If sin 0 + cos = root 2, find value of sin -cos .plss fassssstttttttt December 30, 2021 by Peyton If sin 0 + cos = root 2, find value of sin -cos . plss fassssstttttttt

Answer: [tex][/tex] Force is push or pull acting on a body which tends to change its state of rest or of motion. It is denoted by “F”. Log in to Reply

Answer: 0 Step-by-step explanation: By squaring we get, [tex] {( \sin(x) + \cos(x)) }^{2} = { \sqrt{2} }^{2} [/tex] [tex] { \sin(x) }^{2} + 2 \sin(x) \cos(x) + { \cos(x) }^{2} = 2[/tex] [tex]1 + 2 \sin(x) \cos(x) = 2[/tex] [tex] \sin(x) \cos(x) = \frac{1}{2} [/tex] Now, [tex] {( \sin(x) – \cos(x)) }^{2} = { \sin(x) }^{2} – 2 \sin(x) \cos(x) + { \cos(x) }^{2} [/tex] [tex] { (\sin(x) – \cos(x) ) }^{2} = 1 – 2 \times \frac{1}{2} = 0[/tex] so, [tex] \sin(x) – \cos(x) = 0[/tex] Log in to Reply

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Answer:0

Step-by-step explanation:By squaring we get,

[tex] {( \sin(x) + \cos(x)) }^{2} = { \sqrt{2} }^{2} [/tex]

[tex] { \sin(x) }^{2} + 2 \sin(x) \cos(x) + { \cos(x) }^{2} = 2[/tex]

[tex]1 + 2 \sin(x) \cos(x) = 2[/tex]

[tex] \sin(x) \cos(x) = \frac{1}{2} [/tex]

Now,

[tex] {( \sin(x) – \cos(x)) }^{2} = { \sin(x) }^{2} – 2 \sin(x) \cos(x) + { \cos(x) }^{2} [/tex]

[tex] { (\sin(x) – \cos(x) ) }^{2} = 1 – 2 \times \frac{1}{2} = 0[/tex]

so,

[tex] \sin(x) – \cos(x) = 0[/tex]