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Grade: 11
        
sinA+sin3A+SIN5A+sin7A
 
4 years ago

Answers : (7)

Latika Leekha
askIITians Faculty
165 Points
							Hello student,
Your question seems to be incomplete as it does not state what exactly are we required to prove. Please check and post the complete question again so that we can provide you a meaningful answer.
4 years ago
VAISHNAVI BATHINA
130 Points
							
what is the value of angle A
4 years ago
grenade
2061 Points
							
MANIDEEP
you question literates that this is incomplete u look out the refrence 
else u will have club the first sinA and last sin 7A terms     
APPROVE IF USEFUL
4 years ago
rahul
211 Points
							
take fthe first and last term and the middle first and middle last term to get the answer
4 years ago
Lab Bhattacharjee
121 Points
							
As the angles are in Arithmetic Series,  multiply the numerator & the denominator by 2sin(3A-A)/2
 
and use Werner's formula: 2sinA sinB=cos(A-B)-cos(A+B)
4 years ago
Athiyaman
57 Points
							
 
Hey Manideep,
the answer should be 16 sinA.cos22A.cos2A
you will have to use these 3 formulas in order to get the answer:
1)sinA+sinB=2sin(A+B)/2.cos(A-B)/2
2)cosA+cosB=2cos(A+B)/2.cos(A-B)/2 and lastly,
3)sin2A=2sinA.cosA
4 years ago
Athiyaman
57 Points
							
 
And if you want the answer here it is:
 sinA + sin5A + sin3A + sin7A
= (sinA + sin7A) + (sin5A + sin3A)
After rearranging the terms use identity 1 from the above post,
= 2 sin 4A cos 3A + 2 sin 4A cos A
Here 2sin4A is common
=2 sin 4A [ cos 3A + cos A ]
Use Identity 2 from the above post,
=2 sin 4A * 2 cos 2A cos A
sin4A can be split into 2sin2A.cos2A from identity 3,
= 4 sin 2A cos 2A * 2 cos 2A cos A
Again use sin2A formula
= 8 sin A * cos A *cos 2A * 2 cos 2A cos A
and then just arrange the terms
= 16 sin A cos² A * cos² (2A)
Please approve the answer if you found it helpful
 
4 years ago
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