Here, #f (x)=cos (x)# and the interval is # [-1/2pi,1/2pi]# #cos (x)# is continuous on # [-1/2pi, 1/2pi]# and differentiable on # (-1/2pi,1/2pi)# #f (-1/2pi)=cos (-pi ...

Explanation: #sin^2 (π/8) + sin ^2 3π/8 + sin ^2 5π/8 + sin ^2 7π/8# #color (red) ("using formula "sin^2theta=1/2 (1-cos2theta))# #=1/2 (1-cos (pi/4)) + 1/2 (1 ...

Dec 2, 2016 · cos(tan^(-1) (3/4) ) =cos(sec^(-1) sqrt(1+(3/4)^2 ) ) =cos(sec^(-1) (sqrt((16+9)/4^2)) ) =cos(sec^(-1) (5/4 ) ) =cos(cos^(-1) (4/5 ) )=4/5

cot^2xcos^2x cot^2x-cos^2x cos^2x/sin^2x-cos^2x (cos^2x-sin^2xcos^2x)/sin^2x (cos^2x (1-sin^2x))/sin^2x (cos^2x*cos^2x)/sin^2x cos^2x/sin^2x*cos^2x cot^2xcos^2x

Prequisites : (1):tan^-1x+tan^-1y=tan^_1 ( (x+y)/ (1-xy)); x,y gt 0, xy lt1. (2):cos^-1x=tan^-1 (sqrt (1-x^2)/x), 0 lt x lt 1. (3):tan^-1x=sin^-1 (x/sqrt (1+x^2)), 0ltxlt1.

Oct 1, 2017 · cos10°cos20°cos40° =4/ (8sin10^@)2sin10^@cos10°cos20°cos40° =2/ (8sin10^@)2sin20^@cos20°cos40° =1/ (8sin10^@)2sin40^@cos40° =1/ (8sin10^@)sin80^@ =1/ …

Read below. We have: f (x)= (x+xcos (x))/ (x-xcos (x)) We can simplify this by factoring: f (x)= (x (1+cos (x)))/ (x (1-cos (x)) We can cross out the x's. f (x)= (1+cos (x))/ (1-cos (x)) Now, a function is even if: f …

60^@; 120^@; 240^@; 300^@ 1. 4cos^2 t = 1 cos^2 t = 1/4 cos t = +- 1/2 a. cos t = 1/2. Trig table and unit circle give 2 solutions t = +- 60^@ Note. - 60^@ is co-terminal to (300^@) b. cos t = - 1/2 Trig …

0 Since this is of the indeterminate form 0/0, we can use L'Hospital’s Rule: Differentiate numerator and denominator: d/ (dh) (1-cos (h))^2=2 (1-cos (h))*sin (h ...

Use definition of tangent and Pythagorean's identity Answer: sin^2x Simplify 1-sin^2x/tan^2x Note that tanx=sinx/cosx We can substitute tan^2x=sin^2x/cos^2x =1-sin^2x/(sin^2x/cos^2x) =1 …