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