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How can i prove that 6cos(x+45) cos(x-45) is equal to 3cosx?
kbr1804 said:How can i prove that 6cos(x+45) cos(x-45) is equal to 3cosx?
skeeter said:use the sum/difference identity $\cos(a \pm b) = \cos{a}\cos{b} \mp \sin{a}\sin{b}$
it should be equal to $3\cos(2x)$, not $3\cos{x}$
Three graphs; shouldn't 2 of them have the same graph?kbr1804 said:yeah i think i got it lol thanks alot:)and yeah it was supposed to equal to 3cos2x that was a typo
Evgeny.Makarov said:jonah, in Desmos one should write $\pi/4$ instead of 45.
Evgeny.Makarov said:jonah, in Desmos one should write $\pi/4$ instead of 45.
Well aware of that.skeeter said:One can change to degree mode with the "wrench" button menu
I've often wondered what kind of platform this type of Desmos "quoting" is ever since I saw one of Klaas van Aarsen's post which used the same method. It isn't just a link to Desmos as I found out when I hit the reply tab on my phone. Is it that the TikZ thingamajig I've been seeing a lot lately on this site? I think I remember copying that stuff in another math site but was surprised that it didn't work there.skeeter said:... and on Desmos ...
[DESMOS]{"version":7,"graph":{"viewport":{"xmin":-180,"ymin":-10.762090536086637,"xmax":180,"ymax":10.762090536086637},"degreeMode":true,"squareAxes":false},"randomSeed":"78931067dd5aa2f77a194c669752ab59","expressions":{"list":[{"type":"expression","id":"1","color":"#c74440","latex":"6\\cos\\left(x+45\\right)\\cos\\left(x-45\\right)\\left\\{x>0\\right\\}"},{"type":"expression","id":"2","color":"#2d70b3","latex":"3\\cos\\left(2x\\right)\\left\\{x<0\\right\\}"},{"type":"expression","id":"3","color":"#388c46"}]}}[/DESMOS]
You CAN'T- it's not true! For example if x= 45 degrees this becomes 6 cos(90)cos(0)= 6(0)(1)= 0 but 3 cos(45)= 3sqrt(2)/2.kbr1804 said:How can i prove that 6cos(x+45) cos(x-45) is equal to 3cosx?
Country Boy said:You CAN'T- it's not true! For example if x= 45 degrees this becomes 6 cos(90)cos(0)= 6(0)(1)= 0 but 3 cos(45)= 3sqrt(2)/2.
skeeter said:use the sum/difference identity $\cos(a \pm b) = \cos{a}\cos{b} \mp \sin{a}\sin{b}$
it should be equal to $3\cos(2x)$, not $3\cos{x}$