English

Prove that cos 570° sin 510° + sin (−330°) cos (−390°) = 0

Advertisements
Advertisements

Question

Prove that cos 570° sin 510° + sin (−330°) cos (−390°) = 0
Sum
Advertisements

Solution

LHS = cos(570°) sin(510°) + sin(−330°) cos(−390°)

= cos(570°) sin(510°) + [−sin(330°)]cos(390°)

= cos(570°) sin(510°) −sin(330°)

= cos(90° × 6 + 30°) sin(90° × 5 + 60°) − sin(90° × 3 + 60°) cos(90° × 4 + 30°)

= − cos(30°) cos(60°) − [−cos(60°)] cos(30°)]

= − cos(30°) cos(60°) + cos(30°) sin(60°)

= 0

= RHS

Hence proved .

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Trigonometric Functions - Exercise 5.3 [Page 39]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 5 Trigonometric Functions
Exercise 5.3 | Q 2.5 | Page 39

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


Find the general solution of the equation cos 4 x = cos 2 x


Find the general solution of the equation  sin x + sin 3x + sin 5x = 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 \[6 T_{10} - 15 T_8 + 10 T_6 - 1 = 0\]


Prove that:  tan 225° cot 405° + tan 765° cot 675° = 0


Prove that: cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = \[\frac{1}{2}\]


Prove that: \[\tan\frac{11\pi}{3} - 2\sin\frac{4\pi}{6} - \frac{3}{4} {cosec}^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6} = \frac{3 - 4\sqrt{3}}{2}\]

 


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

\[\frac{cosec(90^\circ + x) + \cot(450^\circ + x)}{cosec(90^\circ - x) + \tan(180^\circ - x)} + \frac{\tan(180^\circ + x) + \sec(180^\circ - x)}{\tan(360^\circ + x) - \sec( - x)} = 2\]

 


Prove that

\[\left\{ 1 + \cot x - \sec\left( \frac{\pi}{2} + x \right) \right\}\left\{ 1 + \cot x + \sec\left( \frac{\pi}{2} + x \right) \right\} = 2\cot x\]

 


Prove that

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

 


Prove that:
\[\sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9} = 2\]

 

If \[\frac{\pi}{2} < x < \frac{3\pi}{2},\text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}}\] is equal to

 


sin6 A + cos6 A + 3 sin2 A cos2 A =


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

 


If \[f\left( x \right) = \cos^2 x + \sec^2 x\], then


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

 

Find the general solution of the following equation:

\[\sin 2x + \cos x = 0\]

Solve the following equation:
\[\sin^2 x - \cos x = \frac{1}{4}\]


Solve the following equation:

\[2 \cos^2 x - 5 \cos x + 2 = 0\]

Solve the following equation:

\[\sin 2x - \sin 4x + \sin 6x = 0\]

Solve the following equation:
 sin x tan x – 1 = tan x – sin x

 


Solve the following equation:
3tanx + cot x = 5 cosec x


If secx cos5x + 1 = 0, where \[0 < x \leq \frac{\pi}{2}\], find the value of x.


Write the set of values of a for which the equation

\[\sqrt{3} \sin x - \cos x = a\] has no solution.

Write the solution set of the equation 

\[\left( 2 \cos x + 1 \right) \left( 4 \cos x + 5 \right) = 0\] in the interval [0, 2π].

If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.


If \[\sqrt{3} \cos x + \sin x = \sqrt{2}\] , then general value of x is


General solution of \[\tan 5 x = \cot 2 x\] is


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:
cos 2θ = `(sqrt(5) + 1)/4`


Solve the following equations:
2cos 2x – 7 cos x + 3 = 0


Choose the correct alternative:
If sin α + cos α = b, then sin 2α is equal to


Solve 2 tan2x + sec2x = 2 for 0 ≤ x ≤ 2π.


Find the general solution of the equation 5cos2θ + 7sin2θ – 6 = 0


In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×