Solveeit Logo
Login

Question

Question: The value of \[\sin 18^\circ \] is A). \(\dfrac{{\sqrt 5 - 1}}{4}\) B). \(\dfrac{{\sqrt 5 + 1}}{...

The value of sin⁡18∘\sin 18^\circ is
A). 5−14\dfrac{{\sqrt 5 - 1}}{4}
B). 5+14\dfrac{{\sqrt 5 + 1}}{4}
C). −5+14\dfrac{{ - \sqrt 5 + 1}}{4}
D). 1−54\dfrac{{1 - \sqrt 5 }}{4}

Explanation

Solution

In this question, remember to take A=18∘A = {18^ \circ } and multiply both side by 5 and use the trigonometric identities such as sin⁡(90∘−θ)=cos⁡θ\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta , using this information we can approach the solution of the problem.

Complete step-by-step solution:
According to the given information, we have to find out the value of sin⁡18∘\sin 18^\circ
Let theA=18∘A = {18^ \circ }, multiply both the sides by 5, we get
Therefore, we can consider, 5A = 90∘{90^ \circ }
Now we can split the 5A=3A+2A5A = 3A + 2A
SO, we have 3A + 2A = 90∘{90^ \circ }
2A = 90∘{90^ \circ } – 3A
Taking sine on both sides, we get
sin⁡2A=sin⁡(90∘−3A)=cos⁡3A\sin 2A = \sin \left( {{{90}^ \circ } - 3A} \right) = \cos 3A
Since we know that sin⁡(90∘−θ)=cos⁡θ\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta and cos⁡3a=4cos⁡3a−3cos⁡a\cos 3a = 4{\cos ^3}a - 3\cos a
Therefore 2sin⁡Acos⁡A=4cos⁡3A−3cos⁡A2\sin A\cos A = 4{\cos ^3}A - 3\cos A
⇒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
Dividing both the sides by cos⁡A=cos⁡18∘≠0\cos A = \cos {18^ \circ } \ne 0, we get
⇒2sin⁡A−4cos⁡2A+3=0\Rightarrow 2\sin A - 4{\cos ^2}A + 3 = 0
Also, we know that cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta
Therefore 2sin⁡A−4(1−sin⁡2A)+3=02\sin A - 4\left( {1 - {{\sin }^2}A} \right) + 3 = 0
4sin⁡2A+2sin⁡A−1=04{\sin ^2}A + 2\sin A - 1 = 0 which is a quadratic in Sin⁡A\operatorname{Sin} A
By the quadratic formula i.e. −b±b2+4ac2a\dfrac{{ - b \pm \sqrt {{b^2} + 4ac} }}{{2a}}
⇒sin⁡A=−2±4−(4)4.(−1)2×4 =−2±4+168 =−2±258 \begin{gathered} \Rightarrow \sin A = \dfrac{{ - 2 \pm \sqrt {4 - \left( 4 \right)4.\left( { - 1} \right)} }}{{2 \times 4}} \\\ =\dfrac{{ - 2 \pm \sqrt {4 + 16} }}{8} \\\ = \dfrac{{ - 2 \pm 2\sqrt 5 }}{8} \\\ \end{gathered}
⇒sin⁡A=−1±54 \Rightarrow \sin A = \dfrac{{ - 1 \pm \sqrt 5 }}{4} \\\
Now since 18∘{18^ \circ } lies in the first quadrant therefore taking sin⁡18∘\sin {18^ \circ } positive value
Therefore,sin⁡A=−1+54\sin A = \dfrac{{ - 1 + \sqrt 5 }}{4}
Hence, “A” is the correct option.

Note: The sine, cosine, and tangent of an angle which we used in the above solution are all defined in terms of trigonometry, but they can also be expressed as functions. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature.