
How to find exact value of #sin(pi/24)#? - Socratic
2017年11月1日 · As #pi/24=180^@/24=(7 1/2)^@#, let us first work out #cos(pi/12)# or #cos15^@#. #cos15^@=cos(45^@-30^@)# = #cos45^@cos30^@+sin45^@sin30^@#
cos(pi/24) = ? A. sin((11pi)/24) B. sin((23pi)/24) C. sec ... - Socratic
2018年4月5日 · A Trigonometric Identity: cos(x)=sin(pi/2-x) SO cos(pi/24)=sin(pi/2-pi/24) cos(pi/24)=sin((12pi)/24-pi/24) cos(pi/24)=sin((11pi)/24) --> A is your answer
1.cos^2 (π/24)+cos^2 ( (19π)/24)+cos^2 ( (31π)/24)+cos^2
2018年5月14日 · Fun. I don't know how to do this one offhand, so we'll just try some things. There don't appear to be complementary or supplementary angles obviously in play, so perhaps our best move is to start with the double angle formula.
How do you find the exact values of cos(13pi/24)sin(31pi/24
2015年8月30日 · How do you find the exact values of cos(13pi/24)sin(31pi/24) using the half angle formula? Trigonometry Trigonometric Identities and Equations Half-Angle Identities 1 Answer
How do you solve tan(2x + pi/3) = 1 from 0 to 2pi? - Socratic
2015年11月6日 · (11pi)/24 ; (23pi)/24 ; (33pi)/24 , (35pi)/24 Trig Table and unit circle --> tan (pi/4) = 1 and tan (pi/4 + pi) = tan ((5pi)/4) = 1 Substitute in the right side of ...
Solving Trigonometric Equations - Trigonometry - Socratic
or #4x- pi/3 = (5 pi)/6+2 pi k# for integer #k#. Adding # pi/3# in the form #(2 pi)/6# to both sides of these equations gives us: #4x = (3 pi)/6+2 pi k = pi/2+2 pi k# for integer #k# or #4x = (7 pi)/6+2 pi k# for integer #k#. Dividing by #4# (multiplying by #1/4#) gets us to: #x= pi/8+(2pi k)/4# or #x=(7 pi)/24+(2 pi k)/4# for integer #k#.
How do you find int_ (pi/24)^ (pi/18) 36 (1+3^ (6sec (6x))) (sec (6x ...
2017年1月17日 · Calculus . Science Anatomy & Physiology
How do you express sin(pi/12) * cos(pi/8 ) without products of ...
2016年12月21日 · How do you express #sin(pi/12) * cos(pi/8 ) # without products of trigonometric functions? Trigonometry Trigonometric Identities and Equations Products, Sums, Linear Combinations, and Applications 1 Answer
A chord with a length of 12 runs from pi/12 to pi/6 radians
The chord AB of length 12 in the above figure runs from #pi/12# to #pi/6# in the circle of radius r and center O, taken as origin .
How to solve #cos(pi/4-x)=sin(pi/6+x)#? - Socratic
2018年6月11日 · We know that, #color(red)((1)sinC-sinD=2cos((C+D)/2)sin((C-D)/2)# #color(blue)((2)cos(pi/2-theta)=sintheta# Here,