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NCERT Exemplar · Q22

Q.Find the value of the expression 3[sin⁡4(3π2−α)+sin⁡4(3π+α)]−2[sin⁡6(π2+α)+sin⁡6(5π−α)]3\left[\sin^4\left(\dfrac{3\pi}{2} - \alpha\right) + \sin^4(3\pi + \alpha)\right] - 2\left[\sin^6\left(\dfrac{\pi}{2} + \alpha\right) + \sin^6(5\pi - \alpha)\right].

Rajasthan RbseLong· 5mImportance★★★★★est
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Reduce each trigonometric term using angle identities to express everything in terms of sin⁡α\sin \alpha and cos⁡α\cos \alpha, then recognize the resulting expression as a polynomial identity that simplifies to a constant.

The heart of this problem lies in understanding how sine behaves under shifts by multiples of π\pi and reflections about the axes. Once we translate each term into basic functions of α\alpha, the algebraic structure reveals itself.

Understanding the angle reductions

Every angle in this expression can be rewritten using the periodicity and symmetry of sine:

  • sin⁡(3π2−α)=sin⁡(3π2)cos⁡α−cos⁡(3π2)sin⁡α=−cos⁡α\sin\left(\frac{3\pi}{2} - \alpha\right) = \sin\left(\frac{3\pi}{2}\right)\cos\alpha - \cos\left(\frac{3\pi}{2}\right)\sin\alpha = -\cos\alpha
  • sin⁡(3π+α)=sin⁡(3π)cos⁡α+cos⁡(3π)sin⁡α=−sin⁡α\sin(3\pi + \alpha) = \sin(3\pi)\cos\alpha + \cos(3\pi)\sin\alpha = -\sin\alpha
  • sin⁡(π2+α)=sin⁡(π2)cos⁡α+cos⁡(π2)sin⁡α=cos⁡α\sin\left(\frac{\pi}{2} + \alpha\right) = \sin\left(\frac{\pi}{2}\right)\cos\alpha + \cos\left(\frac{\pi}{2}\right)\sin\alpha = \cos\alpha
  • sin⁡(5π−α)=sin⁡(5π)cos⁡α−cos⁡(5π)sin⁡α=sin⁡α\sin(5\pi - \alpha) = \sin(5\pi)\cos\alpha - \cos(5\pi)\sin\alpha = \sin\alpha
Tip

For angles of the form nπ±αn\pi \pm \alpha or (2n+1)π2±α\frac{(2n+1)\pi}{2} \pm \alpha, use the co-function and sign rules: sine at odd multiples of π2\frac{\pi}{2} becomes ±cos⁡\pm\cos, and sine at multiples of π\pi becomes ±sin⁡\pm\sin.

Step-by-step simplification

  1. Substitute the reduced angles into the first bracket:

sin⁡4(3π2−α)+sin⁡4(3π+α)=(−cos⁡α)4+(−sin⁡α)4=cos⁡4α+sin⁡4α\sin^4\left(\frac{3\pi}{2} - \alpha\right) + \sin^4(3\pi + \alpha) = (-\cos\alpha)^4 + (-\sin\alpha)^4 = \cos^4\alpha + \sin^4\alpha

  1. Substitute the reduced angles into the second bracket:

sin⁡6(π2+α)+sin⁡6(5π−α)=(cos⁡α)6+(sin⁡α)6=cos⁡6α+sin⁡6α\sin^6\left(\frac{\pi}{2} + \alpha\right) + \sin^6(5\pi - \alpha) = (\cos\alpha)^6 + (\sin\alpha)^6 = \cos^6\alpha + \sin^6\alpha

  1. Rewrite the entire expression: …
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