Question

What is value of sin15

Answer 1

The value of sin(15°)=3√−122√sin(15°)=3−122. We may get this value by any of the following methods:

Using the value of sin 30°.

For all values of the angle A we know that, 
(sin(A/2)+cos(A/2))2=sin2A/2+cos2A/2+2sin(A/2)cos(A/2)=1+sinA(sin(A/2)+cos(A/2))2=sin2A/2+cos2A/2+2sin(A/2)cos(A/2)=1+sinA

Therefore, sin(A/2)+cos(A/2)=±√(1+sinA)sin(A/2)+cos(A/2)=±√(1+sinA), [taking square root on both the sides]

Now, let A = 30° then, A/2=30°/2=15°A/2=30°/2=15° and from the above equation we get,

sin15°+cos15°=±√(1+sin30°)sin15°+cos15°=±√(1+sin30°) ….. (i)

Similarly, for all values of the angle A we know that,

(sin(A/2)−cos(A/2))2=sin(A/2)2+cos(A/2)2−2sin(A/2)cos(A/2)=1−sinA(sin(A/2)−cos(A/2))2=sin(A/2)2+cos(A/2)2−2sin(A/2)cos(A/2)=1−sinA

Therefore, sin(A/2)+cos(A/2)=±√(1−sinA)sin(A/2)+cos(A/2)=±√(1−sinA) , [taking square root on both the sides]

Now, let A = 30° then, A/2=30°/2=15°A/2=30°/2=15° and from the above equation we get,

sin15°−cos15°=±√(1−sin30°)sin15°−cos15°=±√(1−sin30°) …… (ii)

Clearly, sin15°>0sin15°>0 and cos15˚>0cos15˚>0

Therefore, sin15°+cos15°>0sin15°+cos15°>0

Therefore, from (i) we get,

sin15°+cos15°=√(1+sin30°)sin15°+cos15°=√(1+sin30°) ..... (iii)

Again, sin15°−cos15°=√2((1√2)sin15˚−(1√2)cos15˚)sin15°−cos15°=√2((1√2)sin15˚−(1√2)cos15˚)

or, sin15°−cos15°=√2(cos45°sin15˚−sin45°cos15°)sin15°−cos15°=√2(cos45°sin15˚−sin45°cos15°)

or, sin15°−cos15°=√2sin(15˚−45˚)sin15°−cos15°=√2sin(15˚−45˚)

or, sin15°−cos15°=√2sin(−30˚)sin15°−cos15°=√2sin(−30˚)

or, sin15°−cos15°=−√2sin30°sin15°−cos15°=−√2sin30°

or, sin15°−cos15°=−√2/2sin15°−cos15°=−√2/2

or, sin15°−cos15°=−1/√2sin15°−cos15°=−1/√2

Thus,sin15°−cos15°<0sin15°−cos15°<0

Therefore, from (ii) we get, sin15°−cos15°=−√(1−sin30°)sin15°−cos15°=−√(1−sin30°) ..... (iv)

Now, adding (iii) and (iv) we get,

2sin15°=√(1+1/2)−√(1−1/2)2sin15°=√(1+1/2)−√(1−1/2)

2sin15°=((√3)−1)/√22sin15°=((√3)−1)/√2

sin(15°)=3√−122√sin(15°)=3−122

Therefore, sin(15°)=3√−122√.

Answer 2

Answer:

Step-by-step explanation:

sin45--sin 30

1/√2--1/2

=2-√2/2√2