Question

find the value of the given que​

Answer 1

Answer :-

→ sin(45°+θ)*cos(15°+θ) - cos(45°+θ)*sin(15°+θ)

we know That ,

☛ sinA*cosB - cosA*sinB = sin(A - B)

comparing we get,

➼ A = (45°+θ)

➼ B = (15°+θ)

So,

☞ sin(45°+θ)*cos(15°+θ) - cos(45°+θ)*sin(15°+θ)

☞ sin[ (45°+θ) - (15°+θ) ]

☞ sin[ 45° - 15° + θ - θ ]

☞ sin30°

☞ (1/2) (Ans).

Answer 2

To Find:

Value of

$\sf{sin(45^{\circ}+\theta )\times cos(15^{\circ}+\theta )-cos(45^{\circ}+\theta )\times sin(15^{\circ}+\theta )}$

Solution:

Let 45°+θ be P and 15°+θ be Q

So,

⇒ P-Q= 45°+θ-(15°+θ)

⇒ P-Q= 45°+θ-15°-θ

⇒ P-Q= 30° -------------(1)

We know that,

• sin(A-B)= sinAcosB-cosAsinB
• $\bf{sin30^{\circ}=\dfrac{1}{2}}$

So, by applying above identity, we get

sinP×cosQ-cosP×sinQ =sin(P-Q)

From (1)

$\longrightarrow\sf{sinP\times cosQ-cosP\times sinQ =sin(30^{\circ})}$

$\longrightarrow\sf{sinP\times cosQ-cosP\times sinQ =\dfrac{1}{2}}$

So,

$\sf\pink{sin(45^{\circ}+\theta )\times cos(15^{\circ}+\theta )-cos(45^{\circ}+\theta )\times sin(15^{\circ}+\theta ) =\dfrac{1}{2}}$

Hence, the required value is  $\orange{\dfrac{1}{2}}$.