Question

सूत्र cos (A + B) = cos A .cos B – sin A .sin B लेकर cos 75º का मान ज्ञात कीजिए।

Answer 1

Answer:

$\large \bold\red {\frac{ (\sqrt{3} - 1) }{2 \sqrt{2} } }$

Step-by-step explanation:

Given,

• cos (A + B) = cos A .cos B – sin A .sin B

Now,

To find the value of cos 75

Lets assume that,

• A = 45°
• B = 30°

Therefore,

We have,

$= > \cos(75) = \cos(45 + 30) \\ = > \cos(75) = \cos(45) \cos(30) - \sin(45) \sin(30)$

But,

We know that,

• $\cos(45) = \frac{1}{ \sqrt{2} }$
• $\sin(45) = \frac{1}{ \sqrt{2} }$
• $\cos(30) = \frac{ \sqrt{3} }{2}$
• $\sin(30) = \frac{1}{2}$

Therefore,

Putting the respective values,

We get,

$= > \cos(75) = ( \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2}) - ( \frac{1}{ \sqrt{2} } \times \frac{1}{2} ) \\ \\ = > \cos(75) = \frac{ \sqrt{3} }{2 \sqrt{2} } - \frac{1}{2 \sqrt{2} } \\ \\ = > \cos(75) = \large \boxed { \green{\frac{ (\sqrt{3} - 1) }{2 \sqrt{2} } }}$

Answer 2

Question :------ we have To find value of cos 75° .

using cos (A + B) = cos A .cos B – sin A .sin B ?

values to be used :-----

• cos30° = √3/2
• cos 45° = 1/√2
• sin30° = 1/2
• sin45° = 1/√2

Solution :------

Formula is already Given in Question , so we just have to Put value and solve it,

Let A = 45° , B = 30°

Cos(A+B) = Cos(75°)

cos(A+B) = cos(45+30)

Now,

Cos(45+30) = cos45°cos30° - sin45°sin30°

Further ,

$(\frac{1}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2}) - ( \frac{1}{ \sqrt{2} } \times \frac{1}{2} ) \\ \\ \frac{ \sqrt{3} }{2 \sqrt{2} } - \frac{1}{2 \sqrt{2} } \\ \\ \frac{( \sqrt{3} - 1) }{2 \sqrt{2} } \times \frac{2 \sqrt{2} }{2 \sqrt{2} } \\ \\ \frac{ \cancel2( \sqrt{6} - \sqrt{2} ) }{ \cancel8} \\ \\ \frac{ \sqrt{6} - \sqrt{2} }{4}$

( आशा है कि आपकी सहायता हुई ll )