JEE Main 2021 (Online) 27th July Morning ShiftQuestion 1
If $\sin\theta+\cos\theta=\frac12$, then $16(\sin(2\theta)+\cos(4\theta)+\sin(6\theta))$ is equal to :
Trigonometric Ratio & Identities
Solution
Concept. Squaring gives $\sin2\theta$; the other two terms follow from it. Formulas used. (1) $(\sin\theta+\cos\theta)^2=1+\sin2\theta$(2) $\cos4\theta=1-2\sin^22\theta$, $\sin6\theta=3\sin2\theta-4\sin^32\theta$Step 1 - $\sin2\theta$. By (1), $1+\sin2\theta=\frac14$, so $\sin2\theta=-\frac34$. Step 2 - Other terms. By (2): $\cos4\theta=1-\frac98=-\frac18$ and $\sin6\theta=-\frac94+\frac{27}{16}=-\frac{9}{16}$. Step 3 - Value. $16\left(-\frac{12}{16}-\frac{2}{16}-\frac{9}{16}\right)=-23$. Answer: (B)