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Trigonometric Ratio and Identites question

2022 · 27 Jun · Shift 2 · Q36
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Trigonometric Ratio and Identites question

2022 · 27 Jun · Shift 2 · Q36

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
α=sin⁡36∘\alpha = \sin 36^\circα=sin36∘ is a root of which of the following equation?
  1. A
    16x4−10x2−5=016{x^4} - 10{x^2} - 5 = 016x4−10x2−5=0
  2. B
    16x4+20x2−5=016{x^4} + 20{x^2} - 5 = 016x4+20x2−5=0
  3. C
    16x4−20x2+5=016{x^4} - 20{x^2} + 5 = 016x4−20x2+5=0
  4. D
    4x4−10x2+5=04{x^4} - 10{x^2} + 5 = 04x4−10x2+5=0
View written solutionFree

Correct answer: C

  1. We need to find an equation satisfied by x=α=sin⁡36∘.x=\alpha=\sin 36^\circ.x=α=sin36∘.

  2. Use the double-angle identity: sin⁡36∘=2sin⁡18∘cos⁡18∘.\sin 36^\circ = 2\sin 18^\circ \cos 18^\circ.sin36∘=2sin18∘cos18∘. But the standard exact value we know is sin⁡18∘=5−14.\sin 18^\circ=\frac{\sqrt{5}-1}{4}.sin18∘=45​−1​.

Also, sin⁡236∘=1−cos⁡236∘.\sin^2 36^\circ = 1-\cos^2 36^\circ.sin236∘=1−cos236∘. A more direct standard value is cos⁡36∘=1+54.\cos 36^\circ=\frac{1+\sqrt{5}}{4}.cos36∘=41+5​​. So, sin⁡236∘=1−(1+54)2.\sin^2 36^\circ = 1-\left(\frac{1+\sqrt{5}}{4}\right)^2.sin236∘=1−(41+5​​)2.

  1. Compute it: (1+54)2=1+25+516=6+2516=3+58.\left(\frac{1+\sqrt{5}}{4}\right)^2=\frac{1+2\sqrt{5}+5}{16}=\frac{6+2\sqrt{5}}{16}=\frac{3+\sqrt{5}}{8}.(41+5​​)2=161+25​+5​=166+25​​=83+5​​. Hence, \sin^2 36^\circ = 1-\frac{3+\sqrt{5}}{8}= rac{8-(3+\sqrt{5})}{8}= rac{5-\sqrt{5}}{8}.

Let y=x2=sin⁡236∘=5−58.y=x^2=\sin^2 36^\circ=\frac{5-\sqrt{5}}{8}.y=x2=sin236∘=85−5​​.

  1. Now test which quadratic in yyy is satisfied. From option C: 16x4−20x2+5=0.16x^4-20x^2+5=0.16x4−20x2+5=0. Putting y=x2y=x^2y=x2, this becomes 16y2−20y+5=0.16y^2-20y+5=0.16y2−20y+5=0. Now substitute y=5−58.y=\frac{5-\sqrt{5}}{8}.y=85−5​​.

First,

=\frac{(5-\sqrt{5})^2}{4}.$$ Compute: $$(5-\sqrt{5})^2=25-10\sqrt{5}+5=30-10\sqrt{5}.$$ Thus, $$16y^2=\frac{30-10\sqrt{5}}{4}=\frac{15-5\sqrt{5}}{2}.$$ Also, $$20y=20\cdot \frac{5-\sqrt{5}}{8}=\frac{25-5\sqrt{5}}{2}.$$ Therefore, $$16y^2-20y+5 =\frac{15-5\sqrt{5}}{2}-\frac{25-5\sqrt{5}}{2}+5 =\frac{-10}{2}+5=-5+5=0.$$ So option C is satisfied. 5. Check uniqueness among options briefly: - A gives $16y^2-10y-5$, not satisfied. - B gives $16y^2+20y-5$, not satisfied. - D gives $4y^2-10y+5$, not satisfied. Hence the required equation is $$\boxed{16x^4-20x^2+5=0}. $$ So the correct option is **C**.
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