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Question: Show that \[{{\sin }^{2}}24-{{\sin }^{2}}6=\dfrac{\sqrt{5}-1}{8}\]....

Show that sin⁡224−sin⁡26=5−18{{\sin }^{2}}24-{{\sin }^{2}}6=\dfrac{\sqrt{5}-1}{8}.

Explanation

Solution

In this given question, we are a trigonometric function, hence we have to prove it. In order to solve the questions, we have to use trigonometric formulas to simplify the given trigonometric function and then follow trigonometric formulas or methods to solve. First, we take the LHS and hence solve for the solution, which should be equal to RHS.

Complete step by step solution:
The basic formula for solve this problem which are going to use is
sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B\sin \left( A+B \right)\sin \left( A-B \right)={{\sin }^{2}}A-{{\sin }^{2}}B
Let us solve the given problem,
Given, sin⁡224−sin⁡26=5−18{{\sin }^{2}}24-{{\sin }^{2}}6=\dfrac{\sqrt{5}-1}{8}
Let us consider the LHS,
sin⁡224∘−sin⁡26∘{{\sin }^{2}}{{24}^{\circ }}-{{\sin }^{2}}{{6}^{\circ }}
As we know this formula,sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B\sin \left( A+B \right)\sin \left( A-B \right)={{\sin }^{2}}A-{{\sin }^{2}}B
In this, A= 24 and B=6
Substituting the values of A and B in the known formula, so that we get the equation as
sin⁡(24∘+6∘)sin⁡(24∘−6∘)=sin⁡224∘−sin⁡26∘\sin \left( {{24}^{\circ }}+{{6}^{\circ }} \right)\sin \left( {{24}^{\circ }}-{{6}^{\circ }} \right)={{\sin }^{2}}{{24}^{\circ }}-{{\sin }^{2}}{{6}^{\circ }}
On further solving,
⇒sin⁡30∘sin⁡18∘\Rightarrow \sin {{30}^{\circ }}\sin {{18}^{\circ }}…………………… (1)
We know that the value of sin⁡30∘=12\sin {{30}^{\circ }}=\dfrac{1}{2}
We need to evaluate the sin⁡18∘\sin {{18}^{\circ }}
Let A=18∘A={{18}^{\circ }}
Therefore,5A=90∘5A={{90}^{\circ }}
Further bringing into 2A term,
2A=90∘−3A2A={{90}^{\circ }}-3A
Taking sine on both sides
\sin 2A=\sin \left( {{90}^{\circ }}-3A \right)=\cos 3A$$$$\left( \because \sin \left( {{90}^{\circ }}-A \right)=\cos A \right)
By applying cos3A formula,
⇒2sin⁡Acos⁡A=4cos⁡3A−3cos⁡A\Rightarrow 2\sin A\cos A=4{{\cos }^{3}}A-3\cos A
⇒cos⁡A(2sin⁡A−4cos⁡2A+3)=0\Rightarrow \cos A\left( 2\sin A-4{{\cos }^{2}}A+3 \right)=0.
But cos⁡A=cos⁡18∘≠0\cos A=\cos {{18}^{\circ }}\ne 0, so solving the second part,
(2sin⁡A−4(1−sin⁡2A)+3)=0\left( 2\sin A-4(1-{{\sin }^{2}}A)+3 \right)=0
4sin⁡2A+2sin⁡A−1=04{{\sin }^{2}}A+2\sin A-1=0
Since, we get quadratic equation by applying formula, we get
x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
sin⁡A=−2±4+162(4)\sin A=\dfrac{-2\pm \sqrt{4+16}}{2\left( 4 \right)}
sin⁡18∘=−1±54\sin {{18}^{\circ }}=\dfrac{-1\pm \sqrt{5}}{4}
For this problem, we are considering the positive value, then
sin⁡18∘=5−14\sin {{18}^{\circ }}=\dfrac{\sqrt{5}-1}{4}
From equation (1),
⇒sin⁡30∘sin⁡18∘\Rightarrow \sin {{30}^{\circ }}\sin {{18}^{\circ }}
⇒12×5−14\Rightarrow \dfrac{1}{2}\times \dfrac{\sqrt{5}-1}{4}
=5−18=\dfrac{\sqrt{5}-1}{8}
Which is the RHS.
Hence, sin⁡224−sin⁡26=5−18{{\sin }^{2}}24-{{\sin }^{2}}6=\dfrac{\sqrt{5}-1}{8}
Therefore, we got the above exact value.

Note: In order to solve these kinds 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.