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If $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = $

$\bf{My\; Try::}$ Given $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\Rightarrow 1+\sin \phi\cdot \cos \phi+\sin \phi+\cos \phi = \frac{5}{4}.$

So $\displaystyle \sin \phi+\cos \phi+\sin \phi \cdot \cos \phi = \frac{1}{4}\Rightarrow \left(\sin \phi+\cos \phi\right) = \frac{1}{4}-\sin \phi \cdot \cos \phi.$

Now $\displaystyle \left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = 1-\left(\sin \phi+\cos \phi\right)+\sin \phi \cdot \cos \phi =\frac{3}{4}+\sin 2\phi$

Now How can I calculate $\sin 2\phi.$

Help me, Thanks

juantheron
  • 53,015

3 Answers3

6

Like Sameer Kailasa,

$\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \dfrac{5}{4}\ \ \ \ (1)$

and let $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) =x \ \ \ \ (2)$

$(1)-(2)\implies2(\sin \phi+\cos\phi)=\dfrac{5}{4}-x$

$\implies \sin \phi+\cos\phi=\dfrac{5-4x}8 $

Squaring we get $1+2\sin \phi\cos\phi=\left(\dfrac{5-4x}8\right)^2\iff2\sin \phi\cos\phi=\left(\dfrac{5-4x}8\right)^2-1$

$(1)+(2)\implies2(1+\sin \phi\cos\phi)=x+\dfrac{5}{4}\implies 2\sin \phi\cos\phi=\dfrac{4x-3}4$

Equate the values of $2\sin \phi\cos\phi$

lab bhattacharjee
  • 274,582
5

Let $$A = (1+\sin(\theta))(1+\cos(\theta))$$ and $$B = (1-\sin(\theta))(1-\cos(\theta))$$ Observe that $$A+B = 2+2\sin(\theta)\cos(\theta)$$ $$AB = (\sin(\theta)\cos(\theta))^2$$

Substituting $A = 5/4$, we now have a quadratic in $B$ that is easily solved.

Sameer Kailasa
  • 8,983
2

HINT:

$$\dfrac54=1+\sin x+\cos x+\sin x\cos x=1+\sqrt2\cos\left(\frac\pi4-x\right)+\dfrac{2\cos^2\left(\dfrac\pi4-x\right)-1}2$$

Write $\cos\left(\frac\pi4-x\right)=u$

$$4u^2+4\sqrt2u-3=0$$

We need $-1\le u\le1$

lab bhattacharjee
  • 274,582