Question

How do you find the exact value of $\cos {{48}^{\circ }}$ using the sum and difference, double angle or half angle formulas?

Answer 1

Hint : Here we have to find the exact value of the given trigonometric function by using the sum and difference, half angle and double angle formulas. Firstly we will divide the angle of the trigonometric function into two parts. Then we will use the sum formula and open the value. Finally using double angle, triple angle and square relation formula we will get our desired answer.

Complete step-by-step answer :
We have to find the exact value of the below trigonometric function by using the sum and difference, double angle or half angle formulas:
$\cos {{48}^{\circ }}$
We can write the above value as:
$\cos {{\left( 30+18 \right)}^{\circ }}$
Using sum formula which state that $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ where $A=30$ and $B=18$ we get,
$\cos {{\left( 30+18 \right)}^{\circ }}=\cos {{30}^{\circ }}\cos {{18}^{\circ }}-\sin {{30}^{\circ }}\sin {{18}^{\circ }}$
Now as we know the value of $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ and $\sin {{30}^{\circ }}=\dfrac{1}{2}$ substitute it above,
$\cos {{\left( 30+18 \right)}^{\circ }}=\dfrac{\sqrt{3}}{2}\cos {{18}^{\circ }}-\dfrac{1}{2}\sin {{18}^{\circ }}$….$\left( 1 \right)$
Next we have to find the value of $\cos {{18}^{\circ }}$ and $\sin {{18}^{\circ }}$ .
So let,
$A={{18}^{\circ }}$
Hence we get,
$\Rightarrow 5A={{18}^{\circ }}\times 5$
$\Rightarrow 5A={{90}^{\circ }}$
We can write the left hand side value as follows:
$2A+3A={{90}^{\circ }}$
$\Rightarrow 2A={{90}^{\circ }}-3A$
Take sine function on both sides,
$\Rightarrow \sin \left( 2A \right)=\sin \left( {{90}^{\circ }}-3A \right)$
As we know $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta$ and Double angle formula state that $\sin 2A =2\sin A \cos A$ using it above we get,
$\Rightarrow 2\sin A \cos A =\cos 3A$
Using triple angle formula which state that $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta$ above we get,
$\Rightarrow 2\sin A\cos A=4{{\cos }^{3}}A-3\cos A$
$\Rightarrow 2\sin A\cos A-4{{\cos }^{3}}A+3\cos A=0$
Taking $\cos A$ common we get,
$\Rightarrow \cos A\left( 2\sin A-4{{\cos }^{2}}A+3 \right)=0$
As we know $\cos A=\cos 18$ and $\cos 18\ne 0$ this means the bracket term is equal to zero as follows:
$\Rightarrow 2\sin A-4{{\cos }^{2}}A+3=0$
Using square relation which states that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ above we get,
$\Rightarrow 2\sin A-4\left( 1-{{\sin }^{2}}A \right)+3=0$
$\Rightarrow 2\sin A-4+4{{\sin }^{2}}A+3=0$
Simplifying it further we get,
$\Rightarrow 4{{\sin }^{2}}A+2\sin A-1=0$….$\left( 2 \right)$
Using quadratic formula which state that for any equation $a{{x}^{2}}+bx+c=0$ the value of $x$ is found by the formula:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$…$\left( 3 \right)$
Comparing equation (2) by the general form we get,
$a=4,b=2,c=-1$ and $x=\sin A$
On substituting the above value in equation (3) we get,
$\sin A=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times 4\times -1}}{2\times 4}$
$\Rightarrow \sin A=\dfrac{-2\pm \sqrt{4+16}}{8}$
Simplifying further we get,
$\Rightarrow \sin A=\dfrac{-2\pm \sqrt{20}}{8}$
$\Rightarrow \sin A=\dfrac{-2\pm 2\sqrt{5}}{8}$
Take $2$ common in the numerator,
$\Rightarrow \sin A=\dfrac{2\left( -1\pm \sqrt{5} \right)}{8}$
$\Rightarrow \sin A=\dfrac{-1\pm \sqrt{5}}{4}$
As $A={{18}^{\circ }}$ so $\sin {{18}^{\circ }}$ value can’t be in negative so we will take only the positive value which is:
$\Rightarrow \sin A=\dfrac{-1+\sqrt{5}}{4}$
Put $A={{18}^{\circ }}$ above we get,
$\Rightarrow \sin {{18}^{\circ }}=\dfrac{-1+\sqrt{5}}{4}$….$\left( 4 \right)$
Next we know the square relation which states that
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
$\Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$
Put the value from equation (4) above we get,
$\Rightarrow {{\cos }^{2}}A=1-{{\left( \dfrac{-1+\sqrt{5}}{4} \right)}^{2}}$
$\Rightarrow {{\cos }^{2}}A=1-\dfrac{1+5-2\sqrt{5}}{16}$
Square rooting both side we get,
$\Rightarrow \cos A=\sqrt{1-\dfrac{1+5-2\sqrt{5}}{16}}$
$\Rightarrow \cos A=\sqrt{\dfrac{16-1-5+2\sqrt{5}}{16}}$
On simplifying,
$\Rightarrow \cos A=\sqrt{\dfrac{10+2\sqrt{5}}{16}}$
$\Rightarrow \cos A=\dfrac{1}{4}\sqrt{10+2\sqrt{5}}$
Put $A={{18}^{\circ }}$
$\Rightarrow \cos {{18}^{\circ }}=\dfrac{1}{4}\sqrt{10+2\sqrt{5}}$….$\left( 5 \right)$
Substitute the value from equation (4) and (5) in equation (1) as follows:
$\Rightarrow \cos {{\left( 30+18 \right)}^{\circ }}=\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{10+2\sqrt{5}}}{4}-\dfrac{1}{2}\times \dfrac{\sqrt{5}-1}{4}$
$\Rightarrow \cos {{\left( 30+18 \right)}^{\circ }}=\dfrac{\sqrt{3}\times \sqrt{10+2\sqrt{5}}}{8}-\dfrac{\sqrt{5}-1}{8}$
Take the LCM and solve further,
$\Rightarrow \cos {{\left( 30+18 \right)}^{\circ }}=\dfrac{\sqrt{3}\times \sqrt{10+2\sqrt{5}}-\sqrt{5}+1}{8}$
$\Rightarrow \cos {{\left( 30+18 \right)}^{\circ }}=\dfrac{\sqrt{30+6\sqrt{5}}-\sqrt{5}+1}{8}$
Hence exact value of $\cos {{48}^{\circ }}$ using the sum and difference, double angle or half angle formulas is $\dfrac{\sqrt{30+6\sqrt{5}}-\sqrt{5}+1}{8}$ .
So, the correct answer is “$\dfrac{\sqrt{30+6\sqrt{5}}-\sqrt{5}+1}{8}$”.

Note : Trigonometry is a very important branch of mathematics which has a trigonometry function namely sine, cosine, tangent, cosecant, secant and cotangent. In this type of question we have to simplify the trigonometric function given as much as we can so that we can obtain a quadratic equation which can be solved easily. We have taken that $\sin {{18}^{\circ }}$ value can’t be negative as the function value is positive for angles between ${{0}^{\circ }}\,to\,{{90}^{\circ }}$ .