$\frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} = 4\cos2A \\ \\ LHS = \frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} \\ \\ = \frac{ \sin5A \:\cos A - \cos5A \: \sin A}{\sin A \:\cos A } \\ \\ = \frac{ \sin(5A -A )}{\sin A \:\cos A} \\ \\ = \frac{ \sin 4A}{\sin A \:\cos A} \\ \\ = \frac{ 2\sin 2A \: \cos 2A}{\sin A \:\cos A} \\ \\ = \frac{ 2 \times 2\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A} \\ \\ = \frac{ 4\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A} \\ \\ =4\cos 2A \\ \\ = RHS \\ \\ thus \\ \\ \frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} = 4\cos2A \\ \\ hence \: proved$

5 0

3 years ago

Solve the following linear system by graphing <br> 4x+5y=10<br> 3x-3y=21

Inga [223]

Answer:

y = 2

x = 5

Step-by-step explanation:

4x + 5y = 10 | ×3 |

3x - 3y = 21 | ×4 |

12x + 15y = 30

12x - 12y = 84

____________--

27y = -54

y = -54/27

y = 2

4x + 5y = 10 | ×3 |

3x - 3y = 21 | ×5 |

12x + 15y = 30

15x - 15y = 105

____________+

27x = 135

x = 135/27

x = 5

6 0

3 years ago