Find the range of #sinx(sinx+cosx)#?

Answer 1

Range of #sinx(sinx+cosx)# is #[1/2-1/sqrt2,1/2+1/sqrt2]#

#sinx(sinx+cosx)#
= #sinxsqrt2(sinx1/sqrt2+cosx1/sqrt2)#
= #sqrt2sinx(sinxcos45^@+cosxsin45^@)#
= #sqrt2sinxsin(x+45^@)#
= #sqrt2/2(2sinxsin(x+45^@))#
= #sqrt2/2(2sinxsin(x+45^@))#
= #1/sqrt2[cos(x-(x+45^@)-cos(x+(x+45^@)]#
= #1/sqrt2[cos(-45^@)-cos(2x+45^@)]#
= #1/sqrt2[1/sqrt2-cos(2x+45^@)]#
= #1/2-1/sqrt2cos(2x+45^@)#
As range of #cos(2x+45^@)# is #[-1,1]#
range of #sinx(sinx+cosx)# or #1/2-1/sqrt2cos(2x+45^@)# is
#[1/2-1/sqrt2,1/2+1/sqrt2]#
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Answer 2

To find the range of ( \sin(x)(\sin(x) + \cos(x)) ), first, note that ( \sin(x) ) and ( \cos(x) ) both have a range of ([-1, 1]). Therefore, their sum, ( \sin(x) + \cos(x) ), will also have a range of ([-1, 1]).

Since ( \sin(x) ) ranges from (-1) to (1), multiplying it by a value ranging from ([-1, 1]) will result in a product ranging from (-1) to (1).

Therefore, the range of ( \sin(x)(\sin(x) + \cos(x)) ) is ([-1, 1]).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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