Question

If A = B = 60°, verify that cos (A - B) = cosA cosB + sin A sinB​

Answer 1

Step-by-step explanation:

GIVEN = A,B = 60 DEGREE

TO PROVE = cos (A - B) = cosA cosB + sin A sinB

ANSWER - COS(60-60)

= COS 0 DEGREE = 1

1 = 1/2+1/2+1/√2+1√2

= 1 = 1

hence proved

Answer 2

Given :

• Value of A = 60°
• Value of B = 60°

To Verify :

• cos (A - B) = cosA cosB + sin A sinB

Remember :

• sin60° = √3/2
• sin30° = 1/2

Solution :

⠀

⟶⠀cos(A - B) = cosA cosB + sinA sinB

⟶⠀cos(60 - 60) = cosA cosB + sinA sinB

⟶⠀cos(0) = cos60 cos60 + sin60 sin60

⟶⠀cos(0) = cos²60 + sin²60

⟶⠀1 = (1/2)² + (√3/2)²

⟶⠀1 = (1/4) + (√9/4)

⟶⠀1 = (1/4) + (3/4)

⟶⠀1 = 4/4

⟶⠀1 = 1

⟶⠀LHS = RHS⠀⠀⠀⠀⠀⠀[Hence Verified]

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