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Trigonometric Ratio and Identites question

2019 · 9 Jan · Shift 1 · Q25
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  5. /2019 · 9 Jan · Shift 1 · Q25

Trigonometric Ratio and Identites question

2019 · 9 Jan · Shift 1 · Q25

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
For any θ∈(π4,π2)\theta \in \left( {{\pi \over 4},{\pi \over 2}} \right)θ∈(4π​,2π​), the expression 3(cos⁡θ−sin⁡θ)43{(\cos \theta - \sin \theta )^4}3(cosθ−sinθ)4 +6(sin⁡θ+cos⁡θ)2+4sin⁡6θ+ 6{(\sin \theta + \cos \theta )^2} + 4{\sin ^6}\theta+6(sinθ+cosθ)2+4sin6θ equals :
  1. A
    13 – 4 cos2 θ\thetaθ + 6sin2 θ\thetaθ cos2 θ\thetaθ
  2. B
    13 – 4 cos6 θ\thetaθ
  3. C
    13 – 4 cos2 θ\thetaθ + 6cos2 θ\thetaθ
  4. D
    13 – 4 cos4 θ\thetaθ + 2sin2 θ\thetaθ cos2 θ\thetaθ
View written solutionFree

Correct answer: NONE OF THE OPTIONS A, B, C, D IS CORRECT. THE EXPRESSION SIMPLIFIES TO $9+12\SIN^2\THETA\COS^2\THETA+4\SIN^6\THETA=9+3\SIN^2 2\THETA+4\SIN^6\THETA$.

  1. Let E=3(cos⁡θ−sin⁡θ)4+6(sin⁡θ+cos⁡θ)2+4sin⁡6θ.E=3(\cos\theta-\sin\theta)^4+6(\sin\theta+\cos\theta)^2+4\sin^6\theta.E=3(cosθ−sinθ)4+6(sinθ+cosθ)2+4sin6θ. We simplify each part.

  2. First, use sin⁡θ+cos⁡θ)2=sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=1+sin⁡2θ.\sin\theta+\cos\theta)^2=\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1+\sin2\theta.sinθ+cosθ)2=sin2θ+cos2θ+2sinθcosθ=1+sin2θ. So, 6(sin⁡θ+cos⁡θ)2=6(1+sin⁡2θ)=6+6sin⁡2θ.6(\sin\theta+\cos\theta)^2=6(1+\sin2\theta)=6+6\sin2\theta.6(sinθ+cosθ)2=6(1+sin2θ)=6+6sin2θ.

  3. Next, (cos⁡θ−sin⁡θ)2=cos⁡2θ+sin⁡2θ−2sin⁡θcos⁡θ=1−sin⁡2θ.(\cos\theta-\sin\theta)^2=\cos^2\theta+\sin^2\theta-2\sin\theta\cos\theta=1-\sin2\theta.(cosθ−sinθ)2=cos2θ+sin2θ−2sinθcosθ=1−sin2θ. Hence, (cos⁡θ−sin⁡θ)4=(1−sin⁡2θ)2=1−2sin⁡2θ+sin⁡22θ.(\cos\theta-\sin\theta)^4=(1-\sin2\theta)^2=1-2\sin2\theta+\sin^22\theta.(cosθ−sinθ)4=(1−sin2θ)2=1−2sin2θ+sin22θ. Therefore, 3(cos⁡θ−sin⁡θ)4=3−6sin⁡2θ+3sin⁡22θ.3(\cos\theta-\sin\theta)^4=3-6\sin2\theta+3\sin^22\theta.3(cosθ−sinθ)4=3−6sin2θ+3sin22θ.

  4. Add the first two parts: 3(cos⁡θ−sin⁡θ)4+6(sin⁡θ+cos⁡θ)23(\cos\theta-\sin\theta)^4+6(\sin\theta+\cos\theta)^23(cosθ−sinθ)4+6(sinθ+cosθ)2 =(3−6sin⁡2θ+3sin⁡22θ)+(6+6sin⁡2θ)=(3-6\sin2\theta+3\sin^22\theta)+(6+6\sin2\theta)=(3−6sin2θ+3sin22θ)+(6+6sin2θ) =9+3sin⁡22θ.=9+3\sin^22\theta.=9+3sin22θ. So, E=9+3sin⁡22θ+4sin⁡6θ.E=9+3\sin^22\theta+4\sin^6\theta.E=9+3sin22θ+4sin6θ.

  5. Now express sin⁡6θ\sin^6\thetasin6θ in terms of cos⁡2θ\cos2\thetacos2θ. Using sin⁡2θ=1−cos⁡2θ2,\sin^2\theta=\frac{1-\cos2\theta}{2},sin2θ=21−cos2θ​, we get

=\frac{(1-\cos2\theta)^3}{2}.$$ Expand: $$\frac{(1-\cos2\theta)^3}{2}=\frac{1-3\cos2\theta+3\cos^22\theta-\cos^32\theta}{2}.$$ Also, $$3\sin^22\theta=3(1-\cos^22\theta)=3-3\cos^22\theta.$$ Thus, $$E=9+(3-3\cos^22\theta)+\frac{1-3\cos2\theta+3\cos^22\theta-\cos^32\theta}{2}.$$ Combine: $$E=12-3\cos^22\theta+\frac{1}{2}-\frac{3}{2}\cos2\theta+\frac{3}{2}\cos^22\theta-\frac{1}{2}\cos^32\theta,$$ $$E=\frac{25}{2}-\frac{3}{2}\cos2\theta-\frac{3}{2}\cos^22\theta-\frac{1}{2}\cos^32\theta.$$ This form is not yet matching the options nicely, so let us instead use a better route. 6. Let $$x=\cos\theta-\sin\theta,\qquad y=\sin\theta+\cos\theta.$$ Then, $$x^2=(\cos\theta-\sin\theta)^2=1-\sin2\theta,$$ $$y^2=(\sin\theta+\cos\theta)^2=1+\sin2\theta.$$ Also, $$x^2+y^2=2.$$ Now use $$\sin\theta=\frac{y-x}{2}.$$ So, $$4\sin^6\theta=4\left(\frac{y-x}{2}\right)^6=\frac{(y-x)^6}{16},$$ which is cumbersome. So instead test the trigonometric identity from option B. 7. Option B says $$E=13-4\cos6\theta.$$ We verify this by checking with standard identity: $$\cos6\theta=1-18\sin^2\theta\cos^2\theta+16\sin^6\theta-24\sin^4\theta\cos^2\theta+6\cos^2\theta-?$$ A cleaner identity is obtained via $$\cos6\theta=32\cos^6\theta-48\cos^4\theta+18\cos^2\theta-1.$$ But an easier verification is to rewrite the expression in powers of $s=\sin\theta$, $c=\cos\theta$. 8. Expand directly: $$(c-s)^4=(c^2+s^2-2sc)^2=(1-2sc)^2=1-4sc+4s^2c^2.$$ Thus, $$3(c-s)^4=3-12sc+12s^2c^2.$$ Also, $$(s+c)^2=1+2sc,$$ so $$6(s+c)^2=6+12sc.$$ Hence, $$E=(3-12sc+12s^2c^2)+(6+12sc)+4s^6=9+12s^2c^2+4s^6.$$ Thus, $$E=9+12s^2c^2+4s^6.$$ 9. Now compare with option B: $$13-4\cos6\theta.$$ Take a test value, say $\theta=\frac{\pi}{3}$, which lies in $\left(\frac{\pi}{4},\frac{\pi}{2}\right)$. Then $$s=\sin\frac{\pi}{3}=\frac{\sqrt3}{2},\qquad c=\cos\frac{\pi}{3}=\frac12.$$ So, $$E=9+12\left(\frac34\cdot\frac14\right)+4\left(\frac{\sqrt3}{2}\right)^6 =9+12\cdot\frac{3}{16}+4\cdot\frac{27}{64}$$ $$=9+\frac94+\frac{27}{16}=\frac{144+36+27}{16}=\frac{207}{16}.$$ Now option B gives $$13-4\cos6\cdot\frac{\pi}{3}=13-4\cos2\pi=13-4=9.$$ This does **not** match. So option B is incorrect. 10. Check option A: $$13-4\cos2\theta+6\sin2\theta\cos2\theta.$$ For $\theta=\pi/3$, $$\cos2\theta=\cos\frac{2\pi}{3}=-\frac12, \qquad \sin2\theta=\sin\frac{2\pi}{3}=\frac{\sqrt3}{2}.$$ Then $$13-4\left(-\frac12\right)+6\left(\frac{\sqrt3}{2}\right)\left(-\frac12\right) =15-\frac{3\sqrt3}{2},$$ not equal to $\frac{207}{16}$. 11. Check option C: $$13-4\cos2\theta+6\cos2\theta=13+2\cos2\theta.$$ At $\theta=\pi/3$, $$13+2\left(-\frac12\right)=12,$$ not equal. 12. Check option D: $$13-4\cos4\theta+2\sin2\theta\cos2\theta.$$ At $\theta=\pi/3$, $$\cos4\theta=\cos\frac{4\pi}{3}=-\frac12.$$ So, $$13-4\left(-\frac12\right)+2\left(\frac{\sqrt3}{2}\right)\left(-\frac12\right) =15-\frac{\sqrt3}{2},$$ not equal. 13. Therefore, none of the given options matches the expression. 14. We can simplify the expression correctly: $$E=9+12\sin^2\theta\cos^2\theta+4\sin^6\theta.$$ Using $$12\sin^2\theta\cos^2\theta=3\sin^22\theta,$$ we get $$E=9+3\sin^22\theta+4\sin^6\theta.$$ This is the correct simplified form. So the stored answer $\boxed{B}$ is not correct.
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