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Mathematics

1 answer:

jarptica [38.1K]3 years ago

8 0

Answer:i dont know just jot down something but i dont care let me submit this

Step-by-step explanation:

listen if u are not taking the test just blop something down and they dont look at the awnser if they do then i dont know friendly hint

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Somone please help me asap i need to turn this in its due tomorrow

Novosadov [1.4K]

Hope this helps !
Answer circled on green

6 0

3 years ago

Sin(x+pi/4)-sin(x-pi/4)=1 solve the equation

evablogger [386]

Sin(α+β)=sin(α)cos(β)+cos(α)sin(β)
sin(α-β)=sin(α)cos(β)-cos(α)sin(β)

$sin(x+ \frac{ \pi }{4} )=sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} \\ sin(x- \frac{ \pi }{4} )=sin(x)cos\frac{ \pi }{4} -cos(x)sin\frac{ \pi }{4} \\ \\ \\sin(x+ \frac{ \pi }{4} )-sin(x- \frac{ \pi }{4} ) =1 \\ sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} -(sin(x)cos\frac{ \pi }{4} -cos(x)sin\frac{ \pi }{4} )=1 \\ sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} -sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} =1 \\ cos(x)sin\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} =1 \\$
$2cos(x)sin\frac{ \pi }{4} =1 \\ sin\frac{ \pi }{4} = \frac{ \sqrt{2} }{2} \\ 2cos(x) \frac{ \sqrt{2} }{2} =1 \\ 2 \cdot \frac{ \sqrt{2} }{2}cos(x) =1 \\ \sqrt{2} cos(x)=1 \\$
$cos(x)= \frac{1}{ \sqrt{2} } \\ x=\pm arccos \frac{1}{ \sqrt{2} }+2 \pi k , k \in Z \\ x=\pm \frac{ \pi }{4} +2 \pi k , k \in Z$

6 0

3 years ago

Solve the following question

nydimaria [60]

9514 1404 393

Answer:

π/3

Step-by-step explanation:

The given integral does not exist. We assume there is a typo in the upper limit, and that you want the integral whose upper limit is (√3)/2.

It is convenient to make the substitution ...

x = sin(y) . . . . so, y = arcsin(x)

dx = cos(y)·dy

Then the integral is ...

$\displaystyle\int_0^{\sqrt{3}/2}{\dfrac{\cos{y}}{\sqrt{1-\sin^2{y}}}}\,dy=\int_0^{\sqrt{3}/2}{dy}=\left.\arcsin{x}\right|\limits_0^{\sqrt{3}/2}\\\\=\arcsin{(\sqrt{3}/2)}=\boxed{\dfrac{\pi}{3}}$