Math, asked by aarushi47, 1 year ago

pls pls help me I will mark u brainylists

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Answered by sach33

2 sin - 2 =cos
sin + 2 cos = sin + 4sin- 4= 5 sin -4

aarushi47: hey enna da u will not say right step

sach33: so it means that value of theta could be different

sach33: 5 sin 90-4=5×1-4=1

sach33: try to verify by putting different values

sach33: check my earlier messages

aarushi47: poda kanna puindiy

sach33: english plz

sach33: the equation can not be correct for all values of theta

sach33: LHS= 2 sin 45- cos 45= 2/sqrt2 - 1/sqrt2= 1/sqrt2

sach33: but RHS =2

Answered by rakeshmohata

Hope u like my process
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$= > \bf2 \sin( \theta ) - \cos( \theta ) = 2 \\ \\ \bf \underline{squaring \: \: both \: \: sides \: \: } \\ \\ = > \: {(2 \sin( \theta ) - \cos( \theta ) )}^{2} = {2}^{2} \\ \\ or. \: \: 4 \sin ^{2} ( \theta ) + \cos ^{2} ( \theta ) - 4 \sin( \theta ) \cos( \theta ) = 4 \\ \\ or. \: \: 4(1 - \cos ^{2} ( \theta ) ) + (1 - \sin ^{2} ( \theta ) ) - 4 \sin( \theta) \cos( \theta ) = 4 \\ \\ or. \: \: 4 - 4 \cos ^{2} ( \theta ) + 1 - \sin ^{2} ( \theta ) - 4 \sin( \theta ) \cos( \theta ) = 4 \\ \\ or. \: \: 4 \cos ^{2} ( \theta ) + \sin ^{2} ( \theta ) + 4 \sin( \theta ) \cos( \theta ) = 1 \\ \\ or. \: \: {( \sin( \theta ) + 2\cos( \theta )) }^{2} = {1}^{2} \\ \\ or. \: \: \bf \: \sin( \theta ) + 2 \cos( \theta ) = \sqrt{1} = \underline{1}$
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Hope this is ur required answer

Proud to help you

aarushi47: step

rakeshmohata: simply take all the sine and cosine terms on one side and constants on other sides

aarushi47: 2 cos2 is oming

aarushi47: coming

rakeshmohata: ?? where

rakeshmohata: which step u think is to doubted?

aarushi47: 5 step

aarushi47: hello

rakeshmohata: take the minus terms on other side to make them positive

aarushi47: ok

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