Please tell me the answer of sin1×sin2×sin3. ×sin89?
Solution:
Using Trigonometric Identities:
We know that,
sin(A+B) = sinAcosB + cosAsinB
sin(A-B) = sinAcosB - cosAsinB
Using these identities, we can write the following:
sin1 × sin2 × sin3 × sin4 × ... × sin89
= sin1 × sin89 × sin2 × sin88 × sin3 × sin87 × ... × sin45 × sin45
= [sin1 × sin89] × [sin2 × sin88] × [sin3 × sin87] × ... × [sin45 × sin45]
= [sin(90-89) / 2] × [sin(90-88) / 2] × [sin(90-87) / 2] × ... × [sin(90-45) / 2] × sin45
= [(cos89 / 2) × sin1] × [(cos88 / 2) × sin2] × [(cos87 / 2) × sin3] × ... × [(cos45 / 2) × sin45]
= (cos89 / 2) × (cos88 / 2) × (cos87 / 2) × ... × (cos45 / 2) × sin1 × sin2 × sin3 × ... × sin45
Now, we know that:
cosx = sin(90-x)
Using this identity, we can write:
cos89 = sin(90-89) = sin1
cos88 = sin(90-88) = sin2
cos87 = sin(90-87) = sin3
...
cos45 = sin(90-45) = sin45
Substituting these values in the above equation, we get:
sin1 × sin2 × sin3 × ... × sin89 = (sin1 / 2) × (sin2 / 2) × (sin3 / 2) × ... × (sin44 / 2) × (sin45 / 2) × (cos45 / 2)
Now, we know that:
sinx = cos(90-x)
Using this identity, we can write:
sin45 = cos(90-45) = cos45
Substituting this value in the above equation, we get:
sin1 × sin2 × sin3 × ... × sin89 = (sin1 / 2) × (sin2 / 2) × (sin3 / 2) × ... × (sin44 / 2) × (cos45 / 2) × (cos45 / 2)
We know that cos²x + sin²x = 1
Squaring both sides, we get:
sin²x = 1 - cos²x
Using this identity, we can write:
cos²45 = 1 - sin²45 = 1 - 1/√2² = 1 - 1/2 = 1/2
Substituting this value in the above equation, we get:
sin1 × sin2 × sin3 × ... × sin89 = (sin1 / 2) × (sin2 / 2) × (sin3 / 2) × ... × (