Top 50 Most Repeated TRIGONOMETRIC FUNCTIONS PYQs | JEE MAINS
A curated collection of the most important questions from TRIGONOMETRIC FUNCTIONS, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from TRIGONOMETRIC FUNCTIONS, fully solved with step-by-step concepts to prepare for JEE MAINS.
Using allied angle identities: $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos(150^\circ) = \cos(180^\circ ...
Read Full Step-by-Step Solution →First find cos θ = √(1‑sin²θ) = 4/5 (positive quadrant). Then sin 2θ = 2 sin θ cos θ = 2·(3/5)·(4/5) = 24/25....
Read Full Step-by-Step Solution →Using identity: $\sin A = \sin B \Rightarrow A = B + 2n\pi$ or $A = \pi - B + 2n\pi$. Solving gives $x = \frac{n\pi}{2}$....
Read Full Step-by-Step Solution →Use $\sin(2\theta)$ to find $\cos(2\theta)$, then apply double angle for $\tan(4\theta)$....
Read Full Step-by-Step Solution →$\sin x(1-2\cos x)=0$. Roots: $0, \pi, 2\pi, \pi/3, 5\pi/3$....
Read Full Step-by-Step Solution →Standard solution for $\sin x = \sin \alpha$ gives $x = n\pi + (-1)^n \alpha$....
Read Full Step-by-Step Solution →Foundational check for Trigonometric Functions in Class 11. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Using allied angle identities: $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos(150^\circ) = \cos(180^\circ ...
Read Full Step-by-Step Solution →First find cos θ = √(1‑sin²θ) = 4/5 (positive quadrant). Then sin 2θ = 2 sin θ cos θ = 2·(3/5)·(4/5) = 24/25....
Read Full Step-by-Step Solution →Using identity: $\sin A = \sin B \Rightarrow A = B + 2n\pi$ or $A = \pi - B + 2n\pi$. Solving gives $x = \frac{n\pi}{2}$....
Read Full Step-by-Step Solution →Use $\sin(2\theta)$ to find $\cos(2\theta)$, then apply double angle for $\tan(4\theta)$....
Read Full Step-by-Step Solution →$\sin x(1-2\cos x)=0$. Roots: $0, \pi, 2\pi, \pi/3, 5\pi/3$....
Read Full Step-by-Step Solution →Standard solution for $\sin x = \sin \alpha$ gives $x = n\pi + (-1)^n \alpha$....
Read Full Step-by-Step Solution →Foundational check for Trigonometric Functions in Class 11. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Using allied angle identities: $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos(150^\circ) = \cos(180^\circ ...
Read Full Step-by-Step Solution →First find cos θ = √(1‑sin²θ) = 4/5 (positive quadrant). Then sin 2θ = 2 sin θ cos θ = 2·(3/5)·(4/5) = 24/25....
Read Full Step-by-Step Solution →Using identity: $\sin A = \sin B \Rightarrow A = B + 2n\pi$ or $A = \pi - B + 2n\pi$. Solving gives $x = \frac{n\pi}{2}$....
Read Full Step-by-Step Solution →Use $\sin(2\theta)$ to find $\cos(2\theta)$, then apply double angle for $\tan(4\theta)$....
Read Full Step-by-Step Solution →$\sin x(1-2\cos x)=0$. Roots: $0, \pi, 2\pi, \pi/3, 5\pi/3$....
Read Full Step-by-Step Solution →Standard solution for $\sin x = \sin \alpha$ gives $x = n\pi + (-1)^n \alpha$....
Read Full Step-by-Step Solution →Foundational check for Trigonometric Functions in Class 11. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Using allied angle identities: $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos(150^\circ) = \cos(180^\circ ...
Read Full Step-by-Step Solution →First find cos θ = √(1‑sin²θ) = 4/5 (positive quadrant). Then sin 2θ = 2 sin θ cos θ = 2·(3/5)·(4/5) = 24/25....
Read Full Step-by-Step Solution →Using identity: $\sin A = \sin B \Rightarrow A = B + 2n\pi$ or $A = \pi - B + 2n\pi$. Solving gives $x = \frac{n\pi}{2}$....
Read Full Step-by-Step Solution →Use $\sin(2\theta)$ to find $\cos(2\theta)$, then apply double angle for $\tan(4\theta)$....
Read Full Step-by-Step Solution →$\sin x(1-2\cos x)=0$. Roots: $0, \pi, 2\pi, \pi/3, 5\pi/3$....
Read Full Step-by-Step Solution →Standard solution for $\sin x = \sin \alpha$ gives $x = n\pi + (-1)^n \alpha$....
Read Full Step-by-Step Solution →Foundational check for Trigonometric Functions in Class 11. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Using allied angle identities: $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos(150^\circ) = \cos(180^\circ ...
Read Full Step-by-Step Solution →First find cos θ = √(1‑sin²θ) = 4/5 (positive quadrant). Then sin 2θ = 2 sin θ cos θ = 2·(3/5)·(4/5) = 24/25....
Read Full Step-by-Step Solution →Using identity: $\sin A = \sin B \Rightarrow A = B + 2n\pi$ or $A = \pi - B + 2n\pi$. Solving gives $x = \frac{n\pi}{2}$....
Read Full Step-by-Step Solution →Use $\sin(2\theta)$ to find $\cos(2\theta)$, then apply double angle for $\tan(4\theta)$....
Read Full Step-by-Step Solution →$\sin x(1-2\cos x)=0$. Roots: $0, \pi, 2\pi, \pi/3, 5\pi/3$....
Read Full Step-by-Step Solution →Standard solution for $\sin x = \sin \alpha$ gives $x = n\pi + (-1)^n \alpha$....
Read Full Step-by-Step Solution →Foundational check for Trigonometric Functions in Class 11. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Using allied angle identities: $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos(150^\circ) = \cos(180^\circ ...
Read Full Step-by-Step Solution →First find cos θ = √(1‑sin²θ) = 4/5 (positive quadrant). Then sin 2θ = 2 sin θ cos θ = 2·(3/5)·(4/5) = 24/25....
Read Full Step-by-Step Solution →Using identity: $\sin A = \sin B \Rightarrow A = B + 2n\pi$ or $A = \pi - B + 2n\pi$. Solving gives $x = \frac{n\pi}{2}$....
Read Full Step-by-Step Solution →Use $\sin(2\theta)$ to find $\cos(2\theta)$, then apply double angle for $\tan(4\theta)$....
Read Full Step-by-Step Solution →$\sin x(1-2\cos x)=0$. Roots: $0, \pi, 2\pi, \pi/3, 5\pi/3$....
Read Full Step-by-Step Solution →Standard solution for $\sin x = \sin \alpha$ gives $x = n\pi + (-1)^n \alpha$....
Read Full Step-by-Step Solution →Foundational check for Trigonometric Functions in Class 11. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Using allied angle identities: $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos(150^\circ) = \cos(180^\circ ...
Read Full Step-by-Step Solution →First find cos θ = √(1‑sin²θ) = 4/5 (positive quadrant). Then sin 2θ = 2 sin θ cos θ = 2·(3/5)·(4/5) = 24/25....
Read Full Step-by-Step Solution →Using identity: $\sin A = \sin B \Rightarrow A = B + 2n\pi$ or $A = \pi - B + 2n\pi$. Solving gives $x = \frac{n\pi}{2}$....
Read Full Step-by-Step Solution →Use $\sin(2\theta)$ to find $\cos(2\theta)$, then apply double angle for $\tan(4\theta)$....
Read Full Step-by-Step Solution →$\sin x(1-2\cos x)=0$. Roots: $0, \pi, 2\pi, \pi/3, 5\pi/3$....
Read Full Step-by-Step Solution →Standard solution for $\sin x = \sin \alpha$ gives $x = n\pi + (-1)^n \alpha$....
Read Full Step-by-Step Solution →Foundational check for Trigonometric Functions in Class 11. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →