1.

Write the principal value of `cos^(-1)[cos(680^@)]`

Answer» Principal value of `cos^-1(cos680^@)` will lie between `0` and `180^@`.
Now, `cos(680^@) = cos(360^@+320^@) = cos320^@`
`=cos(360^@-40^@) = cos(-40^@) = cos40^@` ....(As `cos(-theta) = costheta`)
`:. cos^-1[cos680^@] = cos^-1[cos40^@] = 40^@`
So, principal value of given expression is `40^@`.


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