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Trigonometry> Please elaborate the answer. Need complet...

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Please elaborate the answer.
Need complete solution.

Kruthik , 7 Years ago

Grade 11

2 Answers

Arun

Last Activity: 7 Years ago

Dear student

Minimum value of expression is obtained when denominator maximizes.
hence
maximum value of 3 sin@ – 4 cos@ +7 = 5 + 7 = 12
Hence minimum value of original expression is 1/12

Soumendu Majumdar

Last Activity: 7 Years ago

Dear Student,

$-\sqrt{a^2 +b^2}\leq asin\Theta + bcos\Theta \leq \sqrt{a^2 +b^2}$

Here a = 3, b = 4 or – 4 (doesn’t matter since squaring it will give the same result)

Now since the expression is in the denominator so we need to find the maximum value of $3sin\Theta -4cos\Theta$ since greater is the denominator lesser will be the value of the original expression.

So maximum value = $\sqrt{3^2+4^2 } = \sqrt{5^2} = 5$

so now maximum value of denominator = 5 + 7 = 12

Hence, minimum value of the expression = 1/12

Hope it helps!

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