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

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 · 1 year, 10 months ago

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. · 1 year, 2 months ago

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

Yes it is , can you post the method plzz · 1 year, 10 months ago

Thanks , got it :) · 1 year, 10 months ago

No prob. · 1 year, 10 months ago

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

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