Question -
Answer -
Given that f(x) =sin┬аx,┬аg┬а(x) = 2x┬аand┬аh┬а(x) = cos┬аx
We┬аknow┬аthat┬аf:┬аRтЖТ[тИТ1,┬а1]┬аand┬аg:┬аRтЖТ R
Clearly,┬аthe┬аrange┬аof┬аg┬аis┬аa┬аsubset┬аof┬аthe┬аdomain┬аof┬аf.
fog:┬аR┬атЖТ┬аR
Now,┬а(fh)┬а(x) = f┬а(x)h┬а(x)┬а=┬а(sin┬аx)┬а(cos┬аx)┬а= ┬╜ sin┬а(2x)
Domain┬аof┬аfh┬аis┬аR.
Since┬аrange┬аof┬аsin┬аx┬аis┬а[-1,1], тИТ1┬атЙд┬аsin┬а2x┬атЙд┬а1
тЗТ -1/2 тЙд sin x/2 тЙд 1/2
Range┬аof┬аfh┬а= [-1/2, 1/2]
So,┬а(fh):┬аR┬атЖТ [(-1)/2, 1/2]
Clearly,┬аrange┬аof┬аfh┬аis┬аa┬аsubset┬аof┬аg.
тЗТ┬аgo┬а(fh):┬аR┬атЖТ┬аR
тЗТDomains┬аof┬аfog┬аand┬аgo┬а(fh)┬аare┬аthe┬аsame.
So,┬а(fog)┬а(x)= f┬а(g┬а(x))┬а
=┬аf┬а(2x)┬а
=┬аsin┬а(2x)
And┬а(go┬а(fh))┬а(x)┬а=┬аg┬а((f(x). h(x))┬а
=┬аg┬а(sinx┬аcos┬аx)┬а
=┬а2sin┬аx┬аcos┬аx┬а
=┬аsin┬а(2x)
тЗТ┬а(fog)┬а(x)┬а=┬а(go(f h))┬а(x),┬атИАx┬атИИ┬аR
Hence,┬аfog┬а=┬аgo┬а(fh)