2016年10月13日 · sec(cos^-1(1/3))=3 The restriction for the range of arccos x is [0,pi]. Since the argument is positive it means our triangle is in quadrant I with adjacent of 1 and hypotenuse 3. …

2018年3月6日 · = 1 sec (- pi/18) = 1/(cos (-pi/18)) = 1/cos (pi/18) cos ((37pi)/18) = cos (pi/18 + (36pi)/18) = cos (pi/18) The product transform itself to: P = cos (pi/18)/(cos (pi ...

2018年2月19日 · #cos x / (sec x + tan x) = cos x / ((1/cos x) + (sin x/ cos x))# # => cos ^2 x / (1 + sin x) = (1 - sin^2x) / (1 + sin x)#

2016年10月29日 · sec x tan x cos x simplifies to tan x. Recall that sec x = 1/(cos x). We will begin by rearranging the factors so that sec x and cos x are beside each other and then proceed …

2017年9月25日 · sin^2x/cosx Sec x - cos x = (1/cos x) - cos x =(1-cos^2x)/cos x =sin^2x/cos x

2017年11月26日 · Prove: #sec^2x-2secxcosx+cos^2x=tan^2x-sin^2x# I boiled it down to each side being 1 but I don't think I did it all the way correctly.

2016年6月21日 · #(sec x / sin x) - (sin x / cos x)# create common denominator # = (sec x cos x - sin^2 x ) / (sin x cos x)# # = (1 - sin^2 x ) / (sin x cos x)#

2018年2月22日 · It simplifies to 1 Given: sec(-x)cos(x) Use the identity sec(u) = 1/cos(u) where u = -x: cos(x)/cos(-x) Use the even property of the cosine function cos(-x) = cos(x): cos(x)/cos(x) …

2018年2月20日 · If #sec theta = 4#, how do you use the reciprocal identity to find #cos theta#? How do you find the domain and range of sine, cosine, and tangent? What quadrant does #cot …

2016年10月31日 · sec(2x) = 1/(1 - 2sin^2(x)) sec(2x) = 1/cos(2x) = 1/(1 - 2sin^2(x))