Anonymous
Anonymous asked in Science & MathematicsEngineering · 8 years ago

can you please prove these trigonometric identities?

i need your exact solutions and answers,,thanks...

a.) (csc^2) (X) + (sec^2) (X) = (csc^2) (X) (sec^2) (X)

b.) ( ( 1 + sin Y ) / ( cos Y ) ) = ( ( cos Y ) / ( 1-sin Y ) )

c.) cos 2X = (cos^4) (X) - (sin^4) (x)

d.) cot X sin 2X = 1 + cos 2X

I HOPE YOU ANSWER AS BEST AS YOU COULD, I NEED ANSWERS WITHIN 4-6 DAYS,,THANK YOU VERY MUCH!!

Update:

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Update 2:

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3 Answers

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  • 8 years ago
    Favorite Answer

    a)

    csc²x+sec²x= 1/sin²x + 1/cos²x= (sin²x + cos²x)/sin²xcos²x =1/ sin²xcos²x =csc²xsec²x

    b)

    ( 1 + sin Y ) / ( cos Y )= [( 1 + sin Y ) x ( 1 - sin Y ) ] ÷ ( cos Y ) x ( 1 - sin Y )

    =cos²Y ÷ [( cos Y ) x ( 1 - sin Y )]

    =( ( cos Y ) / ( 1-sin Y ) )

    c)

    cos 2X= (cos²x −sin²x ) x 1

    = (cos²x −sin²x ) x (sin²x + cos²x)

    =(cos^4) (X) - (sin^4) (x)

    d) LHS:

    cot X sin 2X=(cosx÷sinx) x (2sinxcosx)

    =2cos²x

    RHS:

    1 + cos 2X=1+cos²x −sin²x

    =2cos²x (since cos²x= 1−sin²x )

    LHS=RHS

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  • 8 years ago

    b) Multiply both sides by (1 - sin Y):

    (1 - sin^2 Y) / (cos Y) = cos Y

    cos^2 Y / cos Y = cos Y. Q.E.D.

    c) Observe that the right hand side = (cos^2 x + sin^2 x) (cos^2 x - sin^2 x) = cos^2 x - sin^2 x = cos (2x)

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  • Anonymous
    8 years ago

    cos 2X = (cos^4) (X) - (sin^4) (x)

    cos(-x) = - cos x (D) sin(x+y) = sinxcosy + cosxsiny

    i think im 100% right

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