Question

Prove that: ${{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =\dfrac{\sqrt{5}+1}{8}$

Answer 1

Hint: First use the formula ${{\sin }^{2}}A-{{\cos }^{2}}B=-\cos \left( A+B \right)\cos \left( A-B \right)$ on the LHS to get ${{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =-\cos \left( 120{}^\circ \right)\cos \left( -36{}^\circ \right)$. Then use the formula $\cos \left( -\theta \right)=\cos \left( \theta \right)$ to get${{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =-\cos \left( 120{}^\circ \right)\cos \left( 36{}^\circ \right)$. Then find the value of $\cos \left( 120{}^\circ \right)$. Substitute this value and the value of \[\cos \left( 36{}^\circ \right)\] in the obtained expression. The resultant will be equal to the RHS.

Complete step-by-step answer:
In this question, we need to prove that ${{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =\dfrac{\sqrt{5}+1}{8}$.
For this, we will simplify the LHS.
LHS $={{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ $
We know that if we have two angles A and B, then:
${{\sin }^{2}}A-{{\cos }^{2}}B=-\cos \left( A+B \right)\cos \left( A-B \right)$
Using this formula on the LHS, we get the following:
LHS $={{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ $
${{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =-\cos \left( 42{}^\circ +78{}^\circ \right)\cos \left( 42{}^\circ -78{}^\circ \right)$
${{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =-\cos \left( 120{}^\circ \right)\cos \left( -36{}^\circ \right)$
Now, we also know that $\cos \left( -\theta \right)=\cos \left( \theta \right)$ as cosine is positive in both the I and the IV quadrant.
Using this property on the above equation, we will get the following:
${{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =-\cos \left( 120{}^\circ \right)\cos \left( -36{}^\circ \right)$
${{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =-\cos \left( 120{}^\circ \right)\cos \left( 36{}^\circ \right)$ …(1)
Now, here we need to calculate $\cos \left( 120{}^\circ \right)$
$\cos \left( 120{}^\circ \right)=\cos \left( 90{}^\circ +30{}^\circ \right)$
Now, we know the property that $\cos \left( 90{}^\circ +\theta \right)=-\sin \left( \theta \right)$
Using this property in the above equation, we will get the following:
\[\cos \left( 120{}^\circ \right)=\cos \left( 90{}^\circ +30{}^\circ \right)\]
\[\cos \left( 120{}^\circ \right)=-\sin \left( 30{}^\circ \right)\]
Now, we will substitute this in the equation (1) to get the following:
${{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =-\cos \left( 120{}^\circ \right)\cos \left( 36{}^\circ \right)$
\[{{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =\sin \left( 30{}^\circ \right)\cos \left( 36{}^\circ \right)\]
Now, we already know that \[sin\left( 30{}^\circ \right)=\dfrac{1}{2}\] and that \[\cos \left( 36{}^\circ \right)=\dfrac{\sqrt{5}+1}{4}\].
We will now substitute these values in the above equation to get the following:
\[{{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =\sin \left( 30{}^\circ \right)\cos \left( 36{}^\circ \right)\]
\[{{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =\dfrac{1}{2}\times \dfrac{\sqrt{5}+1}{4}\]
\[{{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =\dfrac{\sqrt{5}+1}{8}\]
Hence, the LHS \[=\dfrac{\sqrt{5}+1}{8}\]
Now, we will look at the RHS.
RHS \[=\dfrac{\sqrt{5}+1}{8}\]
Hence, the LHS is equal to the RHS.
So, ${{\sin }^{2}}42{}^\circ -{{\cos }^{2}}78{}^\circ =\dfrac{\sqrt{5}+1}{8}$
Hence proved.

Note: In this question, it is important to know about the trigonometric properties like ${{\sin }^{2}}A-{{\cos }^{2}}B=-\cos \left( A+B \right)\cos \left( A-B \right)$, $\cos \left( -\theta \right)=\cos \left( \theta \right)$, and $\cos \left( 90{}^\circ +\theta \right)=-\sin \left( \theta \right)$. Without knowing these properties, you will be unable to solve this kind of problem as it involves a very unconventional measure of the angles.