Question

sin 40+sin 20-cos 10 =0

Answer 1

To prove: $\mathsf{sin40^{\circ}+sin20^{\circ}-cos10^{\circ}=0}$

Rules to know:

• $\mathsf{sinC+sinD=2\:sin\dfrac{C+D}{2}\:cos\dfrac{C-D}{2}}$

• $\mathsf{sinC-sinD=2\:cos\dfrac{C+D}{2}\:sin\dfrac{C-D}{2}}$

• $\mathsf{cosC+cosD=2\:cos\dfrac{C+D}{2}\:cos\dfrac{C-D}{2}}$

• $\mathsf{cosC-cosD=2\:sin\dfrac{C+D}{2}\:sin\dfrac{D-C}{2}}$

• $\mathsf{sin30^{\circ}=\dfrac{1}{2}}$

Proof:

Left Hand Side = $\bold{sin40^{\circ}+sin20^{\circ}-cos10^{\circ}}$

= $\mathsf{(sin40^{\circ}+sin20^{\circ})-cos10^{\circ}}$

= $\mathsf{2\:\sin\dfrac{40^{\circ}+20^{\circ}}{2}\:cos\dfrac{40^{\circ}-20^{\circ}}{2}-cos10^{\circ}}$

= $\mathsf{2\:sin\dfrac{60^{\circ}}{2}\:cos\dfrac{20^{\circ}}{2}-cos10^{\circ}}$

= $\mathsf{2\:sin30^{\circ}\:cos10^{\circ}-cos10^{\circ}}$

= $\mathsf{2\times\dfrac{1}{2}\:cos10^{\circ}-cos10^{\circ}}$

= $\mathsf{cos10^{\circ}-cos10^{\circ}}$

= $\bold{0}$ = Right Hand Side

∴ $\boxed{\mathsf{sin40^{\circ}+sin20^{\circ}-cos10^{\circ}=0}}$

Thus proved.