हिंदी

Prove the following: sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x - Mathematics

Advertisements
Advertisements

प्रश्न

Prove the following:

sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

योग
Advertisements

उत्तर

L.H.S.

= sin (n + 1)x sin (n + 2) x + cos (n + 1)x cos (n + 2)x

Let (n + 2)x = A, (n + 1) x = B

= sin B sin A + cos B cos A

= cos A cos B + sin A sin B

= cos (A – B) = cos [(n + 2) x – (n + 1)x]

[∵ By keeping the values of A and B]

= cos (nx + 2x – nx –x)

= cos x = R.H.S.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometric Functions - EXERCISE 3.3 [पृष्ठ ६७]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 11
अध्याय 3 Trigonometric Functions
EXERCISE 3.3 | Q 10. | पृष्ठ ६७

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Find the value of: sin 75°


Prove the following:

`(cos (pi + x) cos (-x))/(sin(pi - x) cos (pi/2 + x)) =  cot^2 x`


Prove the following:

`(cos9x - cos5x)/(sin17x - sin 3x) = - (sin2x)/(cos 10x)`


Prove that: `(cos x  + cos y)^2 + (sin x - sin y )^2 =  4 cos^2  (x + y)/2`


Prove that: `((sin 7x + sin 5x) + (sin 9x + sin 3x))/((cos 7x + cos 5x) + (cos 9x + cos 3x)) = tan 6x`


 If \[\sin A = \frac{12}{13}\text{ and } \sin B = \frac{4}{5}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
sin (A + B)


 If \[\sin A = \frac{12}{13}\text{ and } \sin B = \frac{4}{5}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
cos (A + B)


If \[\sin A = \frac{3}{5}, \cos B = - \frac{12}{13}\], where A and B both lie in second quadrant, find the value of sin (A + B).


If \[\cos A = - \frac{24}{25}\text{ and }\cos B = \frac{3}{5}\], where π < A < \[\frac{3\pi}{2}\text{ and }\frac{3\pi}{2}\]< B < 2π, find the following:
sin (A + B)


If \[\tan A = \frac{3}{4}, \cos B = \frac{9}{41}\], where π < A < \[\frac{3\pi}{2}\] and 0 < B <\[\frac{\pi}{2}\], find tan (A + B).

 


If \[\sin A = \frac{1}{2}, \cos B = \frac{\sqrt{3}}{2}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
tan (A - B)


If \[\cos A = - \frac{12}{13}\text{ and }\cot B = \frac{24}{7}\], where A lies in the second quadrant and B in the third quadrant, find the values of the following:
sin (A + B)


If \[\cos A = - \frac{12}{13}\text{ and }\cot B = \frac{24}{7}\], where A lies in the second quadrant and B in the third quadrant, find the values of the following:
cos (A + B)


Prove that \[\frac{\tan 69^\circ + \tan 66^\circ}{1 - \tan 69^\circ \tan 66^\circ} = - 1\].


If \[\tan A = \frac{m}{m - 1}\text{ and }\tan B = \frac{1}{2m - 1}\], then prove that \[A - B = \frac{\pi}{4}\].


Prove that:
\[\frac{\sin \left( A - B \right)}{\cos A \cos B} + \frac{\sin \left( B - C \right)}{\cos B \cos C} + \frac{\sin \left( C - A \right)}{\cos C \cos A} = 0\]

 


Prove that:
\[\frac{\tan \left( A + B \right)}{\cot \left( A - B \right)} = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}\]


Prove that:
tan 36° + tan 9° + tan 36° tan 9° = 1


If x lies in the first quadrant and \[\cos x = \frac{8}{17}\], then prove that:

\[\cos \left( \frac{\pi}{6} + x \right) + \cos \left( \frac{\pi}{4} - x \right) + \cos \left( \frac{2\pi}{3} - x \right) = \left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} \right)\frac{23}{17}\]

 


If α, β are two different values of x lying between 0 and 2π, which satisfy the equation 6 cos x + 8 sin x = 9, find the value of sin (α + β).

 

Prove that:

\[\frac{1}{\cos \left( x - a \right) \cos \left( a - b \right)} = \frac{\tan \left( x - b \right) - \tan \left( x - a \right)}{\sin \left( a - b \right)}\]

 


If angle \[\theta\]  is divided into two parts such that the tangents of one part is \[\lambda\] times the tangent of other, and \[\phi\] is their difference, then show that\[\sin\theta = \frac{\lambda + 1}{\lambda - 1}\sin\phi\]

 

If α and β are two solutions of the equation a tan x + b sec x = c, then find the values of sin (α + β) and cos (α + β).

 

Reduce each of the following expressions to the sine and cosine of a single expression: 

cos x − sin 


Prove that \[\left( 2\sqrt{3} + 3 \right) \sin x + 2\sqrt{3} \cos x\]  lies between \[- \left( 2\sqrt{3} + \sqrt{15} \right) \text{ and } \left( 2\sqrt{3} + \sqrt{15} \right)\]


If α + β − γ = π and sin2 α +sin2 β − sin2 γ = λ sin α sin β cos γ, then write the value of λ. 


If A + B + C = π, then \[\frac{\tan A + \tan B + \tan C}{\tan A \tan B \tan C}\] is equal to

 

The value of \[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\] is

 

The maximum value of \[\sin^2 \left( \frac{2\pi}{3} + x \right) + \sin^2 \left( \frac{2\pi}{3} - x \right)\] is


Express the following as the sum or difference of sines and cosines:
 2 cos 7x cos 3x


If α and β are the solutions of the equation a tan θ + b sec θ = c, then show that tan (α + β) = `(2ac)/(a^2 - c^2)`.


If sinθ + cosecθ = 2, then sin2θ + cosec2θ is equal to ______.


The value of sin(45° + θ) - cos(45° - θ) is ______.


State whether the statement is True or False? Also give justification.

If tanA = `(1 - cos B)/sinB`, then tan2A = tanB


State whether the statement is True or False? Also give justification.

If tanθ + tan2θ + `sqrt(3)` tanθ tan2θ = `sqrt(3)`, then θ = `("n"pi)/3 + pi/9`


In the following match each item given under the column C1 to its correct answer given under the column C2:

Column A Column B
(a) sin(x + y) sin(x – y) (i) cos2x – sin2y
(b) cos (x + y) cos (x – y) (ii) `(1 - tan theta)/(1 + tan theta)`
(c) `cot(pi/4 + theta)` (iii) `(1 + tan theta)/(1 - tan theta)`
(d) `tan(pi/4 + theta)` (iv) sin2x – sin2y

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×