\(sin(-\frac{11\pi}{3})=-sin\frac{11\pi}{3}=-sin\frac{11}{3}\times180\)
\(=-sin\frac{11}{3}\times\frac{360}{2}=-sin360(\frac{11}{6})\)
\(=-sin360(2-\frac{1}{6})\)
\(=-[sin(360\times2)-\frac{360}{2}]\)
\(=-(-sin60)=\frac{\sqrt 3}{2}\)

0

sin 150° = sin (90° + 60°)
= cos(60°) = \(\frac{1}{2}\)
(or) sin 150° = sin (180° - 30°)
= sin(30⁰) = \(\frac{1}{2}\)

cos 65^o + cos 15^o = \(2\cos { \left( \frac { 65+15 }{ 2 } \right) } .\cos { \left( \frac { 65-15 }{ 2 } \right) } \)
= 2 cos 40^o cos 25^o

sin 4x cos 2x = \(\frac{1}{2}\) [sin (4x + 2x) + sin (4x - 2x)]
= \(\frac{1}{2}\) [sin 6x + sin 2x]

LHS = \(\sin { 4\alpha } =sin2(2\alpha)\)
= \(2\left( \frac { 2\tan { \alpha } }{ 1+\tan ^{ 2 }{ \alpha } } \right) \left( \frac { 1-\tan ^{ 2 }{ \alpha } }{ 1+\tan ^{ 2 }{ \alpha } } \right) \)
= \(4\tan { \alpha } .\frac { 1-\tan ^{ 2 }{ \alpha } }{ { \left( 1+\tan ^{ 2 }{ \alpha } \right) }^{ 2 } } \) = RHS
Hence proved.

AC=\(\sqrt{2^2+1^2}=\sqrt{5}\)
Since θ lies in the II quadrant, only sinθ and cosec θ are positive.

sin θ = \(\frac{2}{\sqrt{5}}\) 
cos θ = \(\frac{-1}{\sqrt{5}}\)
cosec θ = \(\frac{\sqrt{5}}{2}\) 
sec θ = -√5 and cot θ = \(\frac{-1}{2}\)

Since \(\angle A=\frac { \pi }{ 2 } \) , then B  +C = \(\frac { \pi }{ 2 } \)
LHS = cos²B + cos²C

= \({ cos }^{ 2 }B+{ \left[ cos\left( \frac { \pi }{ 2 } -B \right) \right] }^{ 2 }\)
= cos²B + sin²B = 1 = RHS
Hence Proved.

BC = \(\sqrt{3^2-2^2}=\sqrt{9-4}=\sqrt{5}\)

Since θ lies in the IV quadrant, only cos θ and sec θ are positive.
ஃ cos θ = \(\frac{\sqrt{5}}{3}\) ,
tan θ = \(\frac{-2}{\sqrt{5}}\)  ;
sec θ = \(\frac{3}{\sqrt{5}}\) , 
cosec θ = \(\frac{-3}{2}\) and
cot θ = \(\frac{-\sqrt{5}}{2}\)

Now, cos 15° = cos (45° - 30°)
= cos 45° cos 30° + sin 45° sin 30°
\(=\frac{1}{\sqrt 2}\frac{\sqrt 3}{2}+\frac{1}{\sqrt 2}\frac{1}{2}=\frac{\sqrt 3+1}{2\sqrt 2}\)
Also, note that  \(sin 75^0=\frac{\sqrt 3+1}{2\sqrt 2}\) [try yourself]