Let a and b be rel numbers such that
sin a+sin b=(√2)/2 and cos a+cos b =(√6)/2 then sin(a+b) will be ????
3sorry!! by mistake i have posted this question in ' class 9-10 mathematics ' so i have posted the same
qstn here again.............!!!!!!!!!!!!!
62an alternate proof
\\e^{ia}+e^{ib}=\frac{\sqrt{3}+i}{\sqrt{2}} \\e^{-ia}+e^{-ib}=\frac{\sqrt{3}-i}{\sqrt{2}} \\\frac{e^{ia}+e^{ib}}{e^{i(a+b)}}=\frac{\sqrt{3}-i}{\sqrt{2}} \\e^{ia}+e^{ib}=\frac{\sqrt{3}-i}{\sqrt{2}}\times {e^{i(a+b)}} \\\frac{\sqrt{3}+i}{\sqrt{2}}=\frac{\sqrt{3}-i}{\sqrt{2}}\times {e^{i(a+b)}} \\{e^{i(a+b)}}=\frac{\sqrt{3}+i}{\sqrt{3}-i}=1/2+\sqrt{3}/2i
hence cos (a+b) = 1/2
sin (a+b) = √3/2
[1]
1sin a+sin b= √22
2 sin(a+b)2cos(a-b)2 = √22........(i)
cos a +cos b = √62
2 cos(a+b)2cos(a-b)2 = √62............(ii)
Dividing (i) by (ii)
tan(a+b)2 = 1√3
Taking principal value ( answer match karna hai yaar....nehi to general values liya hota)
a+b2 = 30°
a+b = 60°
sin(a+b) = √32 (Ans.)
62can you comment on Taking principal value ( answer match karna hai yaar....nehi to general values liya hota) ???
1Sir, not only 30° but many values of this expression satisfies the equation.....
else
(a+b)2 = nπ + 30°
a+b = 2nπ + 60 °
But of course sin(2nπ+60)° would have returned the same value for all integer values of n.
sin(2nπ+60)° = √32