Question

if sin(a+b)=sin A cos B +cos A sin B,find the value of sin(60+45)​

Answer 1

Given :

sin(A+B)=sin A cos B +cos A sin B

To find :

sin(60°+45°)

Solution :

⟹ sin(A+B)=sin A cos B +cos A sin B

Now put :-

• A = 60°
• B = 45°

⟹ sin(60°+45°)=sin 60° cos45° +cos60° sin 45°

We know that :-

${\bf{\sin(60 \degree) = \dfrac{ \sqrt{3} }{2} }} \\ \\ {\bf{\sin(45 \degree) = \dfrac{ 1 }{ \sqrt{2} } }} \\ \\ {\bf{\cos(45 \degree) = \dfrac{ 1 }{ \sqrt{2} } }} \\ \\ {\bf{\cos(60 \degree) = \dfrac{ 1 }{ {2} } }}$

⟹ sin(60°+45°)=(√3/2) (1/√2)+(1/2) (1/√2)

⟹ sin(60°+45°)=(√3/2√2) + (1/2√2)

⟹ sin(60°+45°)=(√3+1)/ (2√2)

Answer :

$\dfrac{ \sqrt{3} + 1}{2 \sqrt{2} }$

____________________

Learn more :-

1. Cosθ = base / hypotenuse

2. cossecθ = 1/ sinθ

3. sec θ = 1/cosθ

4. Cotθ = 1/ tanθ

5. Sin²θ+ Cos²θ= 1

6. Sec²θ - tan²θ = 1

7. cosec ²θ - cot²θ = 1

8. sin(90°−θ) = cos θ

9. cos(90°−θ) = sin θ

10. tan(90°−θ) = cot θ

11. cot(90°−θ) = tan θ

12. sec(90°−θ) = cosec θ

13. cosec(90°−θ) = sec θ

14. Sin2θ = 2 sinθ cosθ

15. cos2θ = Cos²θ- Sin²θ

$\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}$

Answer 2

Answer:

(√3+1)/2√2

Step-by-step explanation:

sin(A+B)=sinA.cosB+cosA.sinB

sin(60+45)=sin60.cos45+cos60.sin45

(√3/2).(1/√2)+(1/2).(1/√2)

=(√3/2√2)+(1/2√2)

=(√3+1)/2√2