Problem (1) For what value of #x#, #sinx+cosx=2# Problem (2) For what value of #x#, #sinxcosx=1#?

Answer 1

#(sinx+cosx)# can never be #2#

and #sinxcosx# can never be #1#.

Problem (1)

#sinx+cosx# can never be #2#, the maximum value of #sinx+cosx# can only be #sqrt2# as
#sinx+cosx=sqrt2(sinx xx 1/sqrt2+cosx xx 1/sqrt2)#
= #sqrt2(sinxcos(pi/4)+cosxsin(pi/4))#
= #sqrt2sin(x+pi/4)#
as maximum value of any sine ratio can only be #1#,
maximum value of #sqrt2sin(x+pi/4)# or #sinx+cosx# can only be #sqrt2#.
Hence #sinx+cosx# can never be #2#.

Problem (2)

As #sinxcosx=1/2(2sinxcosx)=1/2xxsin2x#
and again as maximum value of any sine ratio can only be #1#,
maximum value of #sinxcosx# or #1/2sin2x# can only be #1/2#
and it can never be more than #1/2# and hence cannot be #1#.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

Problem (1): For what value of x, (\sin(x) + \cos(x) = 2)?

The equation (\sin(x) + \cos(x) = 2) has no solution because the maximum value of (\sin(x)) is 1 and the maximum value of (\cos(x)) is also 1, so the maximum possible sum of (\sin(x) + \cos(x)) is (1 + 1 = 2), but this only occurs when both (\sin(x)) and (\cos(x)) are at their maximum values, which never happens simultaneously since their maximum values occur at different points in the unit circle ((\sin(x)) at (x = \frac{\pi}{2}) and (\cos(x)) at (x = 0)).

Problem (2): For what value of x, (\sin(x)\cos(x) = 1)?

This equation also has no real solution because the product (\sin(x)\cos(x)) has a maximum value of (\frac{1}{2}). This is derived from the identity (\sin(2x) = 2\sin(x)\cos(x)), implying that (2\sin(x)\cos(x) = \sin(2x)). The maximum value of (\sin(2x)) is 1, so the maximum of (2\sin(x)\cos(x)) is 1, making the maximum of (\sin(x)\cos(x) = \frac{1}{2}), occurring when (x = \frac{\pi}{4}) or (x = \frac{3\pi}{4}), not 1. Therefore, there's no real value of (x) that satisfies (\sin(x)\cos(x) = 1).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7