New General Mathematics for SSS 2 – Exercise 5a Solutions (Number 1)

118⁰ lies on the 2nd quadrant

On the 2nd quadrant;

sin θ = sin (180⁰ − θ)

∴ sin 118⁰ = sin (180⁰ − 118⁰)

= sin 62⁰

= 0.8829

118⁰ lies on the 2nd quadrant

On the 2nd quadrant;

tan θ = − tan (180⁰ − θ)

∴ tan 118⁰ = − tan (180⁰ − 118⁰)= − tan 62⁰

= − 1.881

305⁰ lies on the 4th quadrant

In the 4th quadrant;

tan θ = − tan (360⁰ − θ)

∴ tan 360⁰ = − tan (360⁰ − 305⁰)

= − tan 55⁰

= − 1.428

305⁰ lies on the 4th quadrant

In the 4th quadrant;

cos θ = cos (360⁰ − θ)

∴ cos 360⁰ = − cos (360⁰ − 305⁰)

= cos 55⁰

= 0.5736

214.8⁰ lies in the 3rd quadrant

In the 3rd quadrant;

cos θ = cos (θ − 180⁰)

∴ cos 214.8⁰ = − cos (214.8⁰ − 180⁰)

=  − cos 34.8⁰

= − 0.8211 

315.8⁰ lie in the 4th quadrant.

In the 4th quadrant;

sin = − sin (360⁰ − θ)

sin 315.8⁰ = − sin (360⁰ − 315.8⁰)

= − sin 44.2⁰

= − 0.6972

172.5 lies in the 2nd quadrant

In the 2nd quadrant;

cos θ = cos (180 − θ)

cos 172.5⁰ = − cos (180⁰ − 172.5⁰)

= − cos 7.5⁰

= − 0.9914

241.6⁰ lies on the 3rd quadrant

In the 3rd quadrant;

sin θ = − sin (θ − 180⁰)

sin 241.6⁰ = − sin (241.6⁰ − 180⁰)

= − sin 61.6⁰

= − 0.8796

The quadrant rule can only be applied for positive angle which is usually done in the anti-clockwise manner.

To obtain the positive angle of any give negative angle, simply subtract the given angel from 360⁰.

∴ sin ( − 42)⁰ = sin (360⁰ − 42)⁰

= sin 318⁰

318⁰ lies in the 4th quadrant.

In the 4th quadrant;

sin θ = − sin (360⁰ – θ)

sin 318⁰ = − sin (360⁰ – 318⁰)

= − sin 42⁰

= − 0.6991

Basically for any angle between 0⁰ and 90⁰, we can simply drop the negativity before the trigonometric function.

∴ sin ( − 42)⁰ = − sin 42⁰

= − 0.6691