trignometry confusion

**grgrsanjay** · October 17th 2010, 05:10 AM

1]the equation $1-\cos^ x + \sinx\cosx = k$ then there exists real solution when k belongs to

A) $0<k<\frac{1+\sqrt2}{2}$
B) $2-\sqrt3 < k 2+\sqrt3$
C) $0<k<2-\sqrt3$
D) $\frac{1-\sqrt2}{2} < k < \frac{1+\sqrt2}{2}$

2]let a,b belong to the set [0, $\frac{\pi}{2}$ ] if
$sin^6 a+3\sin^2 a\cos^2 a+\cos^6 b=1$

then $\frac{255a+b}{a+b} =?$

**Unknown008** · October 17th 2010, 05:34 AM

Ok. Is that it?

$1-\cos^2 (x) + \sin (x) \cos (x) = k$

Well...

$1 - \cos^2(x) = \sin^2(x)$

$\sin^2(x)+ \sin (x) \cos (x) = k$

$\dfrac12 \cos(2x) + \dfrac12 + \dfrac12\sin(2x) = k$

$\dfrac12 (1 + \cos(2x) + \sin(2x)) = k$

Now,

$\cos(2x) + \sin(2x) = R\sin(2x + \alpha)$

$R = \sqrt{1^2 + 1^2} = \sqrt2$

$\alpha = \tan^{-1}\left(\dfrac11\right) = \dfrac{\pi}{4}$

$\cos(2x) + \sin(2x) = \sqrt2 \sin(2x + \frac{\pi}{4})$

Hence;

$\dfrac12 (1 + \sqrt2\sin(2x + \frac{\pi}{4})) = k$

$\sin(2x + \frac{\pi}{4}) = \dfrac{2k-1}{\sqrt2}$

So,

$-1 < \dfrac{2k-1}{\sqrt2} < 1$

Can you complete it now?

And your second question is... difficult to understand.

**grgrsanjay** · October 17th 2010, 04:54 PM

in the second question i substituted $\frac{\pi}{2}$ for a and b and got answer as 128

need to prove this is true for all values

**sa-ri-ga-ma** · October 17th 2010, 05:01 PM

You are right.

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