Question

Find the value of \[\cos q\] , if the value of \[\sin q\] is equal to \[\dfrac{a}{b}\].

Answer 1

Hint: Assume a right-angled triangle, \[\Delta ABC\] such that the angle at the vertex C is q and right-angled at B. We know the formula for the sine ratio in a triangle, \[\sin \theta =\dfrac{Perpendicular}{Hypotenuse}\] . It is given that, \[\sin q=\dfrac{a}{b}\] . Now, compare \[\sin \theta =\dfrac{Perpendicular}{Hypotenuse}\] and \[\sin q=\dfrac{a}{b}\] . Then, get the values of height, hypotenuse, and \[\theta \] . Now, use the Pythagoras theorem \[{{\left( Hypotenuse \right)}^{2}}={{\left( Height \right)}^{2}}+{{\left( Base \right)}^{2}}\] and get the value of the base. In \[\Delta ABC\] , we have \[\operatorname{cosq}=\dfrac{Base}{Hypotenuse}\] . Then, put the values of the base and hypotenuse.

Complete step by step solution:
According to the question, we have the value of \[\sin q\] and we have to find the value of \[\cos q\] .
First of all, let us assume a right-angled triangle, \[\Delta ABC\] and the angle at the vertex C be q. The \[\Delta ABC\] is right-angled at B that is, the angle at the vertex B of \[\Delta ABC\] is \[90{}^\circ \] .

It is given that,
\[\sin q=\dfrac{a}{b}\] ………………….(1)
We know the formula for the sine ratio in a triangle, \[\sin \theta =\dfrac{Perpendicular}{Hypotenuse}\] ………………(2)
On comparing equation (1) and equation (2), we get
Perpendicular = a ………………………..(3)
Hypotenuse = b ……………………………(4)
\[\theta =q\] …………………………………(5)
We also know the formula for the cosine ratio in a triangle, \[\cos \theta =\dfrac{Base}{Hypotenuse}\] ………………(6)
From equation (5), we have \[\theta =q\] .
Now, putting \[\theta =q\] in equation (6), we get
\[\operatorname{cosq}=\dfrac{Base}{Hypotenuse}\] ……………………………..(7)
Here, we need the value of the length of the base of \[\Delta ABC\] .
Since \[\Delta ABC\] is a right-angled triangle, we can apply Pythagoras theorem here.
Using Pythagoras theorem,
\[{{\left( Hypotenuse \right)}^{2}}={{\left( Perpendicular \right)}^{2}}+{{\left( Base \right)}^{2}}\] ……………………….(8)
From equation (3) and equation (4), we have the values of the hypotenuse and height of the \[\Delta ABC\] .
Perpendicular = a,
Hypotenuse = b.
Now, putting the values of the height and hypotenuse in equation (8), we get
\[\begin{align}
  & \Rightarrow {{\left( b \right)}^{2}}={{\left( a \right)}^{2}}+{{\left( Base \right)}^{2}} \\
 & \Rightarrow {{b}^{2}}-{{a}^{2}}={{\left( Base \right)}^{2}} \\
\end{align}\]
Takin square root in the above equation, we get
\[\Rightarrow \sqrt{\left( {{b}^{2}}-{{a}^{2}} \right)}=Base\] ……………………(9)
Now, in \[\Delta ABC\] , we have
Hypotenuse = AC = b ………………………..(10)
Perpendicular = AB = a ………………………….(11)
Base = BC = \[\sqrt{\left( {{b}^{2}}-{{a}^{2}} \right)}\] ………………….(12)
From equation (7), we have
\[\operatorname{cosq}=\dfrac{Base}{Hypotenuse}\] .
Now, putting the values of the base from equation (12) and hypotenuse from equation (10), in equation (7), we get
\[\operatorname{cosq}=\dfrac{\sqrt{\left( {{b}^{2}}-{{a}^{2}} \right)}}{b}\] ………………………………..(13)
Hence, the value of \[\cos q\] is \[\dfrac{\sqrt{\left( {{b}^{2}}-{{a}^{2}} \right)}}{b}\] .

Note: We can also solve this question by another method.
It is given that the value of \[\sin q\] is equal to \[\dfrac{a}{b}\] . So,
\[\sin q=\dfrac{a}{b}\] ……………………………(1)
We know the identity, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] …………………………….(2)
Replacing \[\theta \] by q in equation (2), we get
\[{{\sin }^{2}}q+{{\cos }^{2}}q=1\] ……………………………(3)
Now, squaring the LHS and RHS of equation (1), we get
\[{{\left( \sin q \right)}^{2}}={{\left( \dfrac{a}{b} \right)}^{2}}\]
\[\Rightarrow {{\sin }^{2}}q=\dfrac{{{a}^{2}}}{{{b}^{2}}}\] ………………………………(4)
Putting the value of \[{{\sin }^{2}}q\] from equation (4), in equation (3), we get
\[\begin{align}
  & \dfrac{{{a}^{2}}}{{{b}^{2}}}+{{\cos }^{2}}q=1 \\
 & \Rightarrow {{\cos }^{2}}q=1-\dfrac{{{a}^{2}}}{{{b}^{2}}} \\
\end{align}\]
\[\Rightarrow {{\cos }^{2}}q=\dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}}\] …………………………..(5)
Now, taking square root in LHS and RHS of equation (5), we get
\[\Rightarrow {{\left( {{\cos }^{2}}q \right)}^{\dfrac{1}{2}}}=\dfrac{{{\left( {{b}^{2}}-{{a}^{2}} \right)}^{\dfrac{1}{2}}}}{{{\left( {{b}^{2}} \right)}^{\dfrac{1}{2}}}}\]
\[\Rightarrow {{\left( {{\cos }^{2}}q \right)}^{\dfrac{1}{2}}}=\dfrac{\sqrt{\left( {{b}^{2}}-{{a}^{2}} \right)}}{{{\left( {{b}^{2}} \right)}^{\dfrac{1}{2}}}}\] ………………………………(6)
We know the formula, \[{{\left( {{x}^{m}} \right)}^{n}}={{x}^{mn}}\] …………………………..(7)
Now, using the formula shown in equation (7), we can simplify equation (6).
Simplifying equation (7), we get
\[\Rightarrow {{\left( \cos q \right)}^{2\times \dfrac{1}{2}}}=\dfrac{\sqrt{\left( {{b}^{2}}-{{a}^{2}} \right)}}{{{\left( b \right)}^{2\times \dfrac{1}{2}}}}\]
\[\Rightarrow \cos q=\dfrac{\sqrt{\left( {{b}^{2}}-{{a}^{2}} \right)}}{b}\] ………………………………(8)
From equation (8), we have got the value of \[\cos q\] .
Hence, the value of \[\cos q\] is \[\dfrac{\sqrt{\left( {{b}^{2}}-{{a}^{2}} \right)}}{b}\] .