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How to find the exact value of tan 15° using the value of sin 30°?
Solution:
For all values of the angle A we know that, (sin A2 + cos A2)2 = sin2 A2 + cos2 A2 + 2 sin A2 cos A2 = 1 + sin A
Therefore, sin A2 + cos A2 = ± √(1 + sin A), [taking square root on both the sides]
Now, let A = 30° then, A2 = 30°2 = 15° and from the above equation we get,
sin 15° + cos 15° = ± √(1 + sin 30°) ….. (i)
Similarly, for all values of the angle A we know that, (sin A2 - cos A2)2 = sin2 A2 + cos2 A2 - 2 sin A2 cos A2 = 1 - sin A
Therefore, sin A2 - cos A2 = ± √(1 - sin A), [taking square root on both the sides]
Now, let A
= 30° then, A2 = 30°2 = 15° and from the above
equation we get,
sin 15° - cos 15° = ± √(1 - sin 30°) …… (ii)
Clearly, sin 15° > 0 and cos 15˚ > 0
Therefore, sin 15° + cos
15° > 0
Therefore, from (i) we get,
sin 15° + cos 15° = √(1 + sin 30°) ..... (iii)
Again, sin 15° - cos 15° = √2
(1√2 sin 15˚ - 1√2 cos 15˚)
or, sin 15° - cos 15° = √2 (cos 45° sin 15˚
- sin 45° cos 15°)
or, sin 15° - cos 15° = √2 sin (15˚ - 45˚)
or, sin 15° - cos 15° = √2 sin (- 30˚)
or, sin 15° - cos 15° = -√2 sin 30°
or, sin 15° - cos 15° = -√2 ∙ 12
or, sin 15° - cos 15° = - √22
Thus, sin 15° - cos 15° < 0
Therefore, from (ii) we get, sin 15° - cos 15°= -√(1 - sin 30°) ..... (iv)
Now, adding (iii) and (iv) we
get,
2 sin 15° = √1+12−√1−12
2 sin 15° = √3−1√2
sin 15° = √3−12√2
Therefore, sin 15° = √3−12√2
Similarly, subtracting (iv) from (iii) we get,
2 cos 15° = √1+12+√1−12
2 cos 15° = √3+1√2
cos 15° = √3+12√2
Therefore, cos 15° = √3+12√2
Now, tan 15° = sin15°cos15°
= √3−12√2√3+12√2
= √3−1√3+1
Thus, tan 15° = √3−1√3+1
11 and 12 Grade Math
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