11) Show that \( \sin (\theta+x)+\sin (\theta-x)=2 \sin \theta \cos x \) (2) Hence show that \( \sin 43^{\circ}+\sin 17^{\circ}=\cos 13^{\circ} \).

1) To prove the identity sin(θ + x) + sin(θ − x) = 2 sin θ cos x, we first expand each sine term using the sine addition and subtraction formulas:   sin(θ + x) = sin θ cos x + cos θ sin x   sin(θ − x) = sin θ cos x − cos θ sin x Now, adding these two expressions:   sin(θ + x) + sin(θ − x)     = (sin θ cos x + cos θ sin x) + (sin θ cos x − cos θ sin x)     = (sin θ cos x + sin θ cos x) + (cos θ sin x − cos θ sin x)     = 2 sin θ cos x Thus, we have proven the identity. 2) To show that sin 43° + sin 17° = cos 13°, notice that the identity from part (1) suggests writing the two sines in the form sin(θ + x) and sin(θ − x). We set up the following:   θ + x = 43°  and  θ − x = 17° Adding these equations gives:   (θ + x) + (θ − x) = 43° + 17°   2θ = 60°   θ = 30° Then, subtracting the equations gives:   (θ + x) - (θ − x) = 43° - 17°   2x = 26°   x = 13° Now, substituting θ = 30° and x = 13° into the identity, we have:   sin 43° + sin 17° = 2 sin 30° cos 13° Since sin 30° = 1/2, it follows that:   2 sin 30° cos 13° = 2 × (1/2) × cos 13° = cos 13° Thus, we have shown that sin 43° + sin 17° = cos 13°.