How to find the exact value of cot 7½° using the value of cos 15°?
Solution:
7½° lies in the first quadrant.
Therefore, both sin 7½° and cos 7½° is positive.
For all values of the angle A we know that, sin (α - β) = sin α cos β - cos α sin β.
Therefore, sin 15° = sin (45° - 30°)
= \(\frac{1}{√2}\)∙\(\frac{√3}{2}\) - \(\frac{1}{√2}\)∙\(\frac{1}{2}\)
= \(\frac{√3}{2√2}\) - \(\frac{1}{2√2}\)
= \(\frac{√3 - 1}{2√2}\)
Again, for all values of the angle A we know that, cos
(α - β) = cos α cos β + sin α sin β.
Therefore, cos 15° = cos (45° - 30°)
cos 15° = cos 45° cos 30° + sin 45° sin 30°
= \(\frac{1}{√2}\)∙\(\frac{√3}{2}\) + \(\frac{1}{√2}\)∙\(\frac{1}{2}\)
= \(\frac{√3}{2√2}\) + \(\frac{1}{2√2}\)
= \(\frac{√3 + 1}{2√2}\)
Now cot 7½°
= \(\frac{cos 7½°}{sin 7½°}\)
= \(\frac{2 cos 7½° ∙ cos 7½°}{2 sin 7½° ∙ cos 7½°}\)
= \(\frac{2 cos^{2} 7½° }{2 sin 7½° cos 7½°}\)
= \(\frac{1 + cos 15°}{sin 15°}\)
= \(\frac{1 + cos (45° - 30°)}{sin (45° - 30°)}\)
= \(\frac{1 + \frac{√3 + 1}{2√2}}{\frac{√3 - 1}{2√2}}\)
= \(\frac{2√2 + √3 + 1}{√3 - 1}\)
= \(\frac{(2√2 + √3 + 1)(√3 + 1)}{(√3 - 1)(√3 + 1)}\)
= \(\frac{2√6 + 2√2 + 3 + √3 + √3 + 1}{3 - 1}\)
= \(\frac{2√6 + 2√2 + 2√3 + 4}{2}\)
= √6 + √2 + √3 + 2
= 2 + √2 + √3 + √6
11 and 12 Grade Math
From Exact Value of cot 7 and Half Degree to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Dec 04, 23 02:14 PM
Dec 04, 23 01:50 PM
Dec 04, 23 01:49 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.