Question

If $ \tan \theta =\dfrac{a-b}{a+b} $ , then find $ \sin \theta $ in terms of $ a $ and $ b $ . \[\]

Answer 1

Hint: W recall the definition of sine and tangent trigonometric ratios from right triangle ABC. We assume the length of opposite side to the angle $ \theta $ as $ p=a-b $ and length of adjacent side as $ b=a+b $ and then use Pythagoras theorem to find the hypotenuse $ h $ . We find $ \sin \theta =\dfrac{p}{h} $ .\[\]

Complete step by step answer:
We know that in right-angled triangle the side opposite to right-angled triangle is called hypotenuse denoted as $ AC=h $, the vertical side is called perpendicular denoted as $ AB=p $ and the horizontal side is called the base denoted as $ BC=b $ . \[\]

We know from the trigonometric ratios in a right angled triangle the sine of any angle is given by the ratio of side opposite to the angle to the hypotenuse. In the figure the sine of the angle $ \theta $ is given by
\[\sin \theta =\dfrac{AB}{AC}=\dfrac{p}{h}\]
The tangent of the angle is the ratio of opposite side to the adjacent side (excluding hypotenuse) . So we have tangent of the angle of angle $ \theta $
\[\tan \theta =\dfrac{AB}{AC}=\dfrac{p}{b}\]
 We are given in the question
\[\tan \theta =\dfrac{a-b}{a+b}\]
Let us assign $ p=a-b,b=a+b $ .We know from Pythagoras theorem that in a right-angled triangle the square of hypotenuse is sum of squares of other two sides which means
\[\begin{align}
  & {{p}^{2}}+{{b}^{2}}={{h}^{2}} \\
 & \Rightarrow {{\left( a-b \right)}^{2}}+{{\left( a+b \right)}^{2}}={{h}^{2}} \\
 & \Rightarrow 2\left( {{a}^{2}}+{{b}^{2}} \right)={{h}^{2}} \\
\end{align}\]
We take square root both sides to have;
\[\Rightarrow h=\sqrt{2\left( {{a}^{2}}+{{b}^{2}} \right)}\]
So the ratio sine can be expressed as;
\[\begin{align}
  & \sin \theta =\dfrac{p}{h} \\
 & \Rightarrow \sin \theta =\dfrac{a-b}{\sqrt{2\left( {{a}^{2}}+{{b}^{2}} \right)}} \\
\end{align}\]

Note:
 We note that since $ \tan \theta $ is well-defined in the question $ a+b\ne 0 $ and hence $ {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\ne 0 $ . Then our obtained value $ \sin \theta =\dfrac{a-b}{\sqrt{2\left( {{a}^{2}}+{{b}^{2}} \right)}} $ is well defined because $ {{a}^{2}}\ge 0,{{b}^{2}}\ge 0 $ . We must be careful of the confusion between algebraic identities $ {{\left( a+b \right)}^{2}}+{{\left( a-b \right)}^{2}}=2\left( {{a}^{2}}+{{b}^{2}} \right) $ and $ {{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}=4ab $ . We can alternatively solve we know the Pythagorean trigonometric identity $ {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ to have
\[\begin{align}
  & \sec \theta =\sqrt{1+{{\left( \dfrac{a-b}{a+b} \right)}^{2}}} \\
 & \Rightarrow \sec \theta =\sqrt{\dfrac{{{\left( a+b \right)}^{2}}+{{\left( a-b \right)}^{2}}}{{{\left( a+b \right)}^{2}}}} \\
 & \Rightarrow \sec \theta =\sqrt{\dfrac{2\left( {{a}^{2}}+{{b}^{2}} \right)}{{{\left( a+b \right)}^{2}}}} \\
 & \Rightarrow \sec \theta =\dfrac{\sqrt{2\left( {{a}^{2}}+{{b}^{2}} \right)}}{\left( a+b \right)} \\
\end{align}\]
We use reciprocal relationship between secant and cosine that is $ \cos \theta =\dfrac{1}{\sec \theta } $ to have;
\[\begin{align}
  & \cos \theta =\dfrac{1}{\sec \theta }=\dfrac{a+b}{\sqrt{2\left( {{a}^{2}}+{{b}^{2}} \right)}} \\
 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{{{\left( a+b \right)}^{2}}}{2\left( {{a}^{2}}+{{b}^{2}} \right)} \\
\end{align}\]
We use Pythagorean trigonometric identity $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ to find required $ \sin \theta $ as
\[\begin{align}
  & \sin \theta =\sqrt{1-{{\cos }^{2}}\theta } \\
 & \Rightarrow \sin \theta =\sqrt{1-\dfrac{{{\left( a+b \right)}^{2}}}{2\left( {{a}^{2}}+{{b}^{2}} \right)}} \\
 & \Rightarrow \sin \theta =\sqrt{\dfrac{2\left( {{a}^{2}}+{{b}^{2}} \right)-{{\left( a+b \right)}^{2}}}{2\left( {{a}^{2}}+{{b}^{2}} \right)}} \\
\end{align}\]
We use the algebraic identity $ {{\left( a+b \right)}^{2}}+{{\left( a-b \right)}^{2}}=2\left( {{a}^{2}}+{{b}^{2}} \right) $ to have
\[\begin{align}
  & \Rightarrow \sin \theta =\sqrt{\dfrac{{{\left( a-b \right)}^{2}}}{2\left( {{a}^{2}}+{{b}^{2}} \right)}} \\
 & \Rightarrow \sin \theta =\dfrac{a-b}{\sqrt{2\left( {{a}^{2}}+{{b}^{2}} \right)}} \\
\end{align}\]