Verify the identities, manipulating only one side of the equation at a time. Use the abbreviations LHS and RHS for left and right hand side?

42) 1-csc(x)sin^3(x)=cos^2

43) (1+cot(alpha))^2- 2cot(alpha)= 1/ (1-cos(alpha))(1+cos(alpha))

44) 1-sin(x) = 1-sin^2(-x) / 1-sin(-x)

45) sinx / sinx+1 = cscx-1 / cot^2 x

46)cos(x- pi/2) = cosx tan x

47)sin(x- pi/2) = cos x

48) sin^4 (z) + cos^4(z) = 1

49) cos (alpha -beta) / sin(alpha + beta) = 1+ tan alpha tan beta / tan alpha - tan beta

for the life of me, I don't understand how to do these....

2 Answers

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  • 6 years ago
    Favorite Answer

    I'll do a few of them to give you a start!

    42.

    LHS = 1 - csc(x)sin³(x)

    = 1 - 1/sin(x)*sin³(x) [Def. of csc(x)]

    = 1 - sin²(x) [Cancel a sin(x)]

    = cos²(x) [sin²(x) + cos²(x) = 1]

    =RHS

    43.

    LHS = (1 + cot(α))² - 2cot(α)

    = 1 + 2cot(α) + cot²(α) - 2cot(α) [Expand]

    = 1 + cot²(α) [Cancel like terms]

    = csc²(α) [Pythagorean Identity]

    = 1/sin²(α) [Definition of csc(α)]

    = 1/(1-cos²(α)) [Pythagorean Identity]

    = 1/((1+cos(α))(1-cos(α))) [Difference of 2 Squares]

    = RHS

    44.

    RHS = (1-sin²(x))/(1-sin(-x))

    = (1+sin(x))(1-sin(x))/(1-sin(-x)) [Difference of 2 Squares]

    = (1+sin(x))(1-sin(x))/(1+sin(x)) [-sin(x) = sin(-x) odd symmetric function]

    = 1 - sin(x) [Cancel 1+sin(x)]

    = LHS

    Ok, I hope this gives you a start! For some of the other ones, these identities may help,

    sin(x ± y) = sin(x)cos(y) ± sin(y)cos(x)

    cos(x ± y) = cos(x)cos(y) ∓ sin(x)sin(y)

    Please pick for best answer!

  • Anonymous
    6 years ago

    extremely tough matter look on yahoo or google that will might help

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