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Trigonometric Ratio and Identites question

2022 · 25 Jun · Shift 2 · Q33
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Trigonometric Ratio and Identites question

2022 · 25 Jun · Shift 2 · Q33

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
The value of 2sin (12 ∘^\circ∘) −-− sin (72 ∘^\circ∘) is :
  1. A
    5(1−3)4{{\sqrt 5 (1 - \sqrt 3 )} \over 4}45​(1−3​)​
  2. B
    1−58{{1 - \sqrt 5 } \over 8}81−5​​
  3. C
    3(1−5)2{{\sqrt 3 (1 - \sqrt 5 )} \over 2}23​(1−5​)​
  4. D
    3(1−5)4{{\sqrt 3 (1 - \sqrt 5 )} \over 4}43​(1−5​)​
View written solutionFree

Correct answer: D

  1. We need to find 2sin⁡12∘−sin⁡72∘.2\sin 12^\circ - \sin 72^\circ.2sin12∘−sin72∘.

  2. Use the identity sin⁡72∘=cos⁡18∘.\sin 72^\circ = \cos 18^\circ.sin72∘=cos18∘. So the expression becomes 2sin⁡12∘−cos⁡18∘.2\sin 12^\circ - \cos 18^\circ.2sin12∘−cos18∘.

  3. Now use the standard exact values: sin⁡18∘=5−14,cos⁡18∘=10+254.\sin 18^\circ = \frac{\sqrt5-1}{4}, \qquad \cos 18^\circ = \frac{\sqrt{10+2\sqrt5}}{4}.sin18∘=45​−1​,cos18∘=410+25​​​. But this does not directly simplify the expression nicely, so instead we use angle formulas.

  4. Write sin⁡72∘=2sin⁡36∘cos⁡36∘.\sin 72^\circ = 2\sin 36^\circ \cos 36^\circ.sin72∘=2sin36∘cos36∘. Also, 2sin⁡12∘=2sin⁡(30∘−18∘).2\sin 12^\circ = 2\sin(30^\circ-18^\circ).2sin12∘=2sin(30∘−18∘). Using sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b,\sin(a-b)=\sin a\cos b-\cos a\sin b,sin(a−b)=sinacosb−cosasinb, we get

= \cos18^\circ-\sqrt3\sin18^\circ.$$ 5. Therefore, $$2\sin12^\circ-\sin72^\circ = \left(\cos18^\circ-\sqrt3\sin18^\circ\right)-\cos18^\circ = -\sqrt3\sin18^\circ.$$ 6. Substitute $$\sin18^\circ=\frac{\sqrt5-1}{4}.$$ Then $$-\sqrt3\sin18^\circ = -\sqrt3\cdot \frac{\sqrt5-1}{4} = \frac{\sqrt3(1-\sqrt5)}{4}.$$ 7. Compare with the options: - A: $\dfrac{\sqrt5(1-\sqrt3)}{4}$ - B: $\dfrac{1-\sqrt5}{8}$ - C: $\dfrac{\sqrt3(1-\sqrt5)}{2}$ - D: $\dfrac{\sqrt3(1-\sqrt5)}{4}$ Hence the correct option is $$\boxed{D}.$$
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