Category: Trigonometry

Question Number 18523 by Tinkutara last updated on 23/Jul/17 $$\mathrm{Match}\mathrm{the}\mathrm{following}$$$$\mathbf{Column}-\mathbf{I}\left(\mathbf{Trigonometric}\mathbf{equation}\right)$$$$\left(\mathrm{A}\right)\sin 9\theta =\cos (\frac{\pi }{2}-\theta )$$$$\left(\mathrm{B}\right)\sin 5\theta =\sin (\frac{\pi }{2}+2\theta )$$$$\left(\mathrm{C}\right)\cos 11\theta =\cos 3\theta$$$$\left(\mathrm{D}\right)3\tan (\theta -15°)=\tan (\theta +15°)$$$$\mathbf{Column}-\mathbf{II}\left(\mathbf{Family}\mathbf{of}\mathbf{solutions}\right)$$$$\left(\mathrm{p}\right)\:\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\frac{\pi}{\mathrm{10}},\:{n}\:\in\:{Z} \…

Question Number 18501 by Tinkutara last updated on 22/Jul/17 $$\mathrm{In}\mathrm{a}\mathrm{triangle}\mathrm{ABC}$$$$\left(1\right)\sin A.\sin B.\sin C=\frac{\mathrm{\Delta }}{2R^2}$$$$\left(2\right)\sin A.\sin B.\sin C=\frac{r}{2R}(\sin A+\sin B+\sin C)$$$$\left(3\right)a\cos A+b\cos B+c\cos C=\frac{abc}{2R^2}$$$$\left(4\right)\sin A.\sin B.\sin C=\frac{R}{2r}(\sin A+\sin B+\sin C)$$ Commented by b.e.h.i.8.3.417@gmail.com…

Question Number 18474 by Tinkutara last updated on 22/Jul/17 $$\mathbf{Assertion}-\mathbf{Reason}\mathbf{Type}\mathbf{Question}$$$$\mathrm{STATEMENT}-1:f\left(x\right)={\log }_{\cos x}\sin x\mathrm{is}$$$$\mathrm{well}\mathrm{defined}\mathrm{in}(0,\frac{\pi }{2}).$$$$\mathbf{and}$$$$\mathrm{STATEMENT}-2:\sin x\mathrm{and}\cos x\mathrm{are}$$$$\mathrm{positive}\mathrm{in}(0,\frac{\pi }{2}).$$ Answered by…

Question Number 18457 by Tinkutara last updated on 21/Jul/17 $$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{equation}$$$$\sin \theta +\cos \theta =1+\sin \theta \cos \theta \mathrm{in}\mathrm{the}$$$$\mathrm{interval}[0,4\pi ]\mathrm{is}$$ Answered by mrW1 last updated on 21/Jul/17 $$\mathrm{sin}\:\theta\:+\:\mathrm{cos}\:\theta\:=\:\mathrm{1}\:+\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta \…

Question Number 18456 by Tinkutara last updated on 21/Jul/17 $$\mathrm{The}\mathrm{complete}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{equation}$$$$\sin 2x-12(\sin x-\cos x)+12=0\mathrm{is}$$$$\mathrm{given}\mathrm{by}$$$$\left(1\right)x=2n\pi +\frac{\pi }{2},(2n-1)\frac{\pi }{4},n\in Z$$$$\left(2\right)x=n\pi +\frac{\pi }{2},(2n+1)\pi ,n\in Z$$$$\left(3\right)x=2n\pi +\frac{\pi }{2},(2n+1)\pi ,n\in Z$$$$\left(4\right)x=n\pi +\frac{\pi }{2},(2n-1)\pi ,n\in Z$$ Answered…