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Find the minimum value of expression \(6\sin x - 8\cos x\)

Help please!

Note by Nitesh Chaudhary
1 year, 10 months ago

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the answer is -10 as the minimum value of the expression Asinx + Bcosx +C = 0 is C - (A^2 + B^2)^1/2 Anurag Patnaikuni · 1 year, 10 months ago

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A very very easy question. The answer is \(-\sqrt{6^2 + {(-8)}^2} = -\sqrt{100} = -10\). @Nitesh Chaudhary lets just forget our disputes and be friends. Youe choice now. Ashish Siva · 1 year, 2 months ago

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Is the answer -10? Harsh Shrivastava · 1 year, 10 months ago

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@Harsh Shrivastava Yes it is , can you post the method plzz Nitesh Chaudhary · 1 year, 10 months ago

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@Nitesh Chaudhary This may help you Harsh Shrivastava · 1 year, 10 months ago

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@Harsh Shrivastava Thanks , got it :) Nitesh Chaudhary · 1 year, 10 months ago

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@Nitesh Chaudhary No prob. Harsh Shrivastava · 1 year, 10 months ago

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@Nitesh Chaudhary Thanks for mentioning me man. I think that you've got your answer, but still

\(min(asin(x)+bcos(x)+c=0)= c- \sqrt {a^2+b^2}\) Mehul Arora · 1 year, 10 months ago

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