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If sin4x/a+cos4x/b = 1/a+b. Prove that sin8x/a³+cos8x/b³ = (1/a+b)³ |
| Answer» Given : Sin⁴x/a + cos⁴x/b = 1/a+bTo prove :sin⁸x/a³ + cos⁸x/b³ =?Solution:Sin⁴x/a + cos⁴x/b = 1/a+bAs we know that Sin²x +Cos²x=1∴Cos²x=1-sin²xCos⁴x=(Cos²x)²=(1-sin²x)²=1+sin⁴x-2sin²x⇒Sin⁴x/a +[1+sin⁴x-2sin²x]/b =1/a+b⇒Sin⁴x/a+1/b +\xa0sin⁴x/b -2sin²x/b=(1/a+b)⇒Sin⁴x/a+\xa0sin⁴x/b -2sin²x/b=(1/a+b)-1/b⇒Sin⁴x(1/a +1/b)-2sin²xa/ab=[b-a-b]/b(a+b)⇒Sin⁴x(1/a +1/b)-2asin²x/ab=-a/b(a+b)⇒Sin⁴x(a+b)/ab-2asin²x/ab=-a/b(a+b)⇒(a+b)²(Sin²x)² - 2a(a+b)sin²x= - a²⇒(a+b)²(Sin²x)² - 2a(a+b)sin²x + a² =0This equation is similar to a²-2ab+b²=0\xa0⇒(a-b)²[(a+b) Sin²x-a]²=0sin²x=a/a+bCos²x=1-sin²x= 1- a/(a+b)= b/a+b∴sin²x=a/a+b and Cos²x=\xa0b/a+bNow :sin⁸x=a⁴/(a+b)⁴ and Cos⁸x= b⁴/(a+b)⁴sin⁸x/a³= a/(a+b)⁴ and Cos⁸x/b³=b/(a+b)⁴Now\xa0sin⁸x/a³ + Cos⁸x/b³ = a/(a+b)⁴ + b/(a+b)⁴=a+b/(a+b)⁴⇒sin⁸x/a³ + Cos⁸x/b³=1/(a+b)³ | |