Maths-
General
Easy
Question
The value of
- 4
- 2
- -2
- -4
Hint:
To find the value of the given function we will first simplify the function using trigonometric identities then at last we will put the value of the required angle to get the answer.
The correct answer is: -4
Related Questions to study
Maths-
For any real , the maximum value of
is
As cos is a decreasing function and sin is an increasing function so the function
is maximum when
=
.
So, the maximum value of
= 
So, the maximum value of
For any real , the maximum value of
is
Maths-General
As cos is a decreasing function and sin is an increasing function so the function
is maximum when
=
.
So, the maximum value of
= 
So, the maximum value of
chemistry-
Identify A and B
With H2 / pd / CaCO3 - is addition of ‘H’ know is oxidizing agent
Identify A and B
chemistry-General
With H2 / pd / CaCO3 - is addition of ‘H’ know is oxidizing agent
Maths-
If
then sin (A-B) =
GIVEN-
√2 cos A = cos B + cos3 B..... equation 1
∴ cos A = 1/√2 (cos B + cos3 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 - sin3 B) * cos B] - [1/√2 (cos B + cos3 B) * sin B] ...... (From Equations i & ii)
∴ Sin (A - B) = [1/√2 * (sin B Cos B - sin3 B Cos B)] - [1/√2 * (sin B cos B + cos3 B * sin B)]
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin3 B Cos B - (1/√2 sin B cos B + 1/√2 cos3 B * sin B)
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin3 B Cos B - 1/√2 sin B cos B - 1/√2 cos3 B * sin B
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin3 B Cos B - 1/√2 sin B cos B - 1/√2 cos3 B * sin B
∴ Sin (A - B) = -1/√2 sin3 B Cos B - 1/√2 cos3 B * sin B
∴ Sin (A - B) = -1/√2 sin B Cos B (sin2 B +cos2 B)
= -1/2√2 sin2B
Now squaring and adding equation 1 and 2 we get

√2 cos A = cos B + cos3 B..... equation 1
∴ cos A = 1/√2 (cos B + cos3 B) ........ (Dividing both sides by √2)
√2 sin A = sin B - sin3 B..... equation 2
∴ sin A = 1/√2 (sin B - sin3 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 - sin3 B) * cos B] - [1/√2 (cos B + cos3 B) * sin B] ...... (From Equations i & ii)
∴ Sin (A - B) = [1/√2 * (sin B Cos B - sin3 B Cos B)] - [1/√2 * (sin B cos B + cos3 B * sin B)]
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin3 B Cos B - (1/√2 sin B cos B + 1/√2 cos3 B * sin B)
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin3 B Cos B - 1/√2 sin B cos B - 1/√2 cos3 B * sin B
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin3 B Cos B - 1/√2 sin B cos B - 1/√2 cos3 B * sin B
∴ Sin (A - B) = -1/√2 sin3 B Cos B - 1/√2 cos3 B * sin B
∴ Sin (A - B) = -1/√2 sin B Cos B (sin2 B +cos2 B)
= -1/2√2 sin2B
Now squaring and adding equation 1 and 2 we get
If
then sin (A-B) =
Maths-General
GIVEN-
√2 cos A = cos B + cos3 B..... equation 1
∴ cos A = 1/√2 (cos B + cos3 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 - sin3 B) * cos B] - [1/√2 (cos B + cos3 B) * sin B] ...... (From Equations i & ii)
∴ Sin (A - B) = [1/√2 * (sin B Cos B - sin3 B Cos B)] - [1/√2 * (sin B cos B + cos3 B * sin B)]
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin3 B Cos B - (1/√2 sin B cos B + 1/√2 cos3 B * sin B)
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin3 B Cos B - 1/√2 sin B cos B - 1/√2 cos3 B * sin B
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin3 B Cos B - 1/√2 sin B cos B - 1/√2 cos3 B * sin B
∴ Sin (A - B) = -1/√2 sin3 B Cos B - 1/√2 cos3 B * sin B
∴ Sin (A - B) = -1/√2 sin B Cos B (sin2 B +cos2 B)
= -1/2√2 sin2B
Now squaring and adding equation 1 and 2 we get

√2 cos A = cos B + cos3 B..... equation 1
∴ cos A = 1/√2 (cos B + cos3 B) ........ (Dividing both sides by √2)
√2 sin A = sin B - sin3 B..... equation 2
∴ sin A = 1/√2 (sin B - sin3 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 - sin3 B) * cos B] - [1/√2 (cos B + cos3 B) * sin B] ...... (From Equations i & ii)
∴ Sin (A - B) = [1/√2 * (sin B Cos B - sin3 B Cos B)] - [1/√2 * (sin B cos B + cos3 B * sin B)]
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin3 B Cos B - (1/√2 sin B cos B + 1/√2 cos3 B * sin B)
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin3 B Cos B - 1/√2 sin B cos B - 1/√2 cos3 B * sin B
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin3 B Cos B - 1/√2 sin B cos B - 1/√2 cos3 B * sin B
∴ Sin (A - B) = -1/√2 sin3 B Cos B - 1/√2 cos3 B * sin B
∴ Sin (A - B) = -1/√2 sin B Cos B (sin2 B +cos2 B)
= -1/2√2 sin2B
Now squaring and adding equation 1 and 2 we get
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