Recently Active Trignometry Questions

\text{The equation} (1+k)\cdot\frac{\cos x\cdot\cos(2x-\alpha)}{\cos(x-\alpha )} =1+k\cos 2x \; has no repeated root in it's domain of definition, if : a) |k| ≤ |csc α| , k ≠± 1 a) |k| ≤ |csc α| , k ≠- 1 a) |k| ≥ ...

\hspace{-16}$Find Max. and Min. value of $\mathbf{f(x)=\frac{2\cos x+2}{\sin x+\cos x+2}}$ ...

\hspace{-16}$It is known about real $\mathbf{a}$ and $\mathbf{b}$ that the inequality $\mathbf{a\; cosx +b\;cos3x >1}$\\\\ has no real solutions.then Prove that $\mathbf{\mid b\mid \leq 1}. ...

\hspace{-16}$Prove that If $\mathbf{n}$ is an positive Integer Then,\\\\ $\mathbf{\sum_{k=1}^{n}\cos^4\left(\frac{k\pi}{2n+1}\right)=\frac{6n-5}{16}}$ thanks guys now i have edited it ...

1) Please do this sum without using Complex no.(Actually this sum can be solved by pure trigonometry...) 1: if theta= pi/1999 then find the value of cos(theta) x cos(2theta) x cos(3theta) x ..... cos 999(theta) ...

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\hspace{-16}$Find all Integer Roots of the equation\\\\ $\mathbf{\cos\left(\frac{\pi}{10}\left(3x-\sqrt{9x^2+80x-40}\right)\right)=1}$ ...

\hspace{-16}$find minimum value of $f(\theta)=a\sec\theta+b\csc\theta$\\\\ $0 <\theta<\frac{\pi}{2}$ ...

If in a triangle ABC \textup{a }cosA+\textup{b }cosB+\textup{c }cosC = \textup{s } , where symbols have their usual meanings, then prove that the triangle is EQUILATERAL. ...

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the maximum value of the expression *Image* is,whee a and x are reals ...

\hspace{-16}$Calculate value of $2$ Trignometric expression\\\\\\ $\mathbf{(1)\; \cos \left(\frac{\pi}{7} \right)-\cos \left(\frac{2\pi}{7}\right)+\cos\left(\frac{3\pi}{7}\right)=}$\\\\\\ $\mathbf{(2)\; \sin \left(\frac{\pi}{ ...

\hspace{-16}$Show that the equation\\\\ $\mathbf{e^{1-\tan^{-1}x}+\tan^{-1}(e^x-1)=2}$\\\\ has no solution for $x\in\mathbb{R}$ ...

\hspace{-16}$Find value of $\mathbf{m}$ in \\\\ $\mathbf{\begin{Vmatrix} 50\sin^2 t+5m\sin t+(4m-41)=0 \\\\ 50\cos^2 t+5m\cos t+(4m-41)=0 & \end{Vmatrix}}$\\\\\\ and $\mathbf{\tan t\neq o}.$ Then value of $\mathbf{m}$ is ...

ABC is an equilateral triangle. P is any point in it satisfying PA=3, PB=4 & PC=5 units. Find area of the triangle. ...

COS-1x/a + COS-1y/b =d , then x2/a2 -2xy/abcos d +y2/b2 = 1.sin2 d 2.cos2d 3. tan2d 4.cot2d ...

1) tan9 - tan27- tan63 + tan81 =4 2) sin4 π/6 + sin4 3π/6 +sin4 5π/6 + sin4 7π/6 = 3/2 3)find the value of : 4cos20 - √3 cot20 4) find the value of : 2√2 sin10 [ sec 5/2 + cos 40/ cos5 - 2sin35] 5)sin212 + sin221 + si ...

\hspace{-16}$In a $\mathbf{\triangle ABC},$ Angle $\mathbf{A\;,B}$ and $\mathbf{C}$ satisfy the equation \\\\ $\mathbf{\cos 2A+\sqrt{3}.\cos 2B+\sqrt{3}.\cos 2C+\frac{5}{2}=0}$\\\\ Determine the Type of $\mathbf{\triangle}$ ...

Solve for x. 1+ \frac{1}{2\sin(30^0+ x)}= \frac{\sin(\frac{x}{2})}{\sin(\frac{x}{2}+ 60^0)}+ \frac{\sqrt{3}}{2\sin(\frac{x}{2}+ 60^0)} ...

\hspace{-16}$Solve $\mathbf{\sin^8 x+\cos^8 x=\frac{17}{32}}$ ...

we have to calculate the value of 1/x in a calculator, but the key of 1/x function is broken and we can only use the functions sinx , cosx. tanx. sin-1x ,cos-1x ,and tan-1x.how itz possible? ...

$Minimum value of $\mathbf{\frac{\tan\left(x+\frac{\pi}{6}\right)}{\tan x}}$ ...

Is there any non-geometrical proof of sin(A+B)=sinAcosB+cosAsinB ...

1) find the value of : 4cos20 - √3 cot20 2) 2cos40 - cos20/sin20 ...

1) tan9 - tan27- tan63 + tan81 =4 2) sin4 π/6 + sin4 3π/6 +sin4 5π/6 + sin4 7π/6 = 3/2 3)find the value of : 4cos20 - √3 cot20 4) find the value of : 2√2 sin10 [ sec 5/2 + cos 40/ cos5 - 2sin35] 5)sin212 + sin221 + si ...

\hspace{-16}$\textbf{(1)\;\; If $\mathbf{x,y\in\mathbb{R}}$ and $\mathbf{x^2+y^2=1}$. Then Max. value of $\mathbf{\mid x-y \mid +\mid x^3-y^3 \mid=}$ } ...

If Σ (i = 1 to n) cos -1xi = 0 , then Σ(i=1 to n) xi is equal to? (a)n (b)-n (c)0 (d) none of these ...

Solve the equation : cos2x + cos22x + cos23x = 1 for all x ...

if (a+2)sinx + (2a-1)cosx = (2a+1) find tanx. the options were (a) 3/4 (B) 4/3 (C) 2a/(a2+1) (D) 2a/(a2-1) well we can see that option B holds. i want the solution. ...

prove that - tan 2pi/13 + 4sin 6pi/13 = √(13+2√13) ...