = (cos^2 x + sin^2 x)(cos^2 x - sin^2 x)
    cos^2 x + sin^2 x = 1

    The others can be done using cos^2 x + sin^2 x = 1, 1 + tan^2 x = sec^2 x and cot^2 x + 1 = cosec^2 x
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    2) (cosx + secx)^2 = tan^2x + cos^2x + 3

    LHS= cos^2x + 2 + sec^2x
    = cos^2x + 2 + (tan^2x + 1)
    = tan^2x + cos^2x + 3
    = RHS

    i'll have the rest up in bout twenty mins haven't got time just now!

    sam
    x
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    (Original post by Bebop)
    I don't know how to do these trig questions:

    4) cose^2x(tan^2x - sin^2x) = tan^2x

    Thanks in advance, the main problem I have with these is that I don't know what to do when there are powers bigger that 2 in the question, so if anyone had any advice I'd be greatful for it.
    I think there's a typo in that one. If you put in x = π/4, you get ¼ = 1!!

    cos²x(tan²x - sin²x)
    cos²x.sin²x(sec²x - 1)
    cos²x.sin²x.tan²x
    sin^4x
    ====
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    (Original post by Fermat)
    I think there's a typo in that one. If you put in x = π/4, you get ¼ = 1!!

    cos²x(tan²x - sin²x)
    cos²x.sin²x(sec²x - 1)
    cos²x.sin²x.tan²x
    sin^4x
    ====
    I think he means the LHS is: cosec²x(tan²x - sin²x)
    anyway, this is how i would do it:
    exapnd LHS to get:
    cosec²x.tan²x - cosec²x.sin²x

    now simplify:

    (1/sin²x)(sin²x/cos²x) - (1/sin²x)(sin²x)
    (1/cos²x) - 1
    sec²x - 1 = tan²x
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    (Original post by Bebop)
    I don't know how to do these trig questions:

    1) sec^4x - tan^4x = 2sec^2x - 1
    LHS: = (sec²x + tan²x)(sec²x - tan²x)
    = [sec²x + (sec²x -1)] [sec²x - (sec²x +1)]

    = (2sec²x - 1) (1)

    = 2sec²x - 1
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    (Original post by mockel)
    I think he means the LHS is: cosec²x(tan²x - sin²x)anyway, this is how i would do it:exapnd LHS to get:cosec²x.tan²x - cosec²x.sin²x now simplify: (1/sin²x)(sin²x/cos²x) - (1/sin²x)(sin²x)(1/cos²x) - 1sec²x - 1 = tan²x
    How do you get from (1/sin²x)(sin²x/cos²x) - (1/sin²x)(sin²x) to that (1/cos²x) - 1?
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    (Original post by Bebop)
    How do you get from (1/sin²x)(sin²x/cos²x) - (1/sin²x)(sin²x) to that (1/cos²x) - 1?
    the sin²x cancels for each term
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    (Original post by Bebop)
    How do you get from (1/sin²x)(sin²x/cos²x) - (1/sin²x)(sin²x) to that (1/cos²x) - 1?
    Mockel meant that the sin^2 x in the denominator cancels out the stn^2 x
    in the nemerator in the next term.
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    (Original post by Bebop)
    I don't know how to do these trig questions:

    1) sec^4x - tan^4x = 2sec^2x - 1

    2) (cosx + secx)^2 = tan^2x + cos^2x + 3

    3) cos^4x - sec^4x = cos^2x - sin^2x

    4) cose^2x(tan^2x - sin^2x) = tan^2x

    Thanks in advance, the main problem I have with these is that I don't know what to do when there are powers bigger that 2 in the question, so if anyone had any advice I'd be greatful for it.
    ok i can do all of these except for no 3
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    can ne1 do question 3 or is it simply a typo??
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    (Original post by lgs98jonee)
    can ne1 do question 3 or is it simply a typo??
    See my earlier post (#5)
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    (Original post by Bezza)
    Is difference of 2 squares so
    LHS = (cos^2 x + sin^2 x)(cos^2 x - sin^2 x)
    cos^2 x + sin^2 x = 1

    The others can be done using cos^2 x + sin^2 x = 1, 1 + tan^2 x = sec^2 x and cot^2 x + 1 = cosec^2 x
    question invloved sec not sin though!!
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    i assume its a typo then
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    (Original post by lgs98jonee)
    question invloved sec not sin though!!
    Oh yeah. Doesn't work though if it's sec instead of sin. Try x = pi/6 though, cosx = 1/2, secx = 2, sinx = sqrt3/2
    cos^4 x - sec^4 x = 1/16 - 16 = -255/16
    cos^2 x - sin^2 x = cos(2x) = cos(2pi/3) = -1/2
    If you misread it my way then it works!!
 
 
 

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