मराठी

Prove the following: cos2 2x – cos2 6x = sin 4x sin 8x - Mathematics

Advertisements
Advertisements

प्रश्न

Prove the following:

cos2 2x – cos2 6x = sin 4x sin 8x

बेरीज
Advertisements

उत्तर

cos2 2x – cos2 6x = (cos2x + cos6x) (cos 2x - cos 6x)

= `2cos  ((2x + 6x)/2) cos ((2x - 6 x) /2) - 2sin ((2x - 6x)/2) sin ((2x + 6x)/2)`

= (2cos 4x cos2x) (2sin 4x sin 2x)

= (2sin 2x cos2x)(2sin 4x cos4x) = sin 4x sin 8x

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometric Functions - Exercise 3.3 [पृष्ठ ७३]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 11
पाठ 3 Trigonometric Functions
Exercise 3.3 | Q 13 | पृष्ठ ७३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Prove that: `sin^2  pi/6 + cos^2  pi/3 - tan^2  pi/4 = -1/2`


Prove the following:

`cos ((3pi)/4 + x) - cos((3pi)/4 - x) = -sqrt2 sin x`


Prove the following:

sin2 6x – sin2 4x = sin 2x sin 10x


Prove the following:

`(sin x + sin 3x)/(cos x + cos 3x) = tan 2x`


Prove the following:

`(cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x`


Prove the following:

cos 4x = 1 – 8sinx cosx


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


Evaluate the following:
cos 47° cos 13° − sin 47° sin 13°


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

\[\frac{\cos 9^\circ + \sin 9^\circ}{\cos 9^\circ - \sin 9^\circ} = \tan 54^\circ\]

Prove that

\[\frac{\cos 8^\circ - \sin 8^\circ}{\cos 8^\circ + \sin 8^\circ} = \tan 37^\circ\]

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


Prove that:
\[\cos^2 45^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{4}\]


Prove that:
sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.


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


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:
cos2 A + cos2 B − 2 cos A cos B cos (A + B) = sin2 (A + B)


Prove that:
tan 8x − tan 6x − tan 2x = tan 8x tan 6x tan 2x


Prove that:
\[\frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} = \tan 3x \tan x\]


Prove that sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.

 

If tan A + tan B = a and cot A + cot B = b, prove that cot (A + B) \[\frac{1}{a} - \frac{1}{b}\].


If sin α + sin β = a and cos α + cos β = b, show that

\[\sin \left( \alpha + \beta \right) = \frac{2ab}{a^2 + b^2}\]

 


If sin α sin β − cos α cos β + 1 = 0, prove that 1 + cot α tan β = 0.


Show that sin 100° − sin 10° is positive. 


If A + B + C = π, then sec A (cos B cos C − sin B sin C) is equal to


If in ∆ABC, tan A + tan B + tan C = 6, then cot A cot B cot C =


If \[\cos P = \frac{1}{7}\text{ and }\cos Q = \frac{13}{14}\], where P and Q both are acute angles. Then, the value of P − Q is

 


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


If tan 69° + tan 66° − tan 69° tan 66° = 2k, then k =


Express the following as the sum or difference of sines and cosines:
2 sin 4x sin 3x


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


If sinθ + cosθ = 1, then find the general value of θ.


The value of tan3A - tan2A - tanA is equal to ______.


Given x > 0, the values of f(x) = `-3cos sqrt(3 + x + x^2)` lie in the interval ______.


The maximum distance of a point on the graph of the function y = `sqrt(3)` sinx + cosx from x-axis is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×