Question

Given that $\cos \dfrac{x}{2} = \dfrac{{12}}{{13}}$ and $x$ lies in first quadrant, calculate without the use of tables, the values of $\sin x$, $\cos x$ and $\tan x$

Answer 1

Here, we will use a half angle formula to find the value of $\cos x$. Similarly, we will find the value of $\sin x$ by using the half angle formula. Then we will find the value of $\tan x$ by simply dividing them using the property of $\tan x$.

Formula used:
1. $\cos 2\theta = 2{\cos ^2}\theta - 1$
2. ${\sin ^2}\dfrac{x}{2} + {\cos ^2}\dfrac{x}{2} = 1$
3. $\sin 2\theta = 2\sin \theta \cos \theta$
4. $\tan x = \dfrac{{\sin x}}{{\cos x}}$

Complete step-by-step answer:
According to the question, $\cos \dfrac{x}{2} = \dfrac{{12}}{{13}}$.
Now, by using the half angle formula, $\cos 2\theta = 2{\cos ^2}\theta - 1$.
$\cos x = 2{\cos ^2}\dfrac{x}{2} - 1$…………………………………..$\left( 1 \right)$
Therefore, from equation $\left( 1 \right)$, we get
$\cos x = 2{\left( {\dfrac{{12}}{{13}}} \right)^2} - 1 = 2\left( {\dfrac{{144}}{{169}}} \right) - 1$
$\Rightarrow \cos x = \left( {\dfrac{{288 - 169}}{{169}}} \right) = \dfrac{{119}}{{169}}$
Therefore, the value of $\cos x = \dfrac{{119}}{{169}}$…………………………$\left( 2 \right)$
Now, we know that, ${\sin ^2}\dfrac{x}{2} + {\cos ^2}\dfrac{x}{2} = 1$.
$\Rightarrow {\sin ^2}\dfrac{x}{2} = 1 - {\cos ^2}\dfrac{x}{2}$
But, $\cos \dfrac{x}{2} = \dfrac{{12}}{{13}}$, hence,
$\Rightarrow {\sin ^2}\dfrac{x}{2} = 1 - {\left( {\dfrac{{12}}{{13}}} \right)^2} = 1 - \dfrac{{144}}{{169}}$
$\Rightarrow {\sin ^2}\dfrac{x}{2} = \dfrac{{169 - 144}}{{169}} = \dfrac{{25}}{{169}}$
Taking square root on both sides, we get
$\Rightarrow \sin \dfrac{x}{2} = \sqrt {\dfrac{{25}}{{169}}} = \dfrac{5}{{13}}$
Substituting $\sin \dfrac{x}{2} = \dfrac{5}{{13}}$ and $\cos \dfrac{x}{2} = \dfrac{{12}}{{13}}$ in the formula $\sin x = 2\sin \dfrac{x}{2} \times \cos \dfrac{x}{2}$, we get
$\Rightarrow \sin x = 2\left( {\dfrac{5}{{13}}} \right)\left( {\dfrac{{12}}{{13}}} \right)$
Hence, solving further, we get
$\Rightarrow \sin x = \dfrac{{120}}{{169}}$………………………………$\left( 3 \right)$
Therefore, the value of $\sin x = \dfrac{{120}}{{169}}$
Now, we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$.
Hence, from equation $\left( 2 \right)$and $\left( 3 \right)$, we get
$\tan x = \dfrac{{\dfrac{{120}}{{169}}}}{{\dfrac{{119}}{{169}}}} = \dfrac{{120}}{{119}}$
Therefore, without the use of tables, we have calculated the values of $\sin x$, $\cos x$ and $\tan x$as $\dfrac{{120}}{{169}}$,$\dfrac{{119}}{{169}}$and $\dfrac{{120}}{{119}}$ respectively.
This is the required answer.

Note: We know that we can apply the trigonometric identities in a right angled triangle whose:
Hypotenuse$= H$, Perpendicular side $= P$and the Base $= B$.
Hence, an alternate way to solve this question is:
We will first find the value of $\cos x$ in the same way as before.
Hence, by using half angle formula, and from $\left( 1 \right)$and $\left( 2 \right)$, we will get:
$\cos x = \dfrac{{119}}{{169}}$
Now, instead of using formulas further, we will use the right angle formulas.
As we know, in a right angled triangle, $\cos x = \dfrac{B}{H} = \dfrac{{119}}{{169}}$
Also, since, $\cos x = \dfrac{B}{H} = \dfrac{{119}}{{169}}$
Hence, substituting $B = 119$and $H = 169$ in the formula ${H^2} = {P^2} + {B^2}$, we get
$\Rightarrow {\left( {169} \right)^2} - {\left( {119} \right)^2} = {P^2}$
Using the property $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$, we get
$\Rightarrow {P^2} = \left( {169 + 119} \right)\left( {169 - 119} \right)$
$\Rightarrow {P^2} = 288 \times 50 = 14400$
Taking square root on both side, we get
$\Rightarrow P = \sqrt {14400} = 120$
Therefore, $\sin x = \dfrac{P}{H} = \dfrac{{120}}{{169}}$
And, $\tan x = \dfrac{P}{B} = \dfrac{{120}}{{119}}$
Therefore, without the use of tables, we have calculated the values of $\sin x$, $\cos x$ and $\tan x$ as $\dfrac{{120}}{{169}}$,$\dfrac{{119}}{{169}}$ and $\dfrac{{120}}{{119}}$ respectively.