Advertisements
Advertisements
प्रश्न
Prove that: cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = \[\frac{1}{2}\]
Advertisements
उत्तर
LHS =\[ \cos 24^\circ + \cos 55^\circ + \cos 125^\circ + \cos 204^\circ + \cos 300^\circ\]
\[ = \cos 24^\circ + \cos \left( 90^\circ - 35^\circ \right) + \cos \left( 90^\circ e \times 1 + 35^\circ \right) + \cos \left( 90^\circ \times 2 + 24^\circ \right) + \cos \left( 90^\circ \times 3 + 30^\circ \right)\]
\[ = \cos 24^\circ + \sin 35^\circ - \sin 35^\circ e - \cos 24^\circ + \sin 30^\circ \]
\[ = 0 + 0 + \frac{1}{2}\]
\[ = \frac{1}{2}\]
= RHS
Hence proved .
APPEARS IN
संबंधित प्रश्न
If \[x = \frac{2 \sin x}{1 + \cos x + \sin x}\], then prove that
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 \[m^2 \leq 2\]
Prove that: tan 225° cot 405° + tan 765° cot 675° = 0
Prove that:
\[\frac{\cos (2\pi + x) cosec (2\pi + x) \tan (\pi/2 + x)}{\sec(\pi/2 + x)\cos x \cot(\pi + x)} = 1\]
Prove that
Prove that:
\[\sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9} = 2\]
In a ∆ABC, prove that:
If tan x = \[x - \frac{1}{4x}\], then sec x − tan x is equal to
If \[cosec x + \cot x = \frac{11}{2}\], then tan x =
If tan θ + sec θ =ex, then cos θ equals
Which of the following is correct?
Find the general solution of the following equation:
Find the general solution of the following equation:
Find the general solution of the following equation:
Find the general solution of the following equation:
Find the general solution of the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
\[\sin x + \cos x = \sqrt{2}\]
Solve the following equation:
Solve the following equation:
sin x tan x – 1 = tan x – sin x
Write the number of solutions of the equation
\[4 \sin x - 3 \cos x = 7\]
Write the number of points of intersection of the curves
If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.
A solution of the equation \[\cos^2 x + \sin x + 1 = 0\], lies in the interval
The solution of the equation \[\cos^2 x + \sin x + 1 = 0\] lies in the interval
Find the principal solution and general solution of the following:
sin θ = `-1/sqrt(2)`
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
cos 2x = 1 − 3 sin x
Solve the following equations:
cos θ + cos 3θ = 2 cos 2θ
Solve the following equations:
sin 2θ – cos 2θ – sin θ + cos θ = θ
Solve the following equations:
2cos 2x – 7 cos x + 3 = 0
Choose the correct alternative:
If cos pθ + cos qθ = 0 and if p ≠ q, then θ is equal to (n is any integer)
Solve the equation sin θ + sin 3θ + sin 5θ = 0
Find the general solution of the equation 5cos2θ + 7sin2θ – 6 = 0
