Aug 4, 2015 · How do you find the exact values #sin (pi/4)# using the special triangles?

Jul 11, 2011 · Find pi/4 on the unit circle, it's root 2 / 2, that doesn't seem to get the right answer even though that's was supposed to be the method? Physics news on Phys.org D-Wave claims its quantum computers can solve a problem of scientific …

Jul 29, 2016 · (sqrt2 + sqrt6)/4 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) = Trig table of ...

May 8, 2018 · # [3 pi}/4 # is #135^circ = 3 times 45^circ # Of course 30/60/90 and 45/45/90 are the only two triangles students are expected to know "exactly." We have to know the multiples of these triangles in the other quadrant. Here we have 45/45/90 in the second quadrant. We start from #cos(45^circ)=sin(45^circ)=1/sqrt{2}#.

Apr 16, 2015 · 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 ...

How do you prove #sin 5pi/12 = sin pi/6 cos pi/4 + cos pi/6 sin pi/4#? Trigonometry Trigonometric Identities and Equations Sum and Difference Identities 1 Answer

Aug 20, 2016 · sqrt(2 + sqrt3)/2 Trig unit circle --> sin ((7pi)/12) = sin (pi/12 + pi/2) = cos (pi/12) Find cos (pi/12) by using the trig identity: cos 2a = 2cos^2 a - 1 cos ((2pi ...

Feb 28, 2012 · ok i know that sin n*(pi/2) = 1 if n=1,5,9,13... = -1 if n=3,7,11,15... = 0 if n is even cos n*(pi/2) = 0 if n is odd = -1 if n=0,4,8,12 = 1 if n=2,6,10,14... is there a simpler way of expressing this? for example simple way to express cos(n*pi)=cos(-n*pi)=(-1)^n is there a...

May 4, 2018 · How do you know if #sin 30 = sin 150#? How do you show that #(costheta)(sectheta) = 1# if #theta=pi/4#? See all questions in Trigonometric Functions of Any Angle

Jan 25, 2016 · Amplitude=1 Period =4 from the given y=sin (pi/2* theta) it should be taken into account that the above is the same as y=1*sin (pi/2* theta) Take the absolute value of any number you see as numerical coefficient of sine or cosine expression and you get the Amplitude. Amplitude =abs1=1 The Period can be obtain using P=(2pi)/a where a is the number beside theta so that P=(2 pi)/abs(pi/2)=4 Learn ...