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

Hint: 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.