Find the value of #sin(120^@-45^@)#?

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Answer 1 · 2017-03-31T05:20:56

#sin(120^@-45^@)=(sqrt3+1)/(2sqrt2)#

#sin(A-B)=sinAcosB-cosAsinB#

Here if #A=120^@# and #B=45^@#, we can find #sin(120^@-45^@)#,

but we need to know sine and cosine ratios of the two angles.

Now as #sin(180^@-A)=sinA# but #cos(180^@-A)=-cosA#

as such #sin120^@=sin60^@=sqrt3/2# and #cos120^@=-cos60^@=-1/2# and as #sin45^@=cos45^@=1/sqrt2#

Hence #sin(120^@-45^@)#

#=sin120^@cos45^@-cos120^@sin45^@#

#=sqrt3/2xx1/sqrt2-(-1/2)xx1/sqrt2#

#=sqrt3/(2sqrt2)+1/(2sqrt2)#

#=(sqrt3+1)/(2sqrt2)#