Question

proof?
sinx+cosx=√2 sin(x+pi/4)

Answer 1

\[\sin(x+\frac{\pi}{4})=\sin{x}\cos{\frac{\pi}{4}}+\sin{\frac{\pi}{4}}\cos{x}\]

Answer 2

We have sin x+ cos x
multiply the numerator and denominator by \(\sqrt 2\)
\[ \frac{\sqrt 2}{\sqrt 2} (\sin x+ \cos x)\]
We have now
\[\frac{\sqrt 2}{1}(\sin x \frac{1}{\sqrt 2}+\cos x \frac{1}{\sqrt 2})\]
now \[ \sin \pi/4= \cos \pi/4= \frac{1}{\sqrt 2}\]
so we have
\[\sqrt 2( \sin x \cos \pi/4+ \sin \pi/4 \cos x)\]
we know
\[ \sin (A+B)=\sin A \cos A+ \sin B \cos B\]
so we get
\[\sqrt 2( \sin (x +\pi/4))\]

Answer 3

sorry
\[\sin (A+B)=\sin A \cos B+ \sin B \cos A\]

Answer 4

Hey ash2326
\[\sin(\pi/4)=\sqrt{2}/2\]

Answer 5

\[\sin{(x+\frac{\pi}{4})}=\frac{1}{\sqrt{2}}(\sin{x}+\cos{x}) \]

Answer 6

\[\sin \pi /4 =\sqrt 2 / 2 = 1/\sqrt 2\]

Answer 7

and you got the demostrations in two ways. then it's done. =).

Answer 8

ash began from the right to left, and I did from left to right.

Answer 9

I proved it true?
=\[\sqrt{2}\cos(x-\pi/4)\]
=\[\sqrt{2}(cosx.\cos \pi/4 + sinx.\sin \pi/4)\]
\[=cosx + sinx\]
prove it?

Answer 10

Yeah, this is also true

Answer 11

Hooray for me

Answer 12

Yeah, you did great:D

Answer 13

متشکرم
thanks

Answer 14

hey I'm From Iran

Answer 15

@hamidaarab Welcome:D

Answer 16

thank you very much

Answer 17

mr.ash2326
Do you know physics?

Answer 18

find domain & Range of Function?
f(x)=2sin^-1 (x+1)

Answer 19

@hamidaarab don't have to thank so much. We all are friends here. I'll try but you should post your question in Physics group.

Answer 20

How?
Indeed ,find domain & Range of Function?
f(x)=2sin^-1 (x+1)

Answer 21

may you solve it?

Answer 22

Yeah
we have
\[ f(x) = 2\sin^{-1} (x+1)\]
\[ domain\ of\ \sin^{-1} x\ is\ [-1, 1]\]
\[and\ range\ is (-90)\ degrees\ to\ (90)]\ degrees\]
so domain of (x+1) is [-1,1] or
domain of x is [-2,0]
range of \( \sin^{-1} x\) is [-90, 90] or \([-\pi/2, \pi/2]\)
so range of \( 2\sin^{-1} x\) is [-180,180] or \([-\pi, \pi]\)

Answer 23

thank you very much ash2326