Math, asked by puneettyagi86, 1 year ago

5sin^2 -12sincos +10cos^2 maximum and minimum value​

Answers

Answered by GENIUS1223
0

Answer:

Step-by-step explanation:

Lets see if we can simplify this a bit:

[math] 10 \cos^2 x - 6\sin x\cos x + 2\sin^2x \\8 \cos^2 x - 6\sin x\cos x+2 \\ 4(2 \cos^2 x -1) - 3(2\sin x\cos x) + 6\\4\cos 2x - 3\sin 2x +6[/math]

Much better.

Here is a trick. [math] A\cos x + B\sin x = \sqrt {A^2+B^2} \sin (x+\phi)\\\phi =\tan^{-1} \frac AB[/math]

And to our problem.

[math] 4\cos 2x - 3\sin 2x + 6 = 5 sin (x-\phi) + 6[/math]

AS we just need the range, we do not need to solve for [math]\phi[math]

[math] - 5\le\sin(2x - \phi)\le 5[/math]

And the range is [math] [1,11] [/math]


puneettyagi86: brother l didn't ask a Google explaination
puneettyagi86: by the way answer is wrong
Answered by nk1865873
2

my answer of your question.it is correct

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