मराठी

Sin2 π/18 + Sin2 π/9 + Sin2 7π/18 + Sin2 4π/9 = - Mathematics

Advertisements
Advertisements

प्रश्न

sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9 =

पर्याय

  • 1

  • 4

  • 2

  • 0

MCQ

उत्तर

2
We have: 
\[ \sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9}\]
\[ = \sin^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{8\pi}{18}\]
\[ = \sin^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{8\pi}{18} \right)\]
\[ = \sin^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \sin^2 \left( \frac{\pi}{2} - \frac{2\pi}{18} \right) + \sin^2 \left( \frac{\pi}{2} - \frac{\pi}{18} \right)\]
\[ = \sin^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \cos^2 \frac{2\pi}{18} + \cos^2 \frac{\pi}{18}\]
\[ = \sin^2 \frac{\pi}{18} + \cos^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \cos^2 \frac{2\pi}{18}\]
\[ = 1 + 1\]
\[ = 2\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Trigonometric Functions - Exercise 5.5 [पृष्ठ ४२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 5 Trigonometric Functions
Exercise 5.5 | Q 17 | पृष्ठ ४२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the principal and general solutions of the equation  `cot x = -sqrt3`


Find the general solution of the equation cos 3x + cos x – cos 2x = 0


If \[\sin x + \cos x = m\], then prove that \[\sin^6 x + \cos^6 x = \frac{4 - 3 \left( m^2 - 1 \right)^2}{4}\], where \[m^2 \leq 2\]


If \[a = \sec x - \tan x \text{ and }b = cosec x + \cot x\], then shown that  \[ab + a - b + 1 = 0\]


If \[T_n = \sin^n x + \cos^n x\], prove that \[\frac{T_3 - T_5}{T_1} = \frac{T_5 - T_7}{T_3}\]

 


If \[T_n = \sin^n x + \cos^n x\], prove that  \[2 T_6 - 3 T_4 + 1 = 0\]


If \[T_n = \sin^n x + \cos^n x\], prove that \[6 T_{10} - 15 T_8 + 10 T_6 - 1 = 0\]


Prove that:

\[3\sin\frac{\pi}{6}\sec\frac{\pi}{3} - 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4} = 1\]

 


Prove that

\[\frac{\sin(180^\circ + x) \cos(90^\circ + x) \tan(270^\circ - x) \cot(360^\circ - x)}{\sin(360^\circ - x) \cos(360^\circ + x) cosec( - x) \sin(270^\circ + x)} = 1\]

 


In a ∆A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that cos(180° − A) + cos (180° + B) + cos (180° + C) − sin (90° + D) = 0


Prove that:
\[\tan 4\pi - \cos\frac{3\pi}{2} - \sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac{1}{4}\]


Prove that:
\[\sin\frac{13\pi}{3}\sin\frac{8\pi}{3} + \cos\frac{2\pi}{3}\sin\frac{5\pi}{6} = \frac{1}{2}\]


If sec \[x = x + \frac{1}{4x}\], then sec x + tan x = 

 

If \[0 < x < \frac{\pi}{2}\], and if \[\frac{y + 1}{1 - y} = \sqrt{\frac{1 + \sin x}{1 - \sin x}}\], then y is equal to


If x = r sin θ cos ϕ, y = r sin θ sin ϕ and r cos θ, then x2 + y2 + z2 is independent of


If tan \[x = - \frac{1}{\sqrt{5}}\] and θ lies in the IV quadrant, then the value of cos x is

 

If \[\frac{3\pi}{4} < \alpha < \pi, \text{ then }\sqrt{2\cot \alpha + \frac{1}{\sin^2 \alpha}}\] is equal to


If \[cosec x - \cot x = \frac{1}{2}, 0 < x < \frac{\pi}{2},\]

 

\[\sec^2 x = \frac{4xy}{(x + y )^2}\] is true if and only if

 


If x is an acute angle and \[\tan x = \frac{1}{\sqrt{7}}\], then the value of \[\frac{{cosec}^2 x - \sec^2 x}{{cosec}^2 x + \sec^2 x}\] is


If tan A + cot A = 4, then tan4 A + cot4 A is equal to


If A lies in second quadrant 3tan A + 4 = 0, then the value of 2cot A − 5cosA + sin A is equal to


Which of the following is incorrect?


The value of \[\tan1^\circ \tan2^\circ \tan3^\circ . . . \tan89^\circ\] is

 

Find the general solution of the following equation:

\[\sqrt{3} \sec x = 2\]

Find the general solution of the following equation:

\[\cos 3x = \frac{1}{2}\]

Find the general solution of the following equation:

\[\tan px = \cot qx\]

 


Solve the following equation:

\[\tan x + \tan 2x = \tan 3x\]

Solve the following equation:
\[2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}\]


A solution of the equation \[\cos^2 x + \sin x + 1 = 0\], lies in the interval


The general value of x satisfying the equation
\[\sqrt{3} \sin x + \cos x = \sqrt{3}\]


If \[\cos x = - \frac{1}{2}\] and 0 < x < 2\pi, then the solutions are


Solve the following equations:
sin θ + sin 3θ + sin 5θ = 0


Solve the following equations:
`tan theta + tan (theta + pi/3) + tan (theta + (2pi)/3) = sqrt(3)`


Choose the correct alternative:
If tan 40° = λ, then `(tan 140^circ - tan 130^circ)/(1 + tan 140^circ *  tan 130^circ)` =


Choose the correct alternative:
If f(θ) = |sin θ| + |cos θ| , θ ∈ R, then f(θ) is in the interval


Solve the equation sin θ + sin 3θ + sin 5θ = 0


If sin θ and cos θ are the roots of the equation ax2 – bx + c = 0, then a, b and c satisfy the relation ______.


Find the general solution of the equation sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×