हिंदी

Prove That: 2 Sin 5 π 12 Sin π 12 = 1 2 - Mathematics

Advertisements
Advertisements

प्रश्न

Prove that:

\[2\sin\frac{5\pi}{12}\sin\frac{\pi}{12} = \frac{1}{2}\]

 

योग

उत्तर

\[LHS = 2\left( \sin \frac{5\pi}{12} \right) \left( \sin \frac{\pi}{12} \right)\]
\[ = \cos \left( \frac{5\pi}{12} - \frac{\pi}{12} \right) - \cos \left( \frac{5\pi}{12} + \frac{\pi}{12} \right) \left[ \because 2 \sin A \sin B = \cos (A - B) - \cos (A + B) \right]\]
\[ = \cos \frac{\pi}{3} - \cos \frac{\pi}{2}\]
\[ = \frac{1}{2} - 0\]
\[ = \frac{1}{2}\]
\[RHS = \frac{1}{2}\]
Hence, LHS = RHS

shaalaa.com
Transformation Formulae
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Transformation formulae - Exercise 8.1 [पृष्ठ ६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.1 | Q 2.1 | पृष्ठ ६

संबंधित प्रश्न

Show that :

\[\sin 25^\circ \cos 115^\circ = \frac{1}{2}\left( \sin 140^\circ - 1 \right)\]

Prove that:
 sin 20° sin 40° sin 80° = \[\frac{\sqrt{3}}{8}\]

 


Prove that tan 20° tan 30° tan 40° tan 80° = 1.


Prove that 
\[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right) = \tan 3x\]


Express each of the following as the product of sines and cosines:
sin 12x + sin 4x


Express each of the following as the product of sines and cosines:
sin 5x − sin x


Express each of the following as the product of sines and cosines:
 cos 12x - cos 4x


Express each of the following as the product of sines and cosines:
sin 2x + cos 4x


Prove that:
sin 105° + cos 105° = cos 45°


Prove that:
cos 20° + cos 100° + cos 140° = 0


Prove that:
 `sin A + sin 2A + sin 4A + sin 5A = 4 cos (A/2) cos((3A)/2)sin3A`


Prove that:
\[\sin\frac{x}{2}\sin\frac{7x}{2} + \sin\frac{3x}{2}\sin\frac{11x}{2} = \sin 2x \sin 5x .\]

 


Prove that:

\[\frac{\sin A - \sin B}{\cos A + \cos B} = \tan\frac{A - B}{2}\]

Prove that:

\[\frac{\sin A + \sin B}{\sin A - \sin B} = \tan \left( \frac{A + B}{2} \right) \cot \left( \frac{A - B}{2} \right)\]

Prove that:

\[\frac{\cos A + \cos B}{\cos B - \cos A} = \cot \left( \frac{A + B}{2} \right) \cot \left( \frac{A - B}{2} \right)\]

Prove that:

\[\frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A} = \cot 3A\]

 


Prove that:

\[\sin \alpha + \sin \beta + \sin \gamma - \sin (\alpha + \beta + \gamma) = 4 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\beta + \gamma}{2} \right) \sin \left( \frac{\gamma + \alpha}{2} \right)\]

 


Prove that:
 sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0


If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), prove that tan A tan B tan C + tan D = 0.

 

If \[m \sin\theta = n \sin\left( \theta + 2\alpha \right)\], prove that \[\tan\left( \theta + \alpha \right) \cot\alpha = \frac{m + n}{m - n}\]


If cos A = m cos B, then write the value of \[\cot\frac{A + B}{2} \cot\frac{A - B}{2}\].

 

Write the value of the expression \[\frac{1 - 4 \sin 10^\circ \sin 70^\circ}{2 \sin 10^\circ}\]


If A + B = \[\frac{\pi}{3}\] and cos A + cos B = 1, then find the value of cos \[\frac{A - B}{2}\].

 

 


If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), then write the value of tan A tan B tan C.


The value of sin 78° − sin 66° − sin 42° + sin 60° is ______.


cos 35° + cos 85° + cos 155° =


If cos A = m cos B, then \[\cot\frac{A + B}{2} \cot\frac{B - A}{2}\]=

 

Express the following as the sum or difference of sine or cosine:

cos 7θ sin 3θ


Express the following as the product of sine and cosine.

sin 6θ – sin 2θ


Express the following as the product of sine and cosine.

cos 2θ – cos θ


Prove that:

cos 20° cos 40° cos 80° = `1/8`


Prove that:

`(cos 2"A" - cos 3"A")/(sin "2A" + sin "3A") = tan  "A"/2`


Prove that:

`(cos 7"A" +cos 5"A")/(sin 7"A" −sin 5"A")` = cot A


Evaluate-

cos 20° + cos 100° + cos 140°


If cos A + cos B = `1/2` and sin A + sin B = `1/4`, prove that tan `(("A + B")/2) = 1/2`


If sin(y + z – x), sin(z + x – y), sin(x + y – z) are in A.P, then prove that tan x, tan y and tan z are in A.P.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×