sina=1/√5 and sinb=1/√10 find (a+b)
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sina=1/√5 so cosa=2/√5.
sinb=1/√10 so cosb=3/√10.
see in image.....
now we have....
sin(a+b)=sina.cosb +cosa.sinb
or sin(a+b)=(1/√5)(2/√5) +(1/√10)(3/√10)
=> sin(a+b)=2/5 + 3/10
=> sin(a+b)=7/10
=> a+b=sin^-1(7/10)
sinb=1/√10 so cosb=3/√10.
see in image.....
now we have....
sin(a+b)=sina.cosb +cosa.sinb
or sin(a+b)=(1/√5)(2/√5) +(1/√10)(3/√10)
=> sin(a+b)=2/5 + 3/10
=> sin(a+b)=7/10
=> a+b=sin^-1(7/10)
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amitabhapal196p3tutz:
theres an easier way for getting the correct answer
what is that?
the derived values of sina,cosa,sinb n cosb will remain da same but at the end u have to use cos(a+b) formula
this is almost the same way .
yes but its better to express the value of some angle in trigonometry in degrees or radians
you are Right Amitabh.
thnx the broly!
btw m unable to msg u
can u plz txt me so dat in future we can discuss problms
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