Question

Evaluate $\sin 24{}^\circ +\cos 60{}^\circ =?$

Answer 1

Hint: We will be using the concept of trigonometric functions to solve the problem. We will be using the trigonometric identities like,
$\begin{aligned} & \sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\\ & \sin 2A=2\sin A\cos A \\\ & \sin \left( 90-A \right)=\cos A \\\ & \cos 3A=4{{\cos }^{3}}A-3\cos A \\\ \end{aligned}$

Complete step-by-step answer:
Now, we have to find the value of,
$\sin 24{}^\circ +\cos 60{}^\circ$
Now, we know that $\sin \left( 90-6 \right)=\cos 6$.
$\sin 24{}^\circ +\cos 84{}^\circ$
Now, we know that,
$\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
So, we have,
$\begin{aligned} & 2\sin \left( \dfrac{24{}^\circ +84{}^\circ }{2} \right)\cos \left( \dfrac{24{}^\circ -84{}^\circ }{2} \right) \\\ & 2\sin \left( \dfrac{108{}^\circ }{2} \right)\cos \left( \dfrac{60{}^\circ }{2} \right) \\\ & 2\sin \left( 54{}^\circ \right)\cos 30{}^\circ \\\ \end{aligned}$
Now, we know that $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$. So,
$\begin{aligned} & 2\times \dfrac{\sqrt{3}}{2}\sin \left( 54{}^\circ \right) \\\ & \sin 24{}^\circ +\cos 60{}^\circ =\sqrt{3}\sin \left( 54{}^\circ \right)..........\left( 1 \right) \\\ \end{aligned}$
Now, we know that,

& \cos \left( 90{}^\circ -54{}^\circ \right)=\sin \left( 54{}^\circ \right) \\\ & \Rightarrow \sin 54{}^\circ =\cos 36{}^\circ \\\ \end{aligned}$$ Also, we know that, ${{\cos }^{2}}A=1-2{{\sin }^{2}}A$ So, let $\begin{aligned} & 2A=36 \\\ & A=18{}^\circ \\\ \end{aligned}$ So, we have, $\begin{aligned} & \sin 54{}^\circ ={{\cos }^{2}}36{}^\circ =1-2{{\sin }^{2}}18{}^\circ ...........\left( 2 \right) \\\ & \cos 2A=1-2{{\sin }^{2}}A \\\ \end{aligned}$ Now, we have, $\begin{aligned} & A=18 \\\ & 5A=90 \\\ & 2A+3A=90{}^\circ \\\ & 2A=90-3A \\\ \end{aligned}$ Now, we will take sin on both sides, so we have, $\begin{aligned} & \sin \left( 2A \right)=\sin \left( 90-3A \right) \\\ & \sin \left( 2A \right)=\cos \left( 3A \right) \\\ \end{aligned}$ Now, we know that, $\begin{aligned} & \sin 2A=2\sin A\cos A \\\ & \cos 3A=4{{\cos }^{3}}A-3\cos A \\\ \end{aligned}$ So, we have, $\begin{aligned} & 2\sin A\cos A=4{{\cos }^{3}}A-3\cos A \\\ & 2\sin A\cos A-4{{\cos }^{3}}A+3\cos A=0 \\\ & \cos A\left( 2\sin A-4{{\cos }^{2}}A+3 \right)=0 \\\ \end{aligned}$ Now, we know that, $\begin{aligned} & \cos A=\cos \left( 18 \right)\ne 0 \\\ & \therefore 2\sin A-4{{\cos }^{2}}A+3=0 \\\ \end{aligned}$ Now, we substitute ${{\cos }^{2}}A=1-{{\sin }^{2}}A$. So, we have, $\begin{aligned} & 2\sin A-4\left( 1-{{\sin }^{2}}A \right)+3=0 \\\ & 2\sin A-4+4{{\sin }^{2}}A+3=0 \\\ & 4{{\sin }^{2}}A+2\sin A-1=0 \\\ \end{aligned}$ Now, we will use quadratic formula to find sin A. So, we have, $\begin{aligned} & \sin A=\dfrac{-2\pm \sqrt{20}}{2\times 4} \\\ & =\dfrac{-1\pm \sqrt{5}}{4} \\\ \end{aligned}$ Now, as we have $A=18{}^\circ $and therefore, $\sin \left( 18{}^\circ \right)>0$. So, we reject $\sin A\ne \dfrac{-1-\sqrt{5}}{4}$. $\sin 18{}^\circ =\dfrac{-1+\sqrt{5}}{4}$ Now, from (2) we have, $$\begin{aligned} & \sin 54{}^\circ =1-2{{\sin }^{2}}18 \\\ & =1-2{{\left( \dfrac{-1+\sqrt{5}}{4} \right)}^{2}} \\\ & =1-\dfrac{2}{16}\left( 1+5-2\sqrt{5} \right) \\\ & =1-\dfrac{1}{8}\left( 6-2\sqrt{5} \right) \\\ & =1-\dfrac{1}{4}\left( 3-\sqrt{5} \right) \\\ & =1-\dfrac{3}{4}+\dfrac{\sqrt{5}}{4} \\\ & =\dfrac{1+\sqrt{5}}{4} \\\ \end{aligned}$$ So, now from (1), we have, $\begin{aligned} & \sin 24+\cos 60=\sqrt{3}\sin 54{}^\circ \\\ & =\sqrt{3}\dfrac{\left( 1+\sqrt{5} \right)}{4} \\\ \end{aligned}$ Note: To solve these types of questions it is important to note how we have found the value of $\sin 54{}^\circ $. Also, it is important to remember identities like, $\begin{aligned} & \sin 2A=2\sin A\cos A \\\ & \cos 3A=4{{\cos }^{3}}A-3\cos A \\\ \end{aligned}$