Question

if sinA+sinB=1/2 and cosA+cosB=5/4 so tanA+tanB=?
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Answer 1

Step-by-step explanation:

Do you know these identities:

cos(x + y) = cos(x).cos(y) - sin(x).sin(y) ← remarkable identity (1)

cos(x - y) = cos(x).cos(y) + sin(x).sin(y) ← remarkable identity (2)

sin(x + y) = sin(x).cos(y) + cos(x).sin(y) ← remarkable identity (3)

sin(x - y) = sin(x).cos(y) - cos(x).sin(y) ← remarkable identity (4)

Given:

(1) : cos(A) + cos(B) = 1/2

(1) : sin(A) + sin(B) = 1/4

Let: A = x + y

Let: B = x - y

Then:

(1') : cos(x + y) + cos(x - y) = 1/2

(2') : sin(x + y) + sin(x - y) = 1/4

Restart with (1') and transform it into (1'')

(1') : cos(x + y) + cos(x - y) = 1/2 → using identities (1) & (2)

(1'') : [cos(x).cos(y) - sin(x).sin(y)] + [cos(x).cos(y) + sin(x).sin(y)] = 1/2

(1'') : cos(x).cos(y) - sin(x).sin(y) + cos(x).cos(y) + sin(x).sin(y) = 1/2

(1'') : 2.cos(x).cos(y) = 1/2

(1'') : cos(x).cos(y) = 1/4

Restart with (2') and transform it into (2'')

(2') : sin(x + y) + sin(x - y) = 1/4 → using identities (3) & (4)

(2'') : [sin(x).cos(y) + cos(x).sin(y)] + [sin(x).cos(y) - cos(x).sin(y)] = 1/4

(2'') : sin(x).cos(y) + cos(x).sin(y) + sin(x).cos(y) - cos(x).sin(y) = 1/4

(2'') : 2.sin(x).cos(y) = 1/4

(2'') : sin(x).cos(y) = 1/8

You calculate: (2'') / (1'')

[sin(x).cos(y)] / [cos(x).cos(y)] = (1/8) / (1/4)

sin(x)/cos(x) = (1/8) * 4

tan(x) = 1/2 ← memorize this result

Recall:

Let: A = x + y

Let: B = x - y

A + B = (x + y) + (x - y)

A + B = x + y + x - y

A + B = 2x

x = (A/2) + (B/2)

Recall the memorized result:

tan(x) = 1/2 → we've just seen that: x = (A/2) + (B/2)

tan[(A/2) + (B/2)] = 1/2

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