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

Hint: 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