Question

sin^2 12+ sin^2 21 + sin ^2 39 + sin ^2 48 - sin ^2 9 - sin ^2 18 equals to ?​

Answer 1

Answer:

Sin^2 12+ sin^2 21 + sin ^2 39 + sin ^2 48 - sin ^2 9 - sin ^2 18 equals to ?​

Sin²12°+sin²21°+sin²39°+sin²48°-sin²9°-sin²18°

sin²12°+sin²21°+(sin²38°-sin²9°)+(sin²48°-sin²18°)

=>sin²12°+sin²21°+sin(48°)sin(30°)+sin(66°)sin(30°)

=>sin²12°+sin²21°+½sin48°+½sin66°

=>½(1-cos24°)+½(1-cos42°)+½cos42°+½cos24°

=>1

Answer 2

Answer:

....

You need to remember and use the half angle formula, sin^2 alpha = (1 - cos 2 alpha)/2 ,such that:

sin^212^o = (1 - cos 24^o)/2

sin^2 21^o = (1 - cos 42^o)/2

sin^2 39^o = (1 - cos 78^o)/2

sin^2 48^o = (1 - cos 96^o)/2

sin^2 9^o = (1 - cos 18^o)/2

sin^2 18^o = (1 - cos 36^o)/2

1 - cos 24^o + 1 - cos 42^o +1 - cos 78^o + 1 - cos 96^o = 2 + 1 - cos 18^o + 1 - cos 36^o

Reducing like terms yields:

cos 24^o + cos 42^o + cos 78^o + cos 96^o = cos 18^o + cos 36^o

You need to combine the terms such that:

(cos 24^o + cos 96^o) + (cos 42^o + cos 78^o) = (cos 18^o + cos 36^o)

You need to convert the sum into product such that:

cos 24^o + cos 96^o = 2cos((24^o + 96^o)/2)*cos((24^o- 96^o)/2)

cos 24^o + cos 96^o = 2cos 60^o*cos(-36^o)

You need to remember that the cosine function is even, hence cos(-36^o)= cos(36^o)

cos 24^o + cos 96^o = 2cos 60^o*cos(36^o)

You need to convert the sum into product such that:

cos 42^o + cos 78^o = 2cos((42^o + 78^o)/2)*cos((42^o - 78^o)/2)

cos 42^o + cos 78^o = 2cos 60^o*cos(-18^o) cos 42^o + cos 78^o = 2cos 60^o*cos18^o

cos 18^o + cos 36^o = 2cos((18^o + 36^o)/2)*cos((18^o - 36^o)/2)

Substituting the products into the sum above yields:

2cos 60^o*cos(36^o) + 2cos 60^o*cos18^o = 2cos 27^o*cos 9^o

2cos 60^o*(cos 18^o + cos 36^o) = cos 18^o + cos 36^o

Reducing cos 18^o + cos 36^o both sides yields:

2cos 60^o = 1 => cos 60^o = 1/2 => 1/2 = 1/2

Hence, the last line proves that the given expression is an identity.