Prove that:
4(cos310o + sin320o) = 3 (cos10o + sin20o)
We know that
⇒ sin (3× 20°)=cos (3× 10°)
⇒ 3sin 20°–4sin320°=4cos310°–3cos 10°
(as sin 3θ=3sin θ–4sin3 θ and cos 3θ =4cos3θ–3cosθ)
⇒ 4(cos310°+sin320°)=3(sin 20°+cos 10°)
LHS=RHS
Hence proved
2
Prove that:
4(cos310o + sin320o) = 3 (cos10o + sin20o)
We know that
⇒ sin (3× 20°)=cos (3× 10°)
⇒ 3sin 20°–4sin320°=4cos310°–3cos 10°
(as sin 3θ=3sin θ–4sin3 θ and cos 3θ =4cos3θ–3cosθ)
⇒ 4(cos310°+sin320°)=3(sin 20°+cos 10°)
LHS=RHS
Hence proved