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Evaluate |[cos alpha cos beta, cos alph...

Evaluate `|[cos alpha cos beta, cos alpha sin beta, -sin alpha],[ -sin beta, cos beta, 0],[ sin alpha cos beta, sin alpha sin beta, cos alpha]|`

Answer

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Knowledge Check

  • If cos alpha + 2cos beta+ 3 cos gamma = 0, sin alpha + 2sin beta +3 sin gamma= 0 and alpha + beta + gamma= pi , then sin 3alpha + 8 sin 3 beta + 27 sin 3 gamma =

    A
    `-18`
    B
    0
    C
    3
    D
    9

  • ( [(cos alpha- cos beta)+ i (sin alpha - sin beta)]^(n) + [(cos alpha- cos beta)- i (sin alpha- sin beta)]^(n))/("sin"^(n) (alpha-beta)/(2) cos ((n pi)/(2)+ (n(alpha+beta))/(2))) is equal to

    A
    `2^(n)`
    B
    `2^(n+1)`
    C
    `2^(n-1)`
    D
    `2^(2n)`

  • If cosalpha+cosbeta=0=sinalpha+sinbeta,cos2alpha+cos2beta is equal to a) -2sin(alpha+beta) b) 2cos(alpha+beta) c) 2sin(alpha-beta) d) -2cos(alpha+beta)

    A
    `-2sin(alpha+beta)`
    B
    `2cos(alpha+beta)`
    C
    `2sin(alpha-beta)`
    D
    `-2cos(alpha+beta)

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    Consider three points P = (-sin(beta - alpha), -cos beta), Q = (cos(beta - alpha), sin beta) , and R = (cos(beta - alpha + theta), sin (beta-theta)) , where 0 lt alpha, beta, theta lt pi//4 . Then

    If (cos^4 alpha)/(cos^2beta)+(sin^4alpha)/(sin^2beta)=1 ,then prove that sin^4alpha+sin^4beta=2sin^2alphasin^2beta .

    If cos(alpha+beta)=0 , then sin(alpha+2beta) =

    The distance between the points (a cos alpha, a sin alpha) and (a cos beta, a sin beta) is a) 2|sin((alpha-beta)/(2))| b) 2|a sin((alpha-beta)/(2))| c) 2|a cos((alpha-beta)/(2))| d) |a cos((alpha-beta)/(2))|

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