The value of sin-2-sin7-2-sin3-2-sin11-2 -sin 2-sin5- is equal to

The correct option is A 0 We have, #160; sinAsinB = cos(A B) cos(A+B)2 Using this, we can expand the given expression as #160; (cos3θ cos4θ ...

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Standard XII

Mathematics

Cos(A+B)Cos(A-B)

The value of ...

Question

The value of $sin \frac{θ}{2} . sin \frac{7 θ}{2} + sin \frac{3 θ}{2} . sin \frac{11 θ}{2} - sin 2 θ . sin 5 θ$ is equal to

0

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sin 6 θ - cos 5 θ

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sin 6 θ + sin 7 θ

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Solution

The correct option is A $0$
We have,
$sin A sin B = \frac{cos (A - B) - cos (A + B)}{2}$
Using this, we can expand the given expression as
$\frac{(cos 3 θ - cos 4 θ) + (cos 4 θ - cos 7 θ) - (cos 3 θ - cos 7 θ)}{2} = 0$

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\frac{sin 3 θ + sin 5 θ + sin 7 θ + sin 9 θ}{cos 3 θ + cos 5 θ + cos 7 θ + cos 9 θ}

is equal to

Q. If

sin θ + cosec θ = 2

, then the value of

{sin}^{6} θ + {cosec}^{6} θ

is equal to

Find the general solution of the equation $s i n 7 θ = s i n 3 θ + s i n θ$

sin 7 θ + sin 4 θ + sin θ = 0, 0 < θ < \frac{π}{2}

$Solve the following equations : (i) c o s θ + c p s 2 θ + c o s 3 θ = 0 (i i) c o s θ + c o s 3 θ - c o s 2 θ = 0 (i i i) s i n θ + s i n 5 θ = s i n 3 θ (i v) c o s θ c o s 2 θ c o s 3 θ = \frac{1}{4} (v) c o s θ + s i n θ = c o s 2 θ + s i n 2 θ (v i) s i n θ + s i n 2 θ + s i n 3 θ = 0 (v i i) s i n θ + s i n 2 θ + s i n 3 θ + s i n 4 θ = 0 (v i i i) s i n 3 θ - s i n θ = 4 c o s^{2} θ - 2 (i x) s i n 2 θ - s i n 4 θ + s i n 6 θ = 0$

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The value of sinθ2.sin7θ2+sin3θ2.sin11θ2−sin2θ.sin5θ is equal to

The correct option is A 0We have, sinAsinB=cos(A−B)−cos(A+B)2Using this, we can expand the given expression as (cos3θ−cos4θ)+(cos4θ−cos7θ)−(cos3θ−cos7θ)2=0

The value of $sin \frac{θ}{2} . sin \frac{7 θ}{2} + sin \frac{3 θ}{2} . sin \frac{11 θ}{2} - sin 2 θ . sin 5 θ$ is equal to

The correct option is A $0$
We have,
$sin A sin B = \frac{cos (A - B) - cos (A + B)}{2}$
Using this, we can expand the given expression as
$\frac{(cos 3 θ - cos 4 θ) + (cos 4 θ - cos 7 θ) - (cos 3 θ - cos 7 θ)}{2} = 0$