proof? sinx+cosx=√2 sin(x+pi/4)
sum formula sin(a+b) = sina cosb + sinb cosa sin(x+pi/4) = cos(pi/4)sinx + sin(pi/4)cosx cos(pi/4) = sin(pi/4) = sqrt2/2 sin(x+pi/4) = sqrt2/2 (sinx +cosx) multiply by sqrt 2 and there you go
You see it right?
dumbcows derivation seems good.
\[\sqrt{2}\cos(x-\pi/4)=\sqrt{2}(cosx.\cos \pi/4 + sinx.\sin \pi/4) =cosx+sinx\] hey! \[\cos \pi/4=\cos \sqrt{2}/2 \sin \pi/4=si \sqrt{2}/2\]
Please note, it is useful to remember the general identity \[ \large \cos x \pm \sin x = \sqrt2 \sin(\frac{\pi}{4} \pm x) = \sqrt2 \cos( \frac{\pi}{4} \mp x) \]
foolformath, never had to learn that identity
This right?
@dumbdow Useful in competitive mathematics. (saves precious seconds during a contest)
With this software(Matlab) can I solve it?
solve what?
solve means proof
Hey, did you prove it?
i don't know if matlab will prove trig identities ?
Did you use it?
best way... divide and multiply by \[\sqrt{2}\] ...it becomes \[\sqrt{2}(1/\sqrt{2}sinx+ 1/\sqrt{2}cosx)\] = \[\sqrt{2}\](sin ( pie/4 +x)
Akshat did you prove it?
if equation is like asinx + bcosx then divide and multiply by \[\sqrt{a^2 + b^2}\] ...then convert it to the basic form of sin(x+\[\alpha\]
i showed the steps to prove equality in 1st post
Slim's dumbcow
dumbcow! I'm From Iran and I don't type english
Who did it prove?
oh ok , i'll redo it in easier form \[\sqrt{2}\sin(x+\pi/4) = \sin x + \cos x\] \[\sqrt{2}(\sqrt{2}/2\sin x + \sqrt{2}/2\cos x) = \sin x + \cos x\] \[\sqrt{2}*\sqrt{2}/2(\sin x +\cos x) = \sin x +\cos x\] \[\sin x + \cos x = \sin x +\cos x\]
Oh thank you! I have a surprise for you(dumbcow)?
I'll send you a file now
Download It Now
umm what is it...if want to share something just attach it to this post
I wrote some of the photos is Persian
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