Question

2cos78°.sin42 is equal to plz answer my question it's important guys​

Answer 1

Answer:

We know that,

\sf{2\ \cos x\ \sin y=\sin (x+y)-\sin (x-y)}2 cosx siny=sin(x+y)−sin(x−y)

Then,

\begin{gathered}\sf{2\cos 78^{\circ}\sin 42^{\circ}=\sin (78^{\circ}+42^{\circ})-\sin (78^{\circ}-42^{\circ})}\\\\\\\underline {\underline {\sf{2\cos 78^{\circ}\sin 42^{\circ}=\sin 120^{\circ}-\sin 36^{\circ}}}}\end{gathered}

2cos78

∘

sin42

∘

=sin(78

∘

+42

∘

)−sin(78

∘

−42

∘

)

2cos78

∘

sin42

∘

=sin120

∘

−sin36

∘

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

2 cos 78°• sin 42° is in the form of 2 cosA • sin B

we know that 2 cos A • sin B =sin(A+B)-sin(A-B)

So, 2cos78°•2sin42°=sin(78°+42°)-sin(78°-42°)

Here, A=78° and B=42

so, we will get=sin(120°)-sin(36°)

I think this may help you