Question

7. sin (x + y) = sin x cos y + cos x sin y​

Answer 1

The figure shows a rectangle with BG = 1 unit, so that in ∆BFG,

BF = CE = BG sin (x + y)

BF = CE = sin (x + y)

In ∆BDG,

BD = BG sin y

BD = sin y

And,

GD = BG cos y

GD = cos y

In ∆GDE,

DE = GD sin x

DE = sin x cos y

And,

⟨GDE = 180° - (90° + x)

⟨GDE = 90° - x

So, since ⟨BDG = 90°,

⟨BDC = 180° - ( ⟨GDE + ⟨BDG )

⟨BDC = x

Then, in ∆BCD,

CD = BD cos x

CD = cos x sin y

Now,

CE = CD + DE

sin (x + y) = sin x cos y + cos x sin y

Hence Proved!

From this figure it is also possible to prove the identity:

cos (x + y) = cos x cos y - sin x sin y