if sinx+cosx=√3cosx then 2cosx-sinx=
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Answered by
7
sinx+cosx=√3cosx
sinx= √3cosx-cosx
sinx=(√3-1)cosx
Now,
2cosx-sinx
=2cosx-(√3-1)cosx
=2cosx-√3cosx+1cosx
=(3-√3)cosx
sinx= √3cosx-cosx
sinx=(√3-1)cosx
Now,
2cosx-sinx
=2cosx-(√3-1)cosx
=2cosx-√3cosx+1cosx
=(3-√3)cosx
it's not completed.. I hope
yes
if we manage to find tan'(3^2-1) then we can find everything
Answered by
0
sinx+cosx=3^1/2cosx
imply
tanx=(3^1/2)-1
we know 1+tan^2x=sec^2x
cosx=1/(1+tan^x)^1/
sinx (1-cos^2x)^1/2
sinx=(3^1/2-1)/(5-2 (3^1/2))
cosx=1/(5-2 (3^1/2))
2cosx-Sinx
=(2-(3^1/2)+1)/(5-2 (3^1/2))
=(3-(3^1/2))/(5-2 (3^1/2))
imply
tanx=(3^1/2)-1
we know 1+tan^2x=sec^2x
cosx=1/(1+tan^x)^1/
sinx (1-cos^2x)^1/2
sinx=(3^1/2-1)/(5-2 (3^1/2))
cosx=1/(5-2 (3^1/2))
2cosx-Sinx
=(2-(3^1/2)+1)/(5-2 (3^1/2))
=(3-(3^1/2))/(5-2 (3^1/2))
hope this is right
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