Math, asked by shree9148, 1 year ago

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Answered by attinderpaul55225
1

 hello \: friend \\ here \: is \: your \: answer\\ the \: question \: is \: to \: find \: the \\ value \: of \:  \sin(2x)  \\ the \: question \: is \:  \\ cos2x = cos60.cos30 + sin60 \: sin30

we \: know \: that \: the \: relation \\ between \:  \:  \sin( \alpha  ) \:  \:  and \:  \cos( \alpha )  \\   \sin( \alpha )  =  \cos(90 -  \beta )  \\ so \: here \:  \\  \sin(2x)  =  \sin(90 - 2x)  \\  =  \cos(2x)  \\ now \\  \cos(60)  \\  =  \cos(90 - 60)  \\  =  \sin(30)  \\  \cos(30) \\  =  \cos(90 - 60)  \\  =  \sin(30)  \\ so \: we \: can \: write \\ cos2x =  \cos60 \cos30 +  \sin60 \sin30 \\  =   >  \sin2x = sin30sin60 + sin60.sin30 \\  \\  we \: know \: that \: the \: value \: of \\  \sin(30)  =  \frac{1}{2}  \\  \sin(60)  =  \frac{ \sqrt{3} }{2}  \\  \sin \: 30sin60 +  \sin60. \sin30 \\  =  \frac{1}{2}  \times  \frac{ \sqrt{3} }{2}  +  \frac{ \sqrt{3} }{2}  \times  \frac{1}{2}  \\  =  \frac{ \sqrt{3} }{4}  +  \frac{ \sqrt{3} }{4}  \\  =    \frac{ 2\sqrt{3} }{4}  \\  =  \frac{ \sqrt{3} }{2}  \\ so \: we \: get \: that \: the \: vaue \: of \\  \\  =  >  \sin(2x)  =  \frac{ \sqrt{3} }{2}  \\

✨ hope it helps✨✨

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