Problem #8

$sinx+cosx=\frac{1}{\sqrt3}$

or, $1+sin2x = \frac{1}{3}$

or, $sin2x=-\frac{2}{3}$ .

So, $cos2x=\pm \frac{\sqrt{5}}{3}$ .

Now, $(sin4x)^2 = (2\,sin2x\; cos2x)^2 = \frac{80}{81}$ .

So, $n=81, \: m=80$ and $n-m=\boxed{1}$

I have just put the value of 'x' as 15 degree.

so (sin4x)^2

=(sin60)^2

=[sqrt(3)/2]^2

=3/4

so n=4, m=3

Hence, n-m = 4-3 = 1

After All, the answer is 1 !! and the name of set is "EASY PROBLEMS"

We have: $(sin(x) + cos(x))^2=\frac{1}{3}$

$\Rightarrow sin^2(x) + cos^2(x)+2sin(x)cos(x)=\frac{1}{3}$

$\Rightarrow 1+sin(2x)=\frac{1}{3}$

$\Rightarrow sin(2x)=-\frac{2}{3}$

Then we have: $cos(4x)=1-2sin^2(2x)=1-2(-\frac{2}{3})^2=\frac{1}{9}$

From there we get: $(sin^2(4x))=1-cos^2(4x)=1-(\frac{1}{9})^2=\frac{80}{81}$

With $\frac{m}{n}=\frac{80}{81}$ , we finally have $n-m=81-80=\boxed{1}$

sin 4x can be written as 4 sin^2 (2x) cos^2 (2x) . squaring the given condition , we get 1+sin^2 (2x) = 1/3 .so we calculate the value of sin^2 (2x) .also we can find the value of cos^2 (2x)...hence put these values in the above said expression to get the answer.