2015年12月4日 · cos (pi/4)=sqrt(2)/2 see explanation for details. Angle pi/4 rad or 45^@ can be an angle in isosceles right triangle, so its catheti are equal and its hypothenuse is sqrt(2) times longer. For example if the catheti are 1 unit long, then the hypothenuse is sqrt(2) units. Sine is a quotient of the cathetus opposite to the angle and the hypothenuse, so: sin …

2016年5月27日 · sqrt2/2 -pi/4 is the same a (11pi)/4 This is a 45^o with respect to the x-axis. This angle is in the IV quadrant where cos is + The cos of a 45^o and is sqrt2/2

2016年7月29日 · Apply the trig identity: cos (a - b) = cos a.cos b + sin a.sin b #cos (pi/3 - pi/4) = cos (pi/3).cos (pi/4) + sin (pi/3).sin (pi/4) =#

2018年5月8日 · #cos(x-pi/4) + cos(x+pi/4)=1# We'll expand with the difference and sum angle formulas and see where we are. #cos x cos (pi/4) + sin x sin(pi/4) + cos x cos(pi/4) - sin x sin(pi/4) = 1# #2 cos x cos(pi/4) = 1 # # 2 cos x (sqrt{2}/2) = 1# #cos x = 1/sqrt{2} # That's 45/45/90 in the first and fourth quadrant, #x = pi/4 or x = {7pi}/4 # Check:

2016年5月22日 · How do you show that #(costheta)(sectheta) = 1# if #theta=pi/4#? See all questions in Trigonometric Functions of Any Angle Impact of this question

2015年4月16日 · Looking at the angle: 7pi/4 is the same as -pi/4 and cos is an even function so: cos(-x)=cos(x) So: cos(7pi/4)=cos(-pi/4)=cos(pi/4)=sqrt(2)/2

2017年3月11日 · #sin(pi/12)cos(pi/4)+cos(pi/12)sin(pi/4)# we use . #sin(alpha+beta)=sinalphacosbeta+sinbetacosalpha# comparing this with the question we have

2016年4月11日 · How do you apply the fundamental identities to values of #theta# and show that they are true

2018年5月11日 · text{Arc}text{sin}(cos (pi/4) ) = pi/4 arcsin(cos (pi/4)) = pi/4 + pi k, quad integer k Oh my, pi/4 again. What a surprise. Question writers, questions like this are your big chance to get away from the two cliche triangles of trig. I prefer the notation text{Arc}text{sin}(x) for the principal value and arcsin(x) for all values. Let's run the problem both ways. Of course, …

2017年5月22日 · #cos(-9/4pi) = cos((-2pi)-1/4pi)# #=cos(-2pi)cos(1/4pi) + sin(-2pi)sin(1/4pi)# #=(1)(sqrt2/2) + (0)(sqrt2/2)# ...