5sin^2 -12sincos +10cos^2 maximum and minimum value
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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
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my answer of your question.it is correct
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