हिंदी

If Sin α + Sin β = a and Cos α + Cos β = B, Show that Cos ( α + β ) = B 2 − a 2 B 2 + a 2

Advertisements
Advertisements

प्रश्न

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

\[\cos \left( \alpha + \beta \right) = \frac{b^2 - a^2}{b^2 + a^2}\]
संक्षेप में उत्तर
Advertisements

उत्तर

\[a^2 + b^2 = \left( \sin\alpha + \sin\beta \right)^2 + (\cos\alpha + \cos\beta)^2 \] 

\[ = \sin^2 \alpha + \sin^2 \beta + \cos^2 \alpha + \cos^2 \beta + 2\sin\alpha\sin\beta + 2\cos\alpha\cos\beta\]

\[ = 2 + 2 \cos(\alpha - \beta)\]  ........(1)

Now,

\[ \Rightarrow b^2 - a^2 = {(\cos\alpha + \cos\beta)}^2 - \left( \sin\alpha + \sin\beta \right)^2 \]

\[ \Rightarrow b^2 - a^2 = \cos^2 \alpha + \cos^2 \beta - \sin^2 \alpha - \sin^2 \beta + 2\cos\alpha\cos\beta - 2\sin\alpha\sin\beta\]

\[ \Rightarrow b^2 - a^2 = ( \cos^2 \alpha - \sin^2 \beta) + ( \cos^2 \beta - \sin^2 \alpha) - 2\cos(\alpha + \beta)\]

\[ \Rightarrow b^2 - a^2 = 2\cos(\alpha + \beta)\cos(\alpha - \beta) + 2\cos(\alpha - \beta)\]

\[ \Rightarrow b^2 - a^2 = \cos(\alpha + \beta)(2 + 2 \cos(\alpha - \beta)) \]  .........(2)

From (1) and (2), we have

\[ \Rightarrow b^2 - a^2 = \cos(\alpha + \beta)\left( a^2 + b^2 \right) \]   

\[\frac{b^2 - a^2}{a^2 + b^2} = \cos(\alpha + \beta)\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.1 [पृष्ठ २१]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 28.2 | पृष्ठ २१

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Prove the following: `cos (pi/4 xx x) cos (pi/4 - y) - sin (pi/4 -  x)sin (pi/4  - y) =  sin (x + y)`


Prove the following:

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


Prove the following:

`cos ((3pi)/ 2 + x ) cos(2pi + x) [cot ((3pi)/2 - x) + cot (2pi + x)]= 1`


Prove the following:

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


Prove the following:

sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x


Prove the following:

`(sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x`


Prove the following:

cos 4x = 1 – 8sinx cosx


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


 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 \[\sin A = \frac{3}{5}, \cos B = - \frac{12}{13}\], where A and B both lie in second quadrant, find the value of 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 \[\sin A = \frac{1}{2}, \cos B = \frac{12}{13}\], where \[\frac{\pi}{2}\]< A < π and \[\frac{3\pi}{2}\] < B < 2π, find tan (A − B).


Evaluate the following:
sin 78° cos 18° − cos 78° sin 18°


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)


Prove that:

\[\sin\left( \frac{3\pi}{8} - 5 \right)\cos\left( \frac{\pi}{8} + 5 \right) + \cos\left( \frac{3\pi}{8} - 5 \right)\sin\left( \frac{\pi}{8} + 5 \right) = 1\]

 


Prove that:

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

 


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


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 \[\tan\theta = \frac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}\] , then show that \[\sin\alpha + \cos\alpha = \sqrt{2}\cos\theta\].


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


If A + B = C, then write the value of tan A tan B tan C.


The value of \[\sin^2 \frac{5\pi}{12} - \sin^2 \frac{\pi}{12}\] 


If \[\tan A = \frac{a}{a + 1}\text{ and } \tan B = \frac{1}{2a + 1}\] 


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

 


\[\frac{\cos 10^\circ + \sin 10^\circ}{\cos 10^\circ - \sin 10^\circ} =\]

 


Show that 2 sin2β + 4 cos (α + β) sin α sin β + cos 2(α + β) = cos 2α


If `(sin(x + y))/(sin(x - y)) = (a + b)/(a - b)`, then show that `tanx/tany = a/b` [Hint: Use Componendo and Dividendo].


Find the general solution of the equation `(sqrt(3) - 1) costheta + (sqrt(3) + 1) sin theta` = 2

[Hint: Put `sqrt(3) - 1` = r sinα, `sqrt(3) + 1` = r cosα which gives tanα = `tan(pi/4 - pi/6)` α = `pi/12`]


If f(x) = cos2x + sec2x, then ______.

[Hint: A.M ≥ G.M.]


The value of tan 75° - cot 75° is equal to ______.


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


The value of `cot(pi/4 + theta)cot(pi/4 - theta)` is ______.


If sinx + cosx = a, then |sinx – cosx| = ______.


3(sinx – cosx)4 + 6(sinx + cosx)2 + 4(sin6x + cos6x) = ______.


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

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


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×