Sin(40+theta)cos(10+theta)-cos(40+theta)sin(10+theta)=1/2
Answers
Answered by
160
Hi ,
we know that ,
************************************************
sin A cos B - sin B cos A = sin ( A - B )
*************************************************
let A = 40 + θ,
B = 10 + θ,
According to the problem ,
sin( 40 + θ )cos( 10 + θ ) - sin( 10 + θ)cos( 40+θ)
= sin A cos B - sin B cos A
= sin ( A - B )
put A and B
= sin [ 40 + θ - ( 10 + θ ) ]
= sin ( 40 + θ - 10 - θ )
= sin (30)
= 1/2
i hope this will useful to you.
*****
we know that ,
************************************************
sin A cos B - sin B cos A = sin ( A - B )
*************************************************
let A = 40 + θ,
B = 10 + θ,
According to the problem ,
sin( 40 + θ )cos( 10 + θ ) - sin( 10 + θ)cos( 40+θ)
= sin A cos B - sin B cos A
= sin ( A - B )
put A and B
= sin [ 40 + θ - ( 10 + θ ) ]
= sin ( 40 + θ - 10 - θ )
= sin (30)
= 1/2
i hope this will useful to you.
*****
Answered by
43
Hi friend,
given is of the form sinAcosB-cosAsinB=sin(A-B)
here A=40+θ, B=10+θ
So sin(40+θ-10-θ)
=sin30=1/2
Therefore LHS =RHS
given is of the form sinAcosB-cosAsinB=sin(A-B)
here A=40+θ, B=10+θ
So sin(40+θ-10-θ)
=sin30=1/2
Therefore LHS =RHS
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