Question

If $\tan \left(\mathrm{A}\right)=\sqrt{2}-1$ then show that $\sin \left(A\right)·\cos \left(A\right)=\frac{\sqrt{2}}{4}$.

Answer 1

Step 1: Divide the given condition into $\sin$ and $\cos$

Given,

$\begin{array}{cc}& \tan \left(A\right)=\sqrt{2}-1\\ \Rightarrow & \frac{\sin \left(A\right)}{\cos \left(A\right)}=\sqrt{2}-1\\ \Rightarrow & \sin \left(A\right)=\cos \left(A\right)\left(\sqrt{2}-1\right)\end{array}$

To prove: $\sin \left(A\right)·\cos \left(A\right)=\frac{\sqrt{2}}{4}$

Proof:

Put the value of $\sin \left(A\right)$ in ${\sin }^2\left(A\right)+{\cos }^2\left(A\right)=1$

$\begin{array}{cc}\Rightarrow & {\left[\cos \left(A\right)\left(\sqrt{2}-1\right)\right]}^2+{\cos }^2\left(A\right)=1\\ \Rightarrow & {\cos }^2\left(A\right){\left(\sqrt{2}-1\right)}^2+{\cos }^2\left(A\right)=1\\ \Rightarrow & {\cos }^2\left(A\right)\left[2+1-2\sqrt{2}+1\right]=1\\ \Rightarrow & \cos \left(A\right)=\pm \frac{1}{\sqrt{4-2\sqrt{2}}}\end{array}$

Step 2: Simplify to find the value of $\sin \left(A\right)·\cos \left(A\right)$

Put the value of $\cos \left(A\right)$ in the expression of $\sin \left(A\right)$

$\begin{array}{cc}\Rightarrow & \sin \left(A\right)=\pm \frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}}\\ \Rightarrow & \sin \left(A\right)·\cos \left(A\right)=\pm \frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}}\times \pm \frac{1}{\sqrt{4-2\sqrt{2}}}\\ \Rightarrow & \sin \left(A\right)·\cos \left(A\right)=\frac{\sqrt{2}-1}{4-2\sqrt{2}}\\ \Rightarrow & \sin \left(A\right)·\cos \left(A\right)=\frac{\sqrt{2}-1}{2\sqrt{2}\left(\sqrt{2}-1\right)}\\ \Rightarrow & \sin \left(A\right)·\cos \left(A\right)=\frac{1}{2\sqrt{2}}\\ \Rightarrow & \sin \left(A\right)·\cos \left(A\right)=\frac{\sqrt{2}}{4}\end{array}$

Hence Proved.