एतत् पृष्ठम् अपरिष्कृतम् अस्ति
Putting 0-90°, From (2) and (3), AN APPROXIMATE FORMULA FOR RSIN Therefore, from (1) 2xR+8100y=" or or giving y and 2xR= Therefore, Rsin=. The same result may also be derived as follows: Let sinλ= where is in radians and corresponds to 0 degrees. Putting λ=0, a=0. 8100 R Putting λ=t, b+c=0. Therefore, c-- Thus sinλ 40(180-0)R 40500-0(180-0) Since sinλ=sin(->), therefore bλ(π-^)/π bλ(π-2)/T A+BA+CA³ A+B(T-A)+C(π-λ)³ Putting in this λ=tn, 1 4R 40500 4R H sin λ= a+bx+cλ² A+BX+Cx²² C=- A+BX+CX=A+B(x-2)+C(n-1)², B(21-T)=CT(T-21), ba(n-λ)/T A+BX+Cx² - B bλ (π-λ) AT+BN(T-λ) AT + Bn(n-1)=2b. ln(n-1), 5x²B_10x¹b 36 36 or AT + = 209 (3)
