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5. (a) Prove that \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \) - Leaving Cert Mathematics - Question 5 - 2015
Question 5
5. (a) Prove that \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \). (b) Find all the values of \( x \) for which \( \sin(3x) = \frac{\sqrt{3}}{2} \), \... show full transcript
Worked Solution & Example Answer:5. (a) Prove that \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \) - Leaving Cert Mathematics - Question 5 - 2015
Step 1
Prove that \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
Answer
To prove the identity, we start by using the definition of tangent in terms of sine and cosine:
[ \tan(A + B) = \frac{\sin(A + B)}{\cos(A + B)} ]
Using the sine and cosine addition formulas:
[ \sin(A + B) = \sin A \cos B + \cos A \sin B ] [ \cos(A + B) = \cos A \cos B - \sin A \sin B ]
We can rewrite ( \tan(A + B) ) as:
[ \tan(A + B) = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B} ]
Now we will transform the right side of the original equation:
[
\frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{1 - \frac{\sin A}{\cos A} \cdot \frac{\sin B}{\cos B}}
]
Combining into a single fraction:
[ = \frac{\frac{\sin A \cos B + \sin B \cos A}{\cos A \cos B}}{\frac{\cos A \cos B - \sin A \sin B}{\cos A \cos B}} = \frac{\sin A \cos B + \sin B \cos A}{\cos A \cos B - \sin A \sin B} ]
This aligns with the left side as shown:
[ \tan(A + B) = \frac{\sin(A + B)}{\cos(A + B)} ]
Thus, we have proved the identity successfully.
Step 2
Find all the values of \( x \) for which \( \sin(3x) = \frac{\sqrt{3}}{2} \)
Answer
To solve the equation ( \sin(3x) = \frac{\sqrt{3}}{2} ), we recognize the angles where sine is ( \frac{\sqrt{3}}{2} ):
- ( 60^\circ ) or ( 120^\circ ).
So, we can write:
[
3x = 60^\circ + n \cdot 360^\circ , (n \in \mathbb{Z})
]
[
3x = 120^\circ + n \cdot 360^\circ , (n \in \mathbb{Z})
]
Next, solving for ( x ):
From ( 3x = 60^\circ + n \cdot 360^\circ ):
[
x = 20^\circ + n \cdot 120^\circ, n \in \mathbb{Z}
]
From ( 3x = 120^\circ + n \cdot 360^\circ ):
[
x = 40^\circ + n \cdot 120^\circ, n \in \mathbb{Z}
]
Now, we can find values of ( x ) within ( 0 \leq x \leq 360 ):
- For ( x = 20^\circ + n \cdot 120^\circ ):
- ( n = 0 ): ( x = 20^\circ )
- ( n = 1 ): ( x = 140^\circ )
- ( n = 2 ): ( x = 260^\circ )
- For ( x = 40^\circ + n \cdot 120^\circ ):
- ( n = 0 ): ( x = 40^\circ )
- ( n = 1 ): ( x = 160^\circ )
- ( n = 2 ): ( x = 280^\circ )
Thus, the solutions for ( x ) in the specified range are:
( 20^\circ, 40^\circ, 140^\circ, 160^\circ, 260^\circ, 280^\circ ).