Question

How do you find the exact value of $\sin 24$ using the sum and difference, double angle or half angle formulas?

Answer 1

In the given question, we have been asked to find the value of sin24 b using any of the formula i.e. sum and difference formula or double angle formula or half angle formula. In order to solve the question, we will use the identity of $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$. We split the $\sin {{\left( 24 \right)}^{0}}=\sin \left( {{60}^{0}}-{{36}^{0}} \right)$, then by using identity we will solve each trigonometric function and by putting into the identity. We will get the required solution.

Formula used:
Difference formula of trigonometry:
$\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$
Using the trigonometric ratios:
$\sin {{60}^{0}}=\dfrac{\sqrt{3}}{2}$
$\Rightarrow\cos {{60}^{0}}=\dfrac{1}{2}$

Complete step by step answer:
Using the sum and difference formula:
We have given,
$\Rightarrow \sin {{\left( 24 \right)}^{0}}$
$\Rightarrow \sin {{\left( 24 \right)}^{0}}=\sin \left( {{60}^{0}}-{{36}^{0}} \right)$
Using the identity; $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$, we obtain
$\Rightarrow \sin \left( {{60}^{0}}-{{36}^{0}} \right)=\sin {{60}^{0}}\cos {{36}^{0}}-\cos {{60}^{0}}\sin {{36}^{0}}$
As we know that, using the trigonometric ratio,
$\Rightarrow\sin {{60}^{0}}=\dfrac{\sqrt{3}}{2}$
$\Rightarrow\cos {{60}^{0}}=\dfrac{1}{2}$

Now solving,
$\cos {{36}^{0}}$
$\Rightarrow \cos {{36}^{0}}=\cos 2\left( {{18}^{0}} \right)$
Using the identity, $\cos 2x=1-2{{\sin }^{2}}x$
$\Rightarrow \cos 2\left( {{18}^{0}} \right)=1-2\left( {{\sin }^{2}}{{\left( 18 \right)}^{0}} \right)$
Put the value of ${{\sin }^{2}}{{\left( 18 \right)}^{0}}=\dfrac{\sqrt{5}+1}{4}$
$\Rightarrow 1-2{{\left( \dfrac{\sqrt{5}+1}{4} \right)}^{2}}$
Using the identity, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
$\Rightarrow 1-2\left( \dfrac{5+1-2\sqrt{5}}{16} \right)$
Simplifying the above, we get
$\Rightarrow 1-2\left( \dfrac{6-2\sqrt{5}}{16} \right)$

Simplifying further, we obtain
$\Rightarrow \cos {{36}^{0}}=\dfrac{\sqrt{5}+1}{4}$
Now solving,
$\Rightarrow \sin {{36}^{0}}$
$\Rightarrow \sin {{36}^{0}}=\sqrt{1-{{\cos }^{2}}{{\left( 36 \right)}^{0}}}$
Putting the value of $\cos {{36}^{0}}=\dfrac{\sqrt{5}+1}{4}$, which we solved above,
$\Rightarrow \sin {{36}^{0}}=\sqrt{1-{{\left( \dfrac{\sqrt{5}+1}{4} \right)}^{2}}}$
$\Rightarrow \sin {{36}^{0}}=\sqrt{1-\left( \dfrac{5+1+2\sqrt{5}}{16} \right)}$
Simplifying the above, we get
$\Rightarrow \sin {{36}^{0}}=\dfrac{1}{4}\times \sqrt{10-2\sqrt{5}}$
$\therefore \sin {{36}^{0}}=\dfrac{\sqrt{10-2\sqrt{5}}}{4}$

Putting all the plug-in solved values,
$\Rightarrow \sin \left( {{60}^{0}}-{{36}^{0}} \right)=\sin {{60}^{0}}\cos {{36}^{0}}-\cos {{60}^{0}}\sin {{36}^{0}}$
$\Rightarrow \left( \dfrac{\sqrt{3}}{2} \right)\times \left( \dfrac{\sqrt{5}+1}{4} \right)+\left( \dfrac{1}{2} \right)\times \left( \dfrac{\sqrt{10-2\sqrt{5}}}{4} \right)$
Simplifying the above, we get
$\Rightarrow \dfrac{\sqrt{3}}{8}\left( \sqrt{5}+1 \right)-\dfrac{\sqrt{2}}{8}\left( \sqrt{5-\sqrt{5}} \right)$
$\therefore \sin {{24}^{0}}=\dfrac{\sqrt{3}}{8}\left( \sqrt{5}+1 \right)-\dfrac{\sqrt{2}}{8}\left( \sqrt{5-\sqrt{5}} \right)$

Note: In order to solve these types of questions, you should always need to remember the properties of trigonometric and the trigonometric ratios as well. It will make questions easier to solve. It is preferred that while solving these types of questions we should carefully examine the pattern of the given function and then you would apply the formulas according to the pattern observed. As if you directly apply the formula it will create confusion ahead and we will get the wrong answer.