IIT JEE , AIEEE Forum - Mathematics , Trignometry

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If sin2A=cos3A and A is an acute angle,then sin A is equal to?

edison

sin2A=cos3A

or sin2A=sin(

/2 - 3A)

or 2A = (

/2 - 3A)

or 5A =

/2

or A =

/10

so, sinA = sin

/10

karthik_18049200

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5-1/4

ash.harith

sin2A= cos3A
sin2A=sin(90-3A)
2A=90-3A
5A=90
A=18
so
sinA=sin18....and the value of sin18 is available in the clarks table!

jackiejr

please mathematicians help me this question. prove that 2sin3xsinx=1

jackiejr

sin2A=cos3A , sin2A=sin(90-3A), 2A=90-3A , 5A=90, A=18(hence solved)

rsashwinkumar

hi this is rsa,

sin2A=cos3A

cos(pi/2-2A)=cos3A

Since, A is acute and taking the minimum possible value,

A = pi/10 radians

or

A = 18 degrees