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Question: Prove that : $4\sin 27^{\circ} = (5 + \sqrt{5})^{1/2} - (3 - \sqrt{5})^{1/2}$...

Prove that : 4sin⁡27∘=(5+5)1/2−(3−5)1/24\sin 27^{\circ} = (5 + \sqrt{5})^{1/2} - (3 - \sqrt{5})^{1/2}

Answer

LHS = RHS, hence proved.

Explanation

Solution

To prove the identity 4sin⁡27∘=(5+5)1/2−(3−5)1/24\sin 27^{\circ} = (5 + \sqrt{5})^{1/2} - (3 - \sqrt{5})^{1/2}, we will simplify both the Left Hand Side (LHS) and the Right Hand Side (RHS) and show that they are equal.

Step 1: Simplify the Left Hand Side (LHS) LHS =4sin⁡27∘= 4\sin 27^{\circ}

We can write 27∘27^{\circ} as 45∘−18∘45^{\circ} - 18^{\circ}. Using the trigonometric identity sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B) = \sin A \cos B - \cos A \sin B:

sin⁡27∘=sin⁡(45∘−18∘)=sin⁡45∘cos⁡18∘−cos⁡45∘sin⁡18∘\sin 27^{\circ} = \sin(45^{\circ} - 18^{\circ}) = \sin 45^{\circ} \cos 18^{\circ} - \cos 45^{\circ} \sin 18^{\circ}

We know that sin⁡45∘=cos⁡45∘=12\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}.

So, sin⁡27∘=12cos⁡18∘−12sin⁡18∘=12(cos⁡18∘−sin⁡18∘)\sin 27^{\circ} = \frac{1}{\sqrt{2}} \cos 18^{\circ} - \frac{1}{\sqrt{2}} \sin 18^{\circ} = \frac{1}{\sqrt{2}}(\cos 18^{\circ} - \sin 18^{\circ})

Therefore, LHS =4⋅12(cos⁡18∘−sin⁡18∘)=42(cos⁡18∘−sin⁡18∘)=22(cos⁡18∘−sin⁡18∘)= 4 \cdot \frac{1}{\sqrt{2}}(\cos 18^{\circ} - \sin 18^{\circ}) = \frac{4}{\sqrt{2}}(\cos 18^{\circ} - \sin 18^{\circ}) = 2\sqrt{2}(\cos 18^{\circ} - \sin 18^{\circ}).

Step 2: Simplify the Right Hand Side (RHS) RHS =(5+5)1/2−(3−5)1/2=5+5−3−5= (5 + \sqrt{5})^{1/2} - (3 - \sqrt{5})^{1/2} = \sqrt{5 + \sqrt{5}} - \sqrt{3 - \sqrt{5}}

To simplify these nested square roots, we multiply and divide each term by 2\sqrt{2}:

5+5=2(5+5)2=10+252\sqrt{5 + \sqrt{5}} = \frac{\sqrt{2(5 + \sqrt{5})}}{\sqrt{2}} = \frac{\sqrt{10 + 2\sqrt{5}}}{\sqrt{2}}

3−5=2(3−5)2=6−252\sqrt{3 - \sqrt{5}} = \frac{\sqrt{2(3 - \sqrt{5})}}{\sqrt{2}} = \frac{\sqrt{6 - 2\sqrt{5}}}{\sqrt{2}}

Now, we simplify the terms in the numerators:

For 6−25\sqrt{6 - 2\sqrt{5}}: We look for two numbers whose sum is 6 and product is 5. These numbers are 5 and 1.

So, 6−25=(5)2+12−25⋅1=(5−1)2=∣5−1∣\sqrt{6 - 2\sqrt{5}} = \sqrt{(\sqrt{5})^2 + 1^2 - 2\sqrt{5} \cdot 1} = \sqrt{(\sqrt{5} - 1)^2} = |\sqrt{5} - 1|. Since 5>1\sqrt{5} > 1, this simplifies to 5−1\sqrt{5} - 1.

We know the standard value sin⁡18∘=5−14\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}.

Thus, 5−1=4sin⁡18∘\sqrt{5} - 1 = 4\sin 18^{\circ}.

For 10+25\sqrt{10 + 2\sqrt{5}}: We recall the standard value cos⁡18∘=10+254\cos 18^{\circ} = \frac{\sqrt{10 + 2\sqrt{5}}}{4}.

Thus, 10+25=4cos⁡18∘\sqrt{10 + 2\sqrt{5}} = 4\cos 18^{\circ}.

Substitute these simplified forms back into the RHS expression:

RHS =4cos⁡18∘2−4sin⁡18∘2= \frac{4\cos 18^{\circ}}{\sqrt{2}} - \frac{4\sin 18^{\circ}}{\sqrt{2}}

RHS =42(cos⁡18∘−sin⁡18∘)= \frac{4}{\sqrt{2}}(\cos 18^{\circ} - \sin 18^{\circ})

RHS =22(cos⁡18∘−sin⁡18∘)= 2\sqrt{2}(\cos 18^{\circ} - \sin 18^{\circ}).

Step 3: Compare LHS and RHS We found:

LHS =22(cos⁡18∘−sin⁡18∘)= 2\sqrt{2}(\cos 18^{\circ} - \sin 18^{\circ})

RHS =22(cos⁡18∘−sin⁡18∘)= 2\sqrt{2}(\cos 18^{\circ} - \sin 18^{\circ})

Since LHS = RHS, the identity is proven.