JEE Main 2021 — Mathematics Trigonometry

$$\mathrm{sin}\theta +\mathrm{cos}\theta =\frac{1}{2}$$sinθ+cosθ=21​

$${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta +2\mathrm{sin}\theta \mathrm{cos}\theta =\frac{1}{4}\Rightarrow \mathrm{sin}2\theta =-\frac{3}{4}$$sin2θ+cos2θ+2sinθcosθ=41​⇒sin2θ=−43​

Now:

$$\mathrm{cos}4\theta =1-2{\mathrm{sin}}^{2}2\theta$$cos4θ=1−2sin22θ

$$=1-2{(-\frac{3}{4})}^{2}$$=1−2(−43​)2

$$=1-2\times \frac{9}{16}=-\frac{1}{8}$$=1−2×169​=−81​

And $\mathrm{sin}6\theta =3\mathrm{sin}2\theta -4{\mathrm{sin}}^{3}2\theta$sin6θ=3sin2θ−4sin32θ

$$=(3-4{\mathrm{sin}}^{2}2\theta )\cdot \mathrm{sin}2\theta$$=(3−4sin22θ)⋅sin2θ

$$=[3-4(\frac{9}{16})]\cdot (-\frac{3}{4})$$=[3−4(169​)]⋅(−43​)

$$=[\frac{3}{4}]\times (-\frac{3}{4})=-\frac{9}{16}$$=[43​]×(−43​)=−169​

So, $16[\mathrm{sin}2\theta +\mathrm{cos}4\theta +\mathrm{sin}6\theta ]$16[sin2θ+cos4θ+sin6θ]

$$=16(-\frac{3}{4}-\frac{1}{8}-\frac{9}{16})=-23$$=16(−43​−81​−169​)=−23