Trig functions help (1 Viewer)

[WEMG]

solve sin3x=cos2x

 

[random-1006]

solve sin3x=cos2x

use double angle results

sin( A+B) = sinAcosB + cosAsinB

cos ( A +B ) = cosAcosB -sinAsinB


sin3x= 3sinx -4(sinx)^3

cos2x= (cosx)^2 -(sinx)^2= 1-2(sinx)^2

so 3sinx-4(sinx)^3 = 1-2(sinx)^2
4(sinx)^3-2(sinx)^2-3sinx +1 =0, let u= sinx'

4u^3 -2u^2 -3u +1=0


you cannot factor this, donty think you can solve it ( for exact values anyway)

 

[NewiJapper]

are we supposed to know double angle results for 2 unit?

 

[random-1006]

are we supposed to know double angle results for 2 unit?

no

they posted in the wrong forum

all 2 unit needs is SOH CAH TOA { im pretty sure}

sin^2 + cos^2 =1 ( and other two of these identities)

sin rule a/ sinA= b/ sinB

and cosine rule a^2 =b^2 +c^2 -2bccosA

 

[hscishard]

sin3x = sin(2x+x)
Then make everything into sinx

 

[pi-ka-chew]

use double angle results

sin( A+B) = sinAcosB + cosAsinB

cos ( A +B ) = cosAcosB -sinAsinB


sin3x= 3sinx -4(sinx)^3

cos2x= (cosx)^2 -(sinx)^2= 1-2(sinx)^2

so 3sinx-4(sinx)^3 = 1-2(sinx)^2
4(sinx)^3-2(sinx)^2-3sinx +1 =0, let u= sinx'

4u^3 -2u^2 -3u +1=0


you cannot factor this, donty think you can solve it ( for exact values anyway)

that can be factorised into:
(u - 1)(4u^2 + 2u - 1) = 0

then you get u = 1 and the values you get by applying quadratic formula.
sub sinx back into u and solve the x.

 

[random-1006]

that can be factorised into:
(u - 1)(4u^2 + 2u - 1) = 0

then you get u = 1 and the values you get by applying quadratic formula.
sub sinx back into u and solve the x.

lol, i dont know why i didnt see that u=1 was a root!

 

[pi-ka-chew]

lol, i dont know why i didnt see that u=1 was a root!

don't worry, we often don't see the straightforward answers. i normally work around in circles. xD