A Treatise of Algebra Wherein the Principles Are Demonstrated to Which is a
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And that the rectangle of the tangent and co-tangent is equal to the square of the radius.
Of Plane Trigonometry* 245 PROPOSITION III.
The co-sine CF of an arch AE being given ; to find the sine and cosine of half that arch* From the two extremities of the diameter AB draw the subtenses AE and BE ; and let CQ bisect the arch AE in Q and its chord (perpendicularly) in D ; then, since the angle BEA is a right one- (by Euc. 31, 3. ), the triangles ABE and ^^^^ ^ ADC are similar ; and, therefore, A...C being = I AB, AD must be = I AE, and CD = \ BE : b ut AE is = VA B X AF, an d BE = VAB X BF ; therefore AD = |VAB_ K¥ = \ /|AC x A F = the sine ? of ^ CD = IVAB X iiF = VIAC X BF = the co-sine 5 ^E. Hence it is evident, that the sine of the half of any- arch is a mean proportional between the half radius and the versed-sine of the whole arch ; and its co-sine a mean proportional between half the radius and the versed-sine of the supplement of the same arch.
PROPOSITION IV.
The sine AD, and co-sine CD, of an arch AQ being given ; to find EF, the sine of the double of that arch.
What to read after A Treatise of Algebra Wherein the Principles Are Demonstrated to Which is a?
You can find similar books in the "Read Also" column, or choose other free books by Thomas Simpson to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
And that the rectangle of the tangent and co-tangent is equal to the square of the radius.
Of Plane Trigonometry* 245 PROPOSITION III.
The co-sine CF of an arch AE being given ; to find the sine and cosine of half that arch* From the two extremities of the diameter AB draw the subtenses AE and BE ; and let CQ bisect the arch AE in Q and its chord (perpendicularly) in D ; then, since the angle BEA is a right one- (by Euc. 31, 3. ), the triangles ABE and ^^^^ ^ ADC are similar ; and, therefore, A...C being = I AB, AD must be = I AE, and CD = \ BE : b ut AE is = VA B X AF, an d BE = VAB X BF ; therefore AD = |VAB_ K¥ = \ /|AC x A F = the sine ? of ^ CD = IVAB X iiF = VIAC X BF = the co-sine 5 ^E. Hence it is evident, that the sine of the half of any- arch is a mean proportional between the half radius and the versed-sine of the whole arch ; and its co-sine a mean proportional between half the radius and the versed-sine of the supplement of the same arch.
PROPOSITION IV.
The sine AD, and co-sine CD, of an arch AQ being given ; to find EF, the sine of the double of that arch.
What to read after A Treatise of Algebra Wherein the Principles Are Demonstrated to Which is a?
You can find similar books in the "Read Also" column, or choose other free books by Thomas Simpson to read onlineMoreLess
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