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Math Help - [SOLVED] Proving Trigonometric Identities HELP!

  1. #1
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    [SOLVED] Proving Trigonometric Identities HELP!

    i have 3 questions i am stuck on for homework and i have a test on this monday. if someone would be able to solve them I would appreciate it

    cotx(tany+tanx/coty+cotx) = tany

    cosx+cos2x+cos3x = cos2x(1+2cosx)

    sin^3x+sin^3x/sinx+cosx = 1-sinxcosx

    Thanks
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  2. #2
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    for the first question:

    cot x [(tan y + tan x)/(cot x + cot y)]

    remembering the rules:
    cot a = cos a/sin a
    tan a = sin a/cos a

    hence the equation can now be written as

    =(cosx/sinx)[ {(sin y/cos y) + (sin x/cos x)} / {(cos y/sin y) + (cos x/sin x)}]
    =(cosx/sinx)[{(cos x.sin y+cos y.sin x)(sin y.sin x)} / {(cos y.cos.x)/(cos y.sin x + cos x.sin y)}
    = (cos x/sin x)[{sin y.sin x}/{cos y.cos x}]
    = sin y/cos y
    = tan y
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  3. #3
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    thanks for the first one. its really appreciated
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    for question 2:

    required formula:
    2.cos x.cos y = cos(x+y) + cos(x-y)

    now instead from solving from left to right we will solve from right to left

    (cos 2x)(1+2cos x)

    = cos 2x + 2cos 2x.cos x
    = cos 2x + {cos (2x+x) + cos(2x-x)} =====> applying the formula stated previously
    = cos 2x + cos 3x + cos x
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  5. #5
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    thanks for that second one
    i just figured out why i couldn't solve it
    i was never taught that formula that you just used

    thanks, now the third one i looked for on the internet and i could not find a solution but there was plenty of solutions for it if it was sin^3x +cos^3y was at the top on the left. i just want to make sure there is no solution because i don't want to tell my teacher he was wrong just to have him show me that i could not find the solution
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  6. #6
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    hmm i cant seem to be able to solve question 3 for now >.< sorry. you might wanna ask the others around.
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  7. #7
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    Hello, math71321!

    I'm 99.99% sure there's a typo in #3 . . .


    \frac{\sin^3\!x + {\color{blue}\cos^3\!x}}{\sin x + \cos x} \;=\;1 - \sin x\cos x\quad\hdots .This is a classic problem
    The numerator is a difference of cubes . . .

    Factor: . \frac{(\sin x + \cos x)(\sin^2\!x - \sin x\cos x + \cos^2\!x)}{\sin x + \cos x}


    \text{Reduce: }\;\overbrace{\sin^2\!x + \cos^2\!x}^{\text{This is 1}} - \sin x\cos x \;\;=\;\;1 - \sin x\cos x

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    well if that is true soroban they you saved my life T_T spent alot of time trying to solve the original question @_@ got me nowhere.
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  9. #9
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    thanks guys....its really appreciated....
    just out of curiosity what level of work do you think this is
    university or high school
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  10. #10
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    well i learn to solve this stuffs at college.
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  11. #11
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    Quote Originally Posted by math71321 View Post
    thanks guys....its really appreciated....
    just out of curiosity what level of work do you think this is
    university or high school
    Pretty sure its high school work. I do learn this at school...
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