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 $sin(120^@-45^@)$

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

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

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

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

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