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sin (pi/2) = 2 sin(pi/4) cos (pi/4)...

`sin (pi/2) = 2 sin(pi/4) cos (pi/4)`

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sin pi/3 = 2sin pi/6 cos pi/6

Prove that (i) "sin " (7pi)/(12) " cos " (pi)/(2) - "cos " *(7pi)/(12) " sin " (pi)/(4) = (sqrt(3))/(2) (ii) " sin " (pi)/(4) " cos " (pi)/(2) + "cos"(pi)/(4) " sin " (pi)/(12) = (sqrt(3))/(2) (iii) " cos " (2pi)/(3) " cos " (pi)/(4) - " sin " (2pi)/(3) " sin " (pi)/(4) =(-(sqrt(3) +1))/(2sqrt(2))

Knowledge Check

  • sin (pi)/(4) cos (pi)/(12) - cos (pi)/(4) sin (pi)/(12) is equal to

    A
    `(1)/(sqrt3)`
    B
    `sqrt3`
    C
    `(sqrt3)/(2)`
    D
    `(1)/(2)`
  • The least positive integer n such that [{:(cos""(pi)/(4), sin""(pi)/(4)) ,(- sin ""(pi)/(4) , cos ""(pi)/(4)):}]^(n) is an identity matrix of order 2 is

    A
    4
    B
    8
    C
    12
    D
    16
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    Using application of trignometric formulas prove that (i)cos(pi/4+x)+cos(pi/4-x)=sqrt2cos x(i1)sin(7pi/12)cos(pi/4)-cos(7pi/12)sin(pi/4)

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    Prove that (i) " tan"^(2) .(pi)/(3) + 2cos^(2) .(pi)/(4)+ 3 sec^(2).(pi)/(6)+ 4 cos^(2).(pi)/(2)=8 (ii) " sin ".(pi)/(6) " cos 0 + sin ".(pi)/(4) " cos " .(pi)(4) + " sin " .(pi)/(3) "cos " .(pi)/(6) =(7)/(4) (iii) " 4sin " (pi)/(6) " sin"^(2) (pi)/(3) + 3 " cos " .(pi)/(3) " tan ".(pi)/(4) = " cosec"^(2).(pi)/(2)=4

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