Maths-

General

Easy

Question

# If then sin (A-B) =

Hint:

### to find the value of the given equation we will use the formula Sin (A - B) = (Sin A * Cos B) - (Cos A * Sin B) and simplify the equation then using the given equations we will find the value of the required equations using the trigonometric identities.

## The correct answer is:

### GIVEN-

$√2cosA=cosB+cos_{3}B..... equation 1$

$cosAcosB+cos_{3}√2)$

# √2 sin A = sin B - sin^{3} B..... equation 2

∴ sin A = 1/√2 (sin B - sin^{3} B) ........ (Dividing both sides by √2)

TO FIND-

Sin (A - B)

SOLUTION-

We know that-

Sin (A - B) = (Sin A * Cos B) - (Cos A * Sin B)

∴ Sin (A - B) = [1/√2 (sin B - sin^{3} B) * cos B] - [1/√2 (cos B + cos^{3} B) * sin B] ...... (From Equations i & ii)

∴ Sin (A - B) = [1/√2 * (sin B Cos B - sin^{3} B Cos B)] - [1/√2 * (sin B cos B + cos^{3} B * sin B)]

∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin^{3} B Cos B - (1/√2 sin B cos B + 1/√2 cos^{3} B * sin B)

∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin^{3} B Cos B - 1/√2 sin B cos B - 1/√2 cos^{3} B * sin B

∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin^{3} B Cos B - 1/√2 sin B cos B - 1/√2 cos^{3} B * sin B

∴ Sin (A - B) = -1/√2 sin^{3} B Cos B - 1/√2 cos^{3} B * sin B

∴ Sin (A - B) = -1/√2 sin B Cos B (sin^{2} B +cos^{2} B)

= -1/2√2 sin2B

Now squaring and adding equation 1 and 2 we get

^{3}B..... equation 2

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