Find the value of : cos[6π +2tan−1(1) ] +sin[3sin−1(21 ) +2cos−1(21 ) ]

We have;cos[6π +2tan−1(1) ] +sin[3sin−1(21 ) +2cos−1(21 ) ] =cos[6π +42π ] +sin[3×6π +2×3π ] =cos[2π +6π ] +sin[2π +32π ] =−sin(6π )+sin(67π ) =−21+sin(π+6π)=−21−sin(6π)=−21−21=−1Hence;cos[6π +2tan−1(1) ] +sin[3sin−1(21 ) +2cos−1(21 ) ] =−1

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Find the value of : $cos [\frac{π}{6} + 2 tan^{- 1} (1)] + sin [3 sin^{- 1} (\frac{1}{2}) + 2 cos^{- 1} (\frac{1}{2})]$

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Find the value of :

cos [\frac{π}{6} + 2 tan^{- 1} (1)] + sin [3 sin^{- 1} (\frac{1}{2}) + 2 cos^{- 1} (\frac{1}{2})]

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Text solutionVerified

We have;

cos [\frac{π}{6} + 2 t an^{- 1} (1)] + sin [3 sin^{- 1} (\frac{1}{2}) + 2 cos^{- 1} (\frac{1}{2})]

= cos [\frac{π}{6} + \frac{2 π}{4}] + sin [3 \times \frac{π}{6} + 2 \times \frac{π}{3}]

= cos [\frac{π}{2} + \frac{π}{6}] + sin [\frac{π}{2} + \frac{2 π}{3}]

= - sin (\frac{π}{6}) + sin (\frac{7 π}{6})

= - \frac{1}{2} + sin (π + \frac{π}{6})

= - \frac{1}{2} - sin (\frac{π}{6})

= - \frac{1}{2} - \frac{1}{2} = - 1

Hence;

cos [\frac{π}{6} + 2 t an^{- 1} (1)] + sin [3 sin^{- 1} (\frac{1}{2}) + 2 cos^{- 1} (\frac{1}{2})] = - 1

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Question Text	Find the value of : $cos [\frac{π}{6} + 2 tan^{- 1} (1)] + sin [3 sin^{- 1} (\frac{1}{2}) + 2 cos^{- 1} (\frac{1}{2})]$
Answer Type	Text solution:1
Upvotes	150

Find the value of : cos[6π ​+2tan−1(1) ] +sin[3sin−1(21​ ) +2cos−1(21​ ) ]

Text solutionVerified

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Find the value of : $cos [\frac{π}{6} + 2 tan^{- 1} (1)] + sin [3 sin^{- 1} (\frac{1}{2}) + 2 cos^{- 1} (\frac{1}{2})]$