Question

(i) sin 80° - sin 20° = cos 110° + cos 10° prove that ​

Answer 1

Answer:

in80∘−sin20∘=3sin50∘

Hence proved

Step-by-step explanation:

To prove:  \sin80^\circ-\cos110^\circ=\sqrt{3}\sin50^\circsin80∘−cos110∘=3sin50∘

Proof:

Taking left side

\Rightarrow \sin80^\circ-\cos110^\circ⇒sin80∘−cos110∘

\Rightarrow \sin80^\circ-\cos(90^\circ+20^\circ)⇒sin80∘−cos(90∘+20∘) \cos(90+\theta)=-\sin\thetacos(90+θ)=−sinθ

\Rightarrow \sin80^\circ+\sin20^\circ⇒sin80∘+sin20∘ \sin A+\sin B=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})sinA+sinB=2sin(2A+B)cos(2A−B)

\Rightarrow 2\sin(\frac{80^\circ+20^\circ}{2})\cos(\frac{80^\circ-20}{2})⇒2sin(280∘+20∘)cos(280∘−20)

\Rightarrow 2\sin50^\circ\cos30^\circ⇒2sin50∘cos30∘

\Rightarrow 2\sin50^\circ\cdot \dfrac{\sqrt{3}}{2}⇒2sin50∘⋅23

\Rightarrow \sqrt{3}\sin50^\circ⇒3sin50∘

LHS = RHS

Hence proved