Question

solve only B.
please .
Topper's can do it.​

Answer 1

Solution→

cos 165°= -(√3+1/2√2)

So cos 165° can also be written as -

cos(180°) = cos(180°-15°)

Now , we know that cos(π-15°) lies in II quadrant where cos functions negative .So,

cos(180°-15°) = - cos(15°)

Now -cos15° can be written as -

-cos 15° =- cos (45°-30°)

As cos (A-B) = cosAcosB +sinAsinB.

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

= -[(1/√2)(√3/2)+(1/√2)(1/2) ]

=-[( √3/2√2)+ (1/2√2 )]

$= \: - \frac{ \sqrt{3} + 1}{2 \sqrt{2} }$

Hence proved .

hope it helps ..

Answer 2

Answer:

Solution→

cos 165°= -(√3+1/2√2)

So cos 165° can also be written as -

cos(180°) = cos(180°-15°)

Now , we know that cos(π-15°) lies in II quadrant where cos functions negative .So,

cos(180°-15°) = - cos(15°)

Now -cos15° can be written as -

-cos 15° =- cos (45°-30°)

As cos (A-B) = cosAcosB +sinAsinB.

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

= -[(1/√2)(√3/2)+(1/√2)(1/2) ]

=-[( √3/2√2)+ (1/2√2 )]

= \: - \frac{ \sqrt{3} + 1}{2 \sqrt{2} }=−

2

2

3

+1

Hence proved .

hope it helps ..