Question

Can someone tell me how to prove $\sin {{78}^{\circ }}+\cos {{132}^{\circ }}=\dfrac{\left( \sqrt{5}-1 \right)}{4}$, in the easiest method.

Answer 1

In order to prove the given question that is $\sin {{78}^{\circ }}+\cos {{132}^{\circ }}=\dfrac{\left( \sqrt{5}-1 \right)}{4}$ Split the angle of sine with the help of trigonometric properties of sine and cosine. Simply until you get $\sin {{18}^{\circ }}$. Now to find the value of $\sin {{18}^{\circ }}$. Assume $x={{18}^{\circ }}$ and then $5x={{90}^{\circ }}$. With the help of cosine properties expand $\cos \left( 5x-2x \right)$ and use the identity that $\cos 3x=4{{\cos }^{3}}x-3\cos x=\sin 2x$. After that simplify it further and solve the quadratic equation $4{{\sin }^{2}}x+2\sin x-1=0$ with the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find the value of $\sin x$which is nothing but $\sin {{18}^{\circ }}$.

Complete step by step solution:
Given equation which needs to be proves is as follows:
$\sin {{78}^{\circ }}+\cos {{132}^{\circ }}=\dfrac{\left( \sqrt{5}-1 \right)}{4}$
Consider the Left-Hand-Side of the equation and solve it further:
LHS $=\sin {{78}^{\circ }}+\cos {{132}^{\circ }}$
We can write ${{78}^{\circ }}={{90}^{\circ }}-{{12}^{\circ }}$, substituting this angle of sine in the above equation we get:
$\Rightarrow \sin \left( {{90}^{\circ }}-{{12}^{\circ }} \right)+\cos {{132}^{\circ }}$
Apply the trigonometric property of sine and cosine that is $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta$in the above equation we get:
$\Rightarrow \cos {{12}^{\circ }}+\cos {{132}^{\circ }}$
Again, apply another trigonometric property of sine and cosine that is $\cos A-\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ in the above equation we get:
$\Rightarrow 2\cos {{72}^{\circ }}\cos {{60}^{\circ }}$
We can write ${{72}^{\circ }}={{90}^{\circ }}-{{18}^{\circ }}$, substituting this angle of sine in the above equation we get:
$\Rightarrow 2\cos ({{90}^{\circ }}-{{18}^{\circ }})\cdot \dfrac{1}{2}$
Apply the trigonometric property of sine and cosine that is $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta$in the above equation we get:
$\Rightarrow \sin {{18}^{\circ }}$
To find the value of $\sin {{18}^{\circ }}$. We will consider the following:
Let $x={{18}^{\circ }}$
$\Rightarrow 5x={{90}^{\circ }}...\left( 1 \right)$
Now, we can write this in the following form:
$\Rightarrow \cos 3x=\cos \left( 5x-2x \right)$
Substituting the equation $\left( 1 \right)$ in above equation we get:
$\Rightarrow \cos 3x=\cos \left( {{90}^{\circ }}-2x \right)$
After this, use the identity that $\cos 3x=4{{\cos }^{3}}x-3\cos x=\sin 2x$.
$\Rightarrow 4{{\cos }^{3}}x-3\cos x=\sin 2x$
$\Rightarrow 4{{\cos }^{3}}x-3\cos x=2\sin x\cos x$
As we know that $\cos x\ne 0$ therefore we get
$\Rightarrow 4{{\cos }^{2}}x-3=2\sin x$
Again, apply another trigonometric property of sine and cosine that is ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$ in the above equation we get:
$\Rightarrow 4\left( 1-{{\sin }^{2}}x \right)-3=2\sin x$
After simplifying it further we get:
$\Rightarrow 4{{\sin }^{2}}x+2\sin x-1=0$
Clearly you can see above equation is a quadratic equation, to solve this and find the value of $\sin x$, we’ll use the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ we get:
$\Rightarrow \sin x=\dfrac{-2+\sqrt{{{2}^{2}}-4\cdot 4\cdot \left( -1 \right)}}{2\cdot 4}$
$\Rightarrow \sin x=\dfrac{-2+\sqrt{20}}{2\cdot 4}$
$\Rightarrow \sin x=\dfrac{-2+2\sqrt{5}}{2\cdot 4}$
After simplifying it further we get:
$\Rightarrow \sin x=\dfrac{\sqrt{5}-1}{4}$
As we took $x={{18}^{\circ }}$ initially therefore we get:
$\Rightarrow \sin {{18}^{\circ }}=\dfrac{\sqrt{5}-1}{4}...\left( 2 \right)$
Which is equal to the RHS.
Therefore, $\sin {{78}^{\circ }}+\cos {{132}^{\circ }}=\dfrac{\left( \sqrt{5}-1 \right)}{4}$. Hence Proved.

Note:
There’s an alternative way to solve this question, which is as follows:
Consider the Left-Hand-Side of the equation that is:
LHS $=\sin {{78}^{\circ }}+\cos {{132}^{\circ }}$
We can write ${{132}^{\circ }}={{90}^{\circ }}+{{42}^{\circ }}$, substituting this angle of sine in the above equation we get:
$\Rightarrow \sin {{78}^{\circ }}+\cos \left( {{90}^{\circ }}+{{42}^{\circ }} \right)$
Apply the trigonometric property of sine and cosine that is $\cos \left( {{90}^{\circ }}+\theta \right)=-\sin \theta$ in the above equation we get:
$\Rightarrow \sin {{78}^{\circ }}-\sin {{42}^{\circ }}$
Again, apply another trigonometric property of sine and cosine that is $\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ in the above equation we get:
$\Rightarrow 2\cos \left( \dfrac{{{78}^{\circ }}+{{42}^{\circ }}}{2} \right)\sin \left( \dfrac{78-42}{2} \right)$
$\Rightarrow 2\cos \left( {{60}^{\circ }} \right)\sin \left( {{18}^{\circ }} \right)$
From equation $\left( 2 \right)$ we get:
$\Rightarrow 2\cdot \dfrac{1}{2}\cdot \dfrac{\sqrt{5}-1}{4}$
$\Rightarrow \dfrac{\sqrt{5}-1}{4}=$RHS